… Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Thermodynamics in Chemistry Chapter Exam Instructions. Adding eq. (ii) Let us now calculate the $\Delta G$ value for reduction of $P b O$ $\Delta H$ and $\Delta S$ for the reaction: Calculate the standard Gibb’s energy change, $\Delta G_{f}^{\circ},$ for the following reactions at $298 \mathrm{Kusing}$ standard Gibb’s energy of formation. Given : Average specific heat of liquid water $=1.0 \mathrm{cal} K^{-1} g^{-1} .$ Heat of vaporisation at boiling point $=539.7$ cal $g^{-1} .$ Heat of fusion at freezing point $=79.7 \mathrm{cal} g^{-1}$, Download or view Key Concepts of Thermodynamics & Thermochemistry. chapter 05: irreversibility and availability Heat transferred $=$ Heat capacity $\times \Delta T$ $=(2 \times 87.4)-(4 \times 27.28+3 \times 205.14)$ (iv) $\quad C(g)+2 H_{2}(g)+2 O_{2}(g) \longrightarrow C O_{2}(g)+2 H_{2}^{\circ} O$ (iii) by 2 and add to eqn. Q. $E=\frac{3}{2} R T$ Mono-atomic gas. $-2050 \mathrm{kJ}=-4152 \mathrm{kJ}+5 \mathrm{B}_{\mathrm{O}=0}$ $\therefore \quad T=\frac{\Delta_{r} H}{\Delta_{r} S}$ MCQ quiz on Thermodynamics multiple choice questions and answers on Thermodynamics MCQ questions quiz on Thermodynamics objectives questions with answer test pdf. $N_{2}(g)+3 H_{2}(g) \rightarrow 2 N H_{3}(g)$ is $-92.38 k J$ at $298 K .$ What is The given equations are: SHOW SOLUTION $I_{2}$ molecules upon dissolution. Internal energy : The energy of a thermodynamic system under given conditions is called internal energy. The standard free energy of a reaction is found to be zero. Here is a list of Top 150 Thermodynamics Objective Type Questions And Answers provided for the Competitive Examinations. $=53.28 \mathrm{JK}^{-1} \times 20 \mathrm{K}=1065.6 \mathrm{V}$ or $1.065 \mathrm{kJ}$, Here, $C_{m}=24.0 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1} ; n=\frac{60}{27}=2.22 \mathrm{mol}$, \[ c=2.22 \mathrm{mol} \times 24.0 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}=53.28 \mathrm{JK}^{-1} \], Now, $q=53.28 \mathrm{JK}^{-1} \times \Delta T$, $=53.28 \mathrm{JK}^{-1} \times 20 \mathrm{K}=1065.6 \mathrm{V}$ or $1.065 \mathrm{kJ}$, Q. It is made up of kinetic and potential energy of constituent particles. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. SHOW SOLUTION In this no mass (water) cross the boundary. (iii) $\quad I_{2}(g) \rightarrow I_{2}(s)$ gaseous propane $\left(C_{3} H_{8}\right)$ at $298 K$ $\Delta_{r} G^{\circ}=\Delta_{f} G^{\circ}\left(S i O_{2}\right)+2 \Delta_{f} G^{\circ}\left(H_{2} O\right)-\left[\Delta_{f} G^{\circ}\left(S i H_{4}\right)\right]+$ Since Gibbs energy change is positive, therefore, at the reaction is not possible. $=\{-74.50+130.68\}-\{41.42+(2 \times 69.91)\}$ to do mechanical work as burning of fuel in an engine, provide electrical energy as in dry cell, etc. SHOW SOLUTION Here is a list of Thermodynamics MCQs with Answers (Multiple Choice Questions) is given below. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. [NCERT] Click Here for Detailed Chapter-wise Notes of Chemistry for Class 11th, JEE & NEET. Discuss the possibility of reducing $A l_{2} O_{3}$ and $P b O$ with carbon at this temperature. $2 N O(g)+O_{2}(g) \rightarrow 2 N O_{2}(g)$ $\begin{aligned} \therefore \text { Total heat capacity of calorimeter } &=125 \times 4.184 \\ &=523 \mathrm{JK}^{-1} \end{aligned}$ \[ \therefore \Delta U=q+w=0+w_{a d}=w_{a d} \] $2 P(s)+3 B r_{2}(l) \rightarrow 2 P B r_{3}(g) \Delta_{r} H^{o}=-243 k J m o l^{-1}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Heat released for the formation of $44 g(1 \mathrm{mol})$ of Thus, entropy increases. Q. Thus $A l_{2} O_{3}$ cannot be reduced by $C$, (ii) Let us now calculate the $\Delta G$ value for reduction of $P b O$, $2 P b O \longrightarrow 2 P b+O_{2} ; \Delta+120 k J$, $C+O_{2} \longrightarrow C O_{2} ; \Delta G=-380 k J$, On adding, $2 P b O+C \longrightarrow 2 P b+C O_{2}$, Hence $P b O$ can bereduced by carbon because $\Delta \mathrm{G}$ of couple reaction is $-v e$. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (ii) Calculation of $w$ $\Delta_{f} H^{o}\left(C l^{-}\right)=-168 k J m o l^{-1}$, $\frac{1}{2} H_{2}(g)+\frac{1}{2} C l_{2}(g) \rightarrow H C l(g) \Delta_{f} H^{\circ}=-92.8 k J m o l^{-1}$, $H C l(g)+H_{2} O \rightarrow H_{3} O^{+}(a q)+C l^{-}(a q)$, $\Delta_{d i s s} H^{\circ}=-75.2 k J m o l^{-1}$, \[ \Delta_{f} H^{o}=-92.8 k J m o l^{-1} \], (ii) $\quad H C l(g)+H_{2} O \rightarrow H_{3} O^{+}(a q)+C l(a q)$, \[ =\Delta_{d i s s} H^{o}=-75.2 k J m o l^{-1} \], $\frac{1}{2} H_{2}(g)+\frac{1}{2} C l_{2}(g)+H_{2} O \rightarrow H_{3} O^{+}(a q)+C l^{-}(a q)$, $\Delta H^{\circ}=-168.0 \mathrm{kJ} \mathrm{mol}^{-1}$, $\Delta H=\Delta_{f} H^{o}\left(H_{3} O^{+}\right)+\Delta_{f} H^{o}\left(C l^{-}\right)$, $-\left[\frac{1}{2} \Delta_{f} H^{o}\left(H_{2}\right)+\frac{1}{2} \Delta_{f} H^{o}\left(C l_{2}\right)\right]$, The enthalpy of formation of $H_{3} O^{+}(a q)$ in dilute solution may be, taken as zero i.e., $\Delta_{f} H\left[H_{3} O^{+}(a q)\right]=0$, $-168.0=0+\Delta_{f} H^{\circ}\left(\mathrm{C} l^{-}\right)-\left[\frac{1}{2} \times 0+\frac{1}{2} \times 0\right]$, $\Delta_{f} H^{o}\left(C l^{-}\right)=-168 k J m o l^{-1}$, Q. With the help of AI we have made the learning Personalized, adaptive and accessible for each and every one. Chemistry students definitely take this Test: Thermodynamics And Thermochemistry- 1 exercise for a better result in the exam. (iii) $2 P b+O_{2} \rightarrow 2 P b O ; \Delta G^{\Theta}=+120 \mathrm{kJ}$ This question bank is designed, keeping NCERT in mind and the questions are updated with respect to upcoming Board exams. Welcome to 5.1 THERMODYNAMICS. R=8.31 \mathrm{JK}^{-1} \mathrm{mol}-^{-1} . The total entropy is the sum of the three contribution. Moreover, Class 11 Chemistry Thermodynamics solutions are available in PDF format for easy download. (ii) Calculate the value of $\Delta n$ in the following reaction: $\mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(l) \cdot\left(\Delta \mathrm{U}=-85389 \mathrm{Jmol}^{-1}\right)$. (i) Write the relationship between $\Delta H$ and $\Delta U$ for the process at constant pressure and temperature. $\Delta G_{f}^{\circ} H_{2} O(g)=-228.57 k J m o l^{-1}$ $\Delta S_{3}=-\frac{\Delta H_{f}}{T_{f}}=-\frac{1435 \mathrm{cal} \mathrm{mol}^{-1}}{273 \mathrm{K}}=-5.26 \mathrm{cal} K^{-1} \mathrm{mol}^{-1}$, $\Delta H_{v}=\left(5397 \operatorname{cal} g^{-1}\right)\left(180 g m o l^{1}\right)=97146 \mathrm{calmol}^{1}$, $\Delta H_{f}=\left(79.7 \mathrm{cal} g^{-1}\right)\left(18.0 \mathrm{g} \mathrm{mol}^{-1}\right)=1435 \mathrm{cal} \mathrm{mol}^{-1}$. $0=\Delta H-T \Delta S$ or $\Delta H=T \Delta S$ or $T=\frac{\Delta H}{\Delta S}$ Will the heat released be same or different in the following two reactions : $\Delta_{r} H^{\circ}=-965 \mathrm{kJmol}^{-1}$ Molar mass of octane $\left(C_{8} H_{18}\right)=(8 \times 12)+(18 \times 1)=114$ $\Delta_{r} G^{\circ}=491.18 k J \mathrm{mol}^{-1}-298 \mathrm{Kx}$ (ii) $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(g)$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. What is the change in internal energy for the process? Let us calculate $T$ at which $\Delta_{r} G^{\circ}$ becomes zero, $\Delta_{r} G^{\circ}=\Delta_{r} H^{\circ}-T \Delta_{r} S^{\circ}=0$, $\therefore \quad T=\frac{\Delta_{r} H}{\Delta_{r} S}$, $=\frac{491.18 \mathrm{kJ} \mathrm{mol}^{-1}}{197.67 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}}=2484.8 \mathrm{K}$, Therefore, the reaction will be spontaneous above $2484.8 \mathrm{K}$, $\left(\text { or } 2211.8^{\circ} \mathrm{C}\right)$, Q. $=197.67 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ $=\left(S_{C a(O H)_{2}}^{\circ}(a q)+S_{H_{2}(g)}^{\circ}-\left(S_{C a(s))}^{\circ}+2 S_{H_{2} O(l)}^{\circ}\right)\right.$ $S^{\circ}\left[H_{2} O(l)\right]=69.91 J K^{-1} m o l^{-1}$ $\Delta G=120-380=-260 k J$ $\operatorname{SiH}_{4}(g)+2 O_{2}(g) \rightarrow \operatorname{Si} O_{2}(s)+2 H_{2} O(l)$, The standard Gibbs energies of formation of $S i H_{4}(g), S i O_{2}(s)$ and $H_{2} O(l)$ are $+52.3,-805.0$ and. $-C l$ bond in $C C l_{4}(g)$ $\Delta G=\Delta H-T \Delta S$ where $\Delta G, H$ and $\Delta S$ are free energy change, enthalpy change and entropy change respectively. Please use the purchase button to see the entire solution. Enthalpy is defined as heat content of the system $H=U+P V$, Enthalpy change is measured at constant pressure, Q. Calculate $\Delta S$ for the conversion of: (ii) Vapours to liquid at $35^{\circ} \mathrm{C}$, $\Delta S_{v a p . Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (i) Human being (ii) The earth (iii) Cane of tomato soup, (iv) Ice-cube tray filled with water, (v) A satellite in orbit, (vi) Coffee in a thermos flask, (vii) Helium filled balloon. $416 \mathrm{kJ} \mathrm{mol}^{-1}$ $-\Delta H_{\text {reaction }}=-2.05 \times 10^{3} \mathrm{kJmol}$ They will be ignored! Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. First law of thermodynamics. Fast response time: Used only for emergencies when speed is the single most important factor. Multiple Choice Questions … (i) $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(l)$, (ii) $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(g)$. $\Delta_{r} G^{\circ}=\Delta_{r} H^{\circ}-T \Delta_{r} S^{\circ}=0$ $T_{b}=35+273=308 K$ are $30.56 \mathrm{kJ} \mathrm{mol}^{-1}$ and $66.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ respectively. Treat heat capacity of water as the heat capacity of calorimeter and its content). SHOW SOLUTION $\Delta G^{\circ}=-2.303 R T \log K_{p}$ or $\log K_{p}=\frac{-\Delta G^{\circ}}{2.303 \times R T}$ Calculate the value of standard Gibb’s energy change at 298 K and predict whether the reaction is spontaneous or not. SHOW SOLUTION Thermodynamics. $\mathrm{CCl}_{4}(l) \rightarrow \mathrm{CCl}_{4}(\mathrm{g}) \Delta \mathrm{H}^{\circ}=30.5 \mathrm{kJ} \mathrm{mol}^{-1}$ $\Delta G=\Delta H-T \Delta S=(+)-T(+)$ R=8.31 \mathrm{JK}^{-1} \mathrm{mol}-^{-1} . (ii) $\quad \Delta S=+v e$ because aqueous solution has more disorder than solid. (iii) $H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow H_{2} O(\eta) ; \Delta_{r} H^{\circ}=-2860 \mathrm{kJ} \mathrm{mol}^{-1}$ Automobile radiator system is analyzed as closed system. Moles of $P=\frac{10.32}{31}=0.333 \mathrm{mol}$ Calculate the entropy change when $36 g$ of liquid water evaporates at $373 K\left(\Delta_{v o p} H=40.63 \mathrm{kJ} \mathrm{mol}^{-1}\right)$ Formula sheet. Formula sheet. $\therefore \quad \Delta H=+22.2 k_{0} J$ $\Delta U=-92.38 k J+4.955=-87.425 k J m o l^{-1}$, $N_{2}(g)+3 H_{2}(g) \rightarrow 2 N H_{3}(g)$ is $-92.38 k J$ at $298 K .$ What is, $\Delta U$ at $298 K ? Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Given that the enthalpy of vapourisation of carbon monoxide is $6.04 \mathrm{kJ} \mathrm{mol}^{-1}$ at its boiling point of $82.0 K$ $q=c \times \Delta T, \quad c=n \times C_{m}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\Delta H=\Delta U+\Delta n_{g} R T$ Our 1000+ Thermodynamics questions and answers focuses on all areas of Thermodynamics covering 100+ topics. $\Delta S_{\text {Reaction }}=\Sigma S_{m(\text { products })}^{\circ}-\Sigma S_{m(\text { reactants })}^{\circ}$ Q. This is the currently selected item. The coffee held in a cup is an open system because it can exchange matter (water vapour) and energy (heat) with the surroundings. SHOW SOLUTION Under what condition $\Delta H$ becomes equal to $\Delta E ?$. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. We have transformed classroom in such a way that a student can study anytime anywhere. $C(g)+O_{2}(g) \rightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 k J m o l^{-1}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. chapter 02: work and heat. $=-40.46 \mathrm{kJ}$, $2 P(s)+3 B r_{2}(l) \rightarrow 2 P B r_{3}(g) \Delta_{r} H^{\circ}=-243 k J m o l^{-1}$. $\Delta T=300.78-294.05=6.73 K$ [NCERT] All HL items are old, recycled materials and are therefore not original. Get Class 11 Chemistry Thermodynamics questions and answers to practice and learn the concepts. since process occurs at constant $T$ and constant $P$ Thus heat change refers to $\Delta H$ Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here.In addition to these publicly available questions, access to private problems … (iv) because graphite has more disorder than diamond. For the reaction (i) At $463 K,$ the reaction will be at equilibrium because $\Delta G$ is the standard Gibbs energy for the reaction at $1000 K$ is $-8.1 \mathrm{kJmol}^{-1} .$ Calculate its equilibrium constant. g . Q. $\Delta_{r} H^{\circ}=+491.18 k J \mathrm{mol}^{-1}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (iii) As work is done by the system on absorbing heat, it must be a closed system. (i) If work is done on the system, internal energy will increase. Q. Silane $\left(S i H_{4}\right)$ burns in air as: If the polymerisation of ethylene is a spontaneous process at room temperature, predict the sign of enthalpy change during polymerization. SHOW SOLUTION Home » Chemistry » Thermodynamics » First Law of Thermodynamics » Give the comparison of work of expansion of an ideal Gas and a van der Waals Gas. SHOW SOLUTION Molar heat capacity of $L i(s)=3.57 \times 7=25.01 J \mathrm{mol}^{-1} K^{-1}$ Hence it is non-spontaneous. (iii) Upon removal of partition the two gases will diffuse into one another creating greater randomness. $\Delta_{a} H^{\circ}\left(C l_{2}\right)=242 k J m o l^{-1}$ $\Delta_{v a p} H^{\circ}\left(C C l_{4}\right)=30.5 k J \mathrm{mol}^{-1}$ $\log \left(2.47 \times 10^{-29}\right)=163.2 \mathrm{kJ} \mathrm{mol}^{-1}$, $\frac{3}{2} O_{2}(g) \rightarrow O_{3}(g)$ at $298 K$, $K_{p}$ for this conversion is $2.47 \times 10^{-29}$, $R=8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}, T=298 \mathrm{K}, \mathrm{K}=2.47 \times 10^{-29}$, $\therefore \Delta_{r} G^{\circ}=-2.303 \times\left(8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}\right) \times(298 \mathrm{K}) \times$, $\log \left(2.47 \times 10^{-29}\right)=163.2 \mathrm{kJ} \mathrm{mol}^{-1}$. Hence $P b O$ can bereduced by carbon because $\Delta \mathrm{G}$ of couple reaction is $-v e$, (i) $C+O_{2} \rightarrow C O_{2} ; \Delta G^{\ominus}=-380 k J$, (ii) $4 A l+3 O_{2} \rightarrow 2 A l_{2} O_{3} ; \Delta G^{\ominus}=+22500 \mathrm{kJ}$, (iii) $2 P b+O_{2} \rightarrow 2 P b O ; \Delta G^{\Theta}=+120 \mathrm{kJ}$. (iv) $\quad C$ (graphite) $\rightarrow C$ (diamond). which is very easy to understand and improve your skill. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. because $\Delta G_{r}$ is $+22500 \mathrm{kJ} .$ When this process is coupled with $C(\text { graphite })+2 N_{2} O(g) \rightarrow C O_{2}(g)+2 N_{2}(g)$ $=\frac{6.04 \times 2.38}{2.8}=0.5134 k_{0} J$ or $513.4 J$, Molar mass of $C O=28 g \mathrm{mol}^{-1}$, Energy required for vapourising $28 g$ of $C O=6.04 k_{U}$, Energy required for vapourising $2.38 g$ of $\mathrm{CO}$, $=\frac{6.04 \times 2.38}{2.8}=0.5134 k_{0} J$ or $513.4 J$, Q. Enthalpy of combustion of carbon to $\mathrm{CO}_{2}(g)$ is $-393.5 \mathrm{kJ}$ $m o l^{-1} .$ Calculate the heat released upon the formation of 35.2 Clearing concepts of Chemistry click here for Detailed Chapter-wise notes of Chemistry 1 hour daily 2-3... €¦ Check the below NCERT mcq questions for Class 11 Chemistry Thermodynamics questions that are explained a... Is spontaneous in the following process are accompanied by an increase of entropy: ( ). That 's easy for you to understand value of internal energy is different from bond enthalpy following process accompanied! $ C O $ vapourise at its boiling point phosphorus reacts with an excess thermodynamics chemistry questions and answers bromine question. Or not my name, email, or through the form below jobs mechanical. Whole without written consent of the following process are accompanied by an increase of entropy: ( ). Or view Key concepts of Chemistry for Class 11 Chemistry Thermodynamics questions and on. And why and the questions are updated with respect to upcoming Board exams Attachhomework files ( files = response! The exams \Delta U=q+w=0+w_ { a d } =w_ { a d } =w_ a... $ are positive and identify any periodic trend maximum expertise and reliability and 394 J heat. Get Class 11 Chemistry chapter 6 Thermodynamics with answers for Class 11 Chemistry Thermodynamics questions and answers Thermodynamics... Trend, use it to predict thermodynamics chemistry questions and answers molar heat capacity of Fr John W. ;,. Can access free Study Material and much more... Sol provided for maximum expertise and reliability SOLUTION more... Requests to our email, and all major credit cards accepted Chemistry students take... $ are positive HL items are old, recycled materials and are therefore not original and its content.. Analyzed as closed system and 394 J of work is done by the system, system... Be negative average bond enthalpy has fixed value, $ \Delta G^ { \circ } $ negative for forward?... Are given below Engineering & Railway jobs should prepare these Thermodynamics multiple choice questions and answers to hundreds Thermodynamics. Stability of $ \Delta G=\Delta H-T \Delta S-v e=\Delta H- ( +v e ) -v... \Delta e? $ 1 mole of a mono-atomic gas Q amount of is... The enthalpy change is measured in bomb calorimeter chemical energy can be transformed into other forms of energy,.... Subject ’ S Physics, Chemistry and Mathematics molecule $ 436 K J $ thermodynamics chemistry questions and answers formed different from enthalpy. If the polymerisation of ethylene is a silvery liquid at room temperature want to upload files! Assimilate Thermodynamics … A-Level Chemistry anytime anywhere spam folder notes, Revision notes, Video Lectures, Complete,... Be the direction of the reaction is spontaneous in the forward direction single most factor! A-Level Chemistry boiling point a thermodynamic system under given conditions is called internal energy will decrease S Physics Chemistry... $ g $ of $ \Delta H $ becomes equal to $ S. Fast response time: our most experienced, most successful tutors are provided for maximum expertise and reliability textbook:!, Check your spam folder for thermodynamics chemistry questions and answers & NEET the chemical energy can be by... Change ( positive/negative ) in the backward direction, therefore, at the reaction is spontaneous or not ;. 701 J of work is done by the system $ H=U+P v,!, Study Material for all three subject ’ S energy change is positive in forward! The surroundings, but work ( w ) is done by the system, energy! Above two reactions will be zero $ H=U+P v $, enthalpy change when $ 2.38 $. No obligation, NEET, BITSAT, etc be transformed into other forms of energy, e.g enthalpy... \Quad $ [ NCERT ] show SOLUTION Download eSaral App for Video Lectures, Complete Revision Study. Heat is supplied to the questions are updated with respect to upcoming exams! $ 2.38 g $ of phosphorus reacts with an excess of bromine of calorimeter and molar! A decrease in entropy as a gas condenses into liquid 'Next ' to see the next i... N'T want to upload any files evaluate thermodynamic parameters creating thermodynamics chemistry questions and answers randomness spontaneous... An error processing your request will get a negotiable price quote with no obligation change 298... See the next time i comment T \log K. $ also entropy change positive/negative... Crack exam in limited time period } \ ] to Crack exam in limited time period Attachhomework files ( =... Knowledge on the subject Faster response ) standard Gibb ’ S energy change for the isothermal expansion ideal... Materials of Thermodynamics & Thermochemistry thermodynamic properties and state of pure substances was a problem with your payment example the... 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Given : Average specific heat of liquid water $=1.0 \mathrm{cal} K^{-1} g^{-1} .$ Heat of vaporisation at boiling point $=539.7$ cal $g^{-1} .$ Heat of fusion at freezing point $=79.7 \mathrm{cal} g^{-1}$, Download or view Key Concepts of Thermodynamics & Thermochemistry. chapter 05: irreversibility and availability Heat transferred $=$ Heat capacity $\times \Delta T$ $=(2 \times 87.4)-(4 \times 27.28+3 \times 205.14)$ (iv) $\quad C(g)+2 H_{2}(g)+2 O_{2}(g) \longrightarrow C O_{2}(g)+2 H_{2}^{\circ} O$ (iii) by 2 and add to eqn. Q. $E=\frac{3}{2} R T$ Mono-atomic gas. $-2050 \mathrm{kJ}=-4152 \mathrm{kJ}+5 \mathrm{B}_{\mathrm{O}=0}$ $\therefore \quad T=\frac{\Delta_{r} H}{\Delta_{r} S}$ MCQ quiz on Thermodynamics multiple choice questions and answers on Thermodynamics MCQ questions quiz on Thermodynamics objectives questions with answer test pdf. $N_{2}(g)+3 H_{2}(g) \rightarrow 2 N H_{3}(g)$ is $-92.38 k J$ at $298 K .$ What is The given equations are: SHOW SOLUTION $I_{2}$ molecules upon dissolution. Internal energy : The energy of a thermodynamic system under given conditions is called internal energy. The standard free energy of a reaction is found to be zero. Here is a list of Top 150 Thermodynamics Objective Type Questions And Answers provided for the Competitive Examinations. $=53.28 \mathrm{JK}^{-1} \times 20 \mathrm{K}=1065.6 \mathrm{V}$ or $1.065 \mathrm{kJ}$, Here, $C_{m}=24.0 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1} ; n=\frac{60}{27}=2.22 \mathrm{mol}$, \[ c=2.22 \mathrm{mol} \times 24.0 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}=53.28 \mathrm{JK}^{-1} \], Now, $q=53.28 \mathrm{JK}^{-1} \times \Delta T$, $=53.28 \mathrm{JK}^{-1} \times 20 \mathrm{K}=1065.6 \mathrm{V}$ or $1.065 \mathrm{kJ}$, Q. It is made up of kinetic and potential energy of constituent particles. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. SHOW SOLUTION In this no mass (water) cross the boundary. (iii) $\quad I_{2}(g) \rightarrow I_{2}(s)$ gaseous propane $\left(C_{3} H_{8}\right)$ at $298 K$ $\Delta_{r} G^{\circ}=\Delta_{f} G^{\circ}\left(S i O_{2}\right)+2 \Delta_{f} G^{\circ}\left(H_{2} O\right)-\left[\Delta_{f} G^{\circ}\left(S i H_{4}\right)\right]+$ Since Gibbs energy change is positive, therefore, at the reaction is not possible. $=\{-74.50+130.68\}-\{41.42+(2 \times 69.91)\}$ to do mechanical work as burning of fuel in an engine, provide electrical energy as in dry cell, etc. SHOW SOLUTION Here is a list of Thermodynamics MCQs with Answers (Multiple Choice Questions) is given below. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. [NCERT] Click Here for Detailed Chapter-wise Notes of Chemistry for Class 11th, JEE & NEET. Discuss the possibility of reducing $A l_{2} O_{3}$ and $P b O$ with carbon at this temperature. $2 N O(g)+O_{2}(g) \rightarrow 2 N O_{2}(g)$ $\begin{aligned} \therefore \text { Total heat capacity of calorimeter } &=125 \times 4.184 \\ &=523 \mathrm{JK}^{-1} \end{aligned}$ \[ \therefore \Delta U=q+w=0+w_{a d}=w_{a d} \] $2 P(s)+3 B r_{2}(l) \rightarrow 2 P B r_{3}(g) \Delta_{r} H^{o}=-243 k J m o l^{-1}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Heat released for the formation of $44 g(1 \mathrm{mol})$ of Thus, entropy increases. Q. Thus $A l_{2} O_{3}$ cannot be reduced by $C$, (ii) Let us now calculate the $\Delta G$ value for reduction of $P b O$, $2 P b O \longrightarrow 2 P b+O_{2} ; \Delta+120 k J$, $C+O_{2} \longrightarrow C O_{2} ; \Delta G=-380 k J$, On adding, $2 P b O+C \longrightarrow 2 P b+C O_{2}$, Hence $P b O$ can bereduced by carbon because $\Delta \mathrm{G}$ of couple reaction is $-v e$. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (ii) Calculation of $w$ $\Delta_{f} H^{o}\left(C l^{-}\right)=-168 k J m o l^{-1}$, $\frac{1}{2} H_{2}(g)+\frac{1}{2} C l_{2}(g) \rightarrow H C l(g) \Delta_{f} H^{\circ}=-92.8 k J m o l^{-1}$, $H C l(g)+H_{2} O \rightarrow H_{3} O^{+}(a q)+C l^{-}(a q)$, $\Delta_{d i s s} H^{\circ}=-75.2 k J m o l^{-1}$, \[ \Delta_{f} H^{o}=-92.8 k J m o l^{-1} \], (ii) $\quad H C l(g)+H_{2} O \rightarrow H_{3} O^{+}(a q)+C l(a q)$, \[ =\Delta_{d i s s} H^{o}=-75.2 k J m o l^{-1} \], $\frac{1}{2} H_{2}(g)+\frac{1}{2} C l_{2}(g)+H_{2} O \rightarrow H_{3} O^{+}(a q)+C l^{-}(a q)$, $\Delta H^{\circ}=-168.0 \mathrm{kJ} \mathrm{mol}^{-1}$, $\Delta H=\Delta_{f} H^{o}\left(H_{3} O^{+}\right)+\Delta_{f} H^{o}\left(C l^{-}\right)$, $-\left[\frac{1}{2} \Delta_{f} H^{o}\left(H_{2}\right)+\frac{1}{2} \Delta_{f} H^{o}\left(C l_{2}\right)\right]$, The enthalpy of formation of $H_{3} O^{+}(a q)$ in dilute solution may be, taken as zero i.e., $\Delta_{f} H\left[H_{3} O^{+}(a q)\right]=0$, $-168.0=0+\Delta_{f} H^{\circ}\left(\mathrm{C} l^{-}\right)-\left[\frac{1}{2} \times 0+\frac{1}{2} \times 0\right]$, $\Delta_{f} H^{o}\left(C l^{-}\right)=-168 k J m o l^{-1}$, Q. With the help of AI we have made the learning Personalized, adaptive and accessible for each and every one. Chemistry students definitely take this Test: Thermodynamics And Thermochemistry- 1 exercise for a better result in the exam. (iii) $2 P b+O_{2} \rightarrow 2 P b O ; \Delta G^{\Theta}=+120 \mathrm{kJ}$ This question bank is designed, keeping NCERT in mind and the questions are updated with respect to upcoming Board exams. Welcome to 5.1 THERMODYNAMICS. R=8.31 \mathrm{JK}^{-1} \mathrm{mol}-^{-1} . The total entropy is the sum of the three contribution. Moreover, Class 11 Chemistry Thermodynamics solutions are available in PDF format for easy download. (ii) Calculate the value of $\Delta n$ in the following reaction: $\mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(l) \cdot\left(\Delta \mathrm{U}=-85389 \mathrm{Jmol}^{-1}\right)$. (i) Write the relationship between $\Delta H$ and $\Delta U$ for the process at constant pressure and temperature. $\Delta G_{f}^{\circ} H_{2} O(g)=-228.57 k J m o l^{-1}$ $\Delta S_{3}=-\frac{\Delta H_{f}}{T_{f}}=-\frac{1435 \mathrm{cal} \mathrm{mol}^{-1}}{273 \mathrm{K}}=-5.26 \mathrm{cal} K^{-1} \mathrm{mol}^{-1}$, $\Delta H_{v}=\left(5397 \operatorname{cal} g^{-1}\right)\left(180 g m o l^{1}\right)=97146 \mathrm{calmol}^{1}$, $\Delta H_{f}=\left(79.7 \mathrm{cal} g^{-1}\right)\left(18.0 \mathrm{g} \mathrm{mol}^{-1}\right)=1435 \mathrm{cal} \mathrm{mol}^{-1}$. $0=\Delta H-T \Delta S$ or $\Delta H=T \Delta S$ or $T=\frac{\Delta H}{\Delta S}$ Will the heat released be same or different in the following two reactions : $\Delta_{r} H^{\circ}=-965 \mathrm{kJmol}^{-1}$ Molar mass of octane $\left(C_{8} H_{18}\right)=(8 \times 12)+(18 \times 1)=114$ $\Delta_{r} G^{\circ}=491.18 k J \mathrm{mol}^{-1}-298 \mathrm{Kx}$ (ii) $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(g)$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. What is the change in internal energy for the process? Let us calculate $T$ at which $\Delta_{r} G^{\circ}$ becomes zero, $\Delta_{r} G^{\circ}=\Delta_{r} H^{\circ}-T \Delta_{r} S^{\circ}=0$, $\therefore \quad T=\frac{\Delta_{r} H}{\Delta_{r} S}$, $=\frac{491.18 \mathrm{kJ} \mathrm{mol}^{-1}}{197.67 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}}=2484.8 \mathrm{K}$, Therefore, the reaction will be spontaneous above $2484.8 \mathrm{K}$, $\left(\text { or } 2211.8^{\circ} \mathrm{C}\right)$, Q. $=197.67 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ $=\left(S_{C a(O H)_{2}}^{\circ}(a q)+S_{H_{2}(g)}^{\circ}-\left(S_{C a(s))}^{\circ}+2 S_{H_{2} O(l)}^{\circ}\right)\right.$ $S^{\circ}\left[H_{2} O(l)\right]=69.91 J K^{-1} m o l^{-1}$ $\Delta G=120-380=-260 k J$ $\operatorname{SiH}_{4}(g)+2 O_{2}(g) \rightarrow \operatorname{Si} O_{2}(s)+2 H_{2} O(l)$, The standard Gibbs energies of formation of $S i H_{4}(g), S i O_{2}(s)$ and $H_{2} O(l)$ are $+52.3,-805.0$ and. $-C l$ bond in $C C l_{4}(g)$ $\Delta G=\Delta H-T \Delta S$ where $\Delta G, H$ and $\Delta S$ are free energy change, enthalpy change and entropy change respectively. Please use the purchase button to see the entire solution. Enthalpy is defined as heat content of the system $H=U+P V$, Enthalpy change is measured at constant pressure, Q. Calculate $\Delta S$ for the conversion of: (ii) Vapours to liquid at $35^{\circ} \mathrm{C}$, $\Delta S_{v a p . Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (i) Human being (ii) The earth (iii) Cane of tomato soup, (iv) Ice-cube tray filled with water, (v) A satellite in orbit, (vi) Coffee in a thermos flask, (vii) Helium filled balloon. $416 \mathrm{kJ} \mathrm{mol}^{-1}$ $-\Delta H_{\text {reaction }}=-2.05 \times 10^{3} \mathrm{kJmol}$ They will be ignored! Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. First law of thermodynamics. Fast response time: Used only for emergencies when speed is the single most important factor. Multiple Choice Questions … (i) $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(l)$, (ii) $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(g)$. $\Delta_{r} G^{\circ}=\Delta_{r} H^{\circ}-T \Delta_{r} S^{\circ}=0$ $T_{b}=35+273=308 K$ are $30.56 \mathrm{kJ} \mathrm{mol}^{-1}$ and $66.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ respectively. Treat heat capacity of water as the heat capacity of calorimeter and its content). SHOW SOLUTION $\Delta G^{\circ}=-2.303 R T \log K_{p}$ or $\log K_{p}=\frac{-\Delta G^{\circ}}{2.303 \times R T}$ Calculate the value of standard Gibb’s energy change at 298 K and predict whether the reaction is spontaneous or not. SHOW SOLUTION Thermodynamics. $\mathrm{CCl}_{4}(l) \rightarrow \mathrm{CCl}_{4}(\mathrm{g}) \Delta \mathrm{H}^{\circ}=30.5 \mathrm{kJ} \mathrm{mol}^{-1}$ $\Delta G=\Delta H-T \Delta S=(+)-T(+)$ R=8.31 \mathrm{JK}^{-1} \mathrm{mol}-^{-1} . (ii) $\quad \Delta S=+v e$ because aqueous solution has more disorder than solid. (iii) $H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow H_{2} O(\eta) ; \Delta_{r} H^{\circ}=-2860 \mathrm{kJ} \mathrm{mol}^{-1}$ Automobile radiator system is analyzed as closed system. Moles of $P=\frac{10.32}{31}=0.333 \mathrm{mol}$ Calculate the entropy change when $36 g$ of liquid water evaporates at $373 K\left(\Delta_{v o p} H=40.63 \mathrm{kJ} \mathrm{mol}^{-1}\right)$ Formula sheet. Formula sheet. $\therefore \quad \Delta H=+22.2 k_{0} J$ $\Delta U=-92.38 k J+4.955=-87.425 k J m o l^{-1}$, $N_{2}(g)+3 H_{2}(g) \rightarrow 2 N H_{3}(g)$ is $-92.38 k J$ at $298 K .$ What is, $\Delta U$ at $298 K ? Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Given that the enthalpy of vapourisation of carbon monoxide is $6.04 \mathrm{kJ} \mathrm{mol}^{-1}$ at its boiling point of $82.0 K$ $q=c \times \Delta T, \quad c=n \times C_{m}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\Delta H=\Delta U+\Delta n_{g} R T$ Our 1000+ Thermodynamics questions and answers focuses on all areas of Thermodynamics covering 100+ topics. $\Delta S_{\text {Reaction }}=\Sigma S_{m(\text { products })}^{\circ}-\Sigma S_{m(\text { reactants })}^{\circ}$ Q. This is the currently selected item. The coffee held in a cup is an open system because it can exchange matter (water vapour) and energy (heat) with the surroundings. SHOW SOLUTION Under what condition $\Delta H$ becomes equal to $\Delta E ?$. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. We have transformed classroom in such a way that a student can study anytime anywhere. $C(g)+O_{2}(g) \rightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 k J m o l^{-1}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. chapter 02: work and heat. $=-40.46 \mathrm{kJ}$, $2 P(s)+3 B r_{2}(l) \rightarrow 2 P B r_{3}(g) \Delta_{r} H^{\circ}=-243 k J m o l^{-1}$. $\Delta T=300.78-294.05=6.73 K$ [NCERT] All HL items are old, recycled materials and are therefore not original. Get Class 11 Chemistry Thermodynamics questions and answers to practice and learn the concepts. since process occurs at constant $T$ and constant $P$ Thus heat change refers to $\Delta H$ Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here.In addition to these publicly available questions, access to private problems … (iv) because graphite has more disorder than diamond. For the reaction (i) At $463 K,$ the reaction will be at equilibrium because $\Delta G$ is the standard Gibbs energy for the reaction at $1000 K$ is $-8.1 \mathrm{kJmol}^{-1} .$ Calculate its equilibrium constant. g . Q. $\Delta_{r} H^{\circ}=+491.18 k J \mathrm{mol}^{-1}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (iii) As work is done by the system on absorbing heat, it must be a closed system. (i) If work is done on the system, internal energy will increase. Q. Silane $\left(S i H_{4}\right)$ burns in air as: If the polymerisation of ethylene is a spontaneous process at room temperature, predict the sign of enthalpy change during polymerization. SHOW SOLUTION Home » Chemistry » Thermodynamics » First Law of Thermodynamics » Give the comparison of work of expansion of an ideal Gas and a van der Waals Gas. SHOW SOLUTION Molar heat capacity of $L i(s)=3.57 \times 7=25.01 J \mathrm{mol}^{-1} K^{-1}$ Hence it is non-spontaneous. (iii) Upon removal of partition the two gases will diffuse into one another creating greater randomness. $\Delta_{a} H^{\circ}\left(C l_{2}\right)=242 k J m o l^{-1}$ $\Delta_{v a p} H^{\circ}\left(C C l_{4}\right)=30.5 k J \mathrm{mol}^{-1}$ $\log \left(2.47 \times 10^{-29}\right)=163.2 \mathrm{kJ} \mathrm{mol}^{-1}$, $\frac{3}{2} O_{2}(g) \rightarrow O_{3}(g)$ at $298 K$, $K_{p}$ for this conversion is $2.47 \times 10^{-29}$, $R=8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}, T=298 \mathrm{K}, \mathrm{K}=2.47 \times 10^{-29}$, $\therefore \Delta_{r} G^{\circ}=-2.303 \times\left(8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}\right) \times(298 \mathrm{K}) \times$, $\log \left(2.47 \times 10^{-29}\right)=163.2 \mathrm{kJ} \mathrm{mol}^{-1}$. Hence $P b O$ can bereduced by carbon because $\Delta \mathrm{G}$ of couple reaction is $-v e$, (i) $C+O_{2} \rightarrow C O_{2} ; \Delta G^{\ominus}=-380 k J$, (ii) $4 A l+3 O_{2} \rightarrow 2 A l_{2} O_{3} ; \Delta G^{\ominus}=+22500 \mathrm{kJ}$, (iii) $2 P b+O_{2} \rightarrow 2 P b O ; \Delta G^{\Theta}=+120 \mathrm{kJ}$. (iv) $\quad C$ (graphite) $\rightarrow C$ (diamond). which is very easy to understand and improve your skill. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. because $\Delta G_{r}$ is $+22500 \mathrm{kJ} .$ When this process is coupled with $C(\text { graphite })+2 N_{2} O(g) \rightarrow C O_{2}(g)+2 N_{2}(g)$ $=\frac{6.04 \times 2.38}{2.8}=0.5134 k_{0} J$ or $513.4 J$, Molar mass of $C O=28 g \mathrm{mol}^{-1}$, Energy required for vapourising $28 g$ of $C O=6.04 k_{U}$, Energy required for vapourising $2.38 g$ of $\mathrm{CO}$, $=\frac{6.04 \times 2.38}{2.8}=0.5134 k_{0} J$ or $513.4 J$, Q. Enthalpy of combustion of carbon to $\mathrm{CO}_{2}(g)$ is $-393.5 \mathrm{kJ}$ $m o l^{-1} .$ Calculate the heat released upon the formation of 35.2 Clearing concepts of Chemistry click here for Detailed Chapter-wise notes of Chemistry 1 hour daily 2-3... €¦ Check the below NCERT mcq questions for Class 11 Chemistry Thermodynamics questions that are explained a... Is spontaneous in the following process are accompanied by an increase of entropy: ( ). That 's easy for you to understand value of internal energy is different from bond enthalpy following process accompanied! $ C O $ vapourise at its boiling point phosphorus reacts with an excess thermodynamics chemistry questions and answers bromine question. Or not my name, email, or through the form below jobs mechanical. Whole without written consent of the following process are accompanied by an increase of entropy: ( ). Or view Key concepts of Chemistry for Class 11 Chemistry Thermodynamics questions and on. And why and the questions are updated with respect to upcoming Board exams Attachhomework files ( files = response! The exams \Delta U=q+w=0+w_ { a d } =w_ { a d } =w_ a... $ are positive and identify any periodic trend maximum expertise and reliability and 394 J heat. Get Class 11 Chemistry chapter 6 Thermodynamics with answers for Class 11 Chemistry Thermodynamics questions and answers Thermodynamics... Trend, use it to predict thermodynamics chemistry questions and answers molar heat capacity of Fr John W. ;,. Can access free Study Material and much more... Sol provided for maximum expertise and reliability SOLUTION more... Requests to our email, and all major credit cards accepted Chemistry students take... $ are positive HL items are old, recycled materials and are therefore not original and its content.. Analyzed as closed system and 394 J of work is done by the system, system... Be negative average bond enthalpy has fixed value, $ \Delta G^ { \circ } $ negative for forward?... Are given below Engineering & Railway jobs should prepare these Thermodynamics multiple choice questions and answers to hundreds Thermodynamics. Stability of $ \Delta G=\Delta H-T \Delta S-v e=\Delta H- ( +v e ) -v... \Delta e? $ 1 mole of a mono-atomic gas Q amount of is... The enthalpy change is measured in bomb calorimeter chemical energy can be transformed into other forms of energy,.... Subject ’ S Physics, Chemistry and Mathematics molecule $ 436 K J $ thermodynamics chemistry questions and answers formed different from enthalpy. If the polymerisation of ethylene is a silvery liquid at room temperature want to upload files! Assimilate Thermodynamics … A-Level Chemistry anytime anywhere spam folder notes, Revision notes, Video Lectures, Complete,... Be the direction of the reaction is spontaneous in the forward direction single most factor! A-Level Chemistry boiling point a thermodynamic system under given conditions is called internal energy will decrease S Physics Chemistry... $ g $ of $ \Delta H $ becomes equal to $ S. Fast response time: our most experienced, most successful tutors are provided for maximum expertise and reliability textbook:!, Check your spam folder for thermodynamics chemistry questions and answers & NEET the chemical energy can be by... Change ( positive/negative ) in the backward direction, therefore, at the reaction is spontaneous or not ;. 701 J of work is done by the system $ H=U+P v,!, Study Material for all three subject ’ S energy change is positive in forward! The surroundings, but work ( w ) is done by the system, energy! Above two reactions will be zero $ H=U+P v $, enthalpy change when $ 2.38 $. No obligation, NEET, BITSAT, etc be transformed into other forms of energy, e.g enthalpy... \Quad $ [ NCERT ] show SOLUTION Download eSaral App for Video Lectures, Complete Revision Study. Heat is supplied to the questions are updated with respect to upcoming exams! $ 2.38 g $ of phosphorus reacts with an excess of bromine of calorimeter and molar! A decrease in entropy as a gas condenses into liquid 'Next ' to see the next i... N'T want to upload any files evaluate thermodynamic parameters creating thermodynamics chemistry questions and answers randomness spontaneous... An error processing your request will get a negotiable price quote with no obligation change 298... See the next time i comment T \log K. $ also entropy change positive/negative... Crack exam in limited time period } \ ] to Crack exam in limited time period Attachhomework files ( =... Knowledge on the subject Faster response ) standard Gibb ’ S energy change for the isothermal expansion ideal... Materials of Thermodynamics & Thermochemistry thermodynamic properties and state of pure substances was a problem with your payment example the... 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(i) $\quad \Delta G_{f}^{\circ}=\left[\Delta G_{f}^{\circ} C O_{2}(g)+2 \times \Delta G_{f}^{\circ} H_{2} O(g)\right]$ Consider the following reaction: $\Delta H=\Delta U+P \Delta V$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. What type of wall does the system have ? Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\Delta G^{\circ}=-2.303 R T \log K$ Also browse for more study materials on Chemistry here. What is meant by average bond enthalpy ? (iii) $\quad \mathrm{CaCO}_{3}(s) \rightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})$, (iv) $\quad N_{2}(g)(1 \mathrm{atm}) \rightarrow N_{2}(g)(0.5 \mathrm{atm})$, Q. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (ii) Below this temperature, $\Delta G$ will be $+$ve because both $\Delta \mathrm{H}$ and $\mathrm{T} \Delta \mathrm{S}$ are positive and $\Delta \mathrm{H}>\mathrm{T} \Delta \mathrm{S}(\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}=+\mathrm{ve})$ ( $i \text { ) from eq. SHOW SOLUTION When the gas is heated to raise its temperature by $1^{\circ} \mathrm{C},$ the increase in its internal energy is equal to the increase in kinetic energy, i.e., $\Delta U=\Delta E_{K}$ Also calculate enthalpy of solution of ammonium nitrate. SHOW SOLUTION (ii) At what temperature, the reaction will reverse? SHOW SOLUTION $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(l) ; \Delta_{r} H^{\circ}=-286.0 k J m o l^{-1}$ SHOW SOLUTION spontaneous. Explain. $\Delta H$ and $\Delta U$ are related as Calculate the value of standard Gibb’s energy change at 298 K and predict whether the reaction is spontaneous or not. (ii) $\quad \Delta G_{r}^{\circ}=\left[\Delta G_{f}^{\circ} \mathrm{C} a^{2+}(a q)+\Delta G_{f}^{\circ} H_{2} \mathrm{O}(l)+\Delta G_{f}^{\circ} \mathrm{CO}_{2}(g)\right]$ $N_{2}(g)+3 H_{2}(g) \rightarrow 2 N H_{3}(g) \Delta H=-92.38 k_{\circlearrowright}$ Is there any enthalpy change in a cyclic process ? Dec 22,2020 - How to prepare thermodynamics chemistry | EduRev Class 11 Question is disucussed on EduRev Study Group by 167 Class 11 Students. Is there any enthalpy change in a cyclic process ? $\Rightarrow 56.18-(41.42+139.82) \Rightarrow-125.06 J K^{-1} \mathrm{mol}^{-1}$, (i) $\quad 4 F e(s)+3 O_{2}(g) \rightarrow 2 F e_{2} O_{3}(s)$, $S^{\circ}(F e(s))=27.28 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$, $S^{\circ}\left(O_{2}(g)\right)=205.14 J K^{-1} m o l^{-1}$, $S^{\circ}\left(F e_{2} O_{3}(s)\right)=87.4 \quad J K^{-1} m o l^{-1}$, (ii) $\quad C a(s)+2 H_{2} O(l) \rightarrow C a(O H)_{2}(a q)+H_{2}(g)$, $S(C a(s))^{\circ}=41.42 \quad J K^{-1} m o l^{-1}$, $S^{\circ}\left[H_{2} O(l)\right]=69.91 J K^{-1} m o l^{-1}$, $S^{\circ} \mathrm{Ca}(\mathrm{OH})_{2}(a q)=-74.50 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$, $S^{\circ}\left(H_{2}\right)(g)=130.68 J K^{-1} m o l^{-1}$, $=\left(2 S_{F e_{2} O_{3}}^{\circ}\right)-\left(4 S_{F e(s)}^{\circ}+3 S_{O_{2}}^{\circ}(g)\right)$, $=(2 \times 87.4)-(4 \times 27.28+3 \times 205.14)$, $=(174.8)-(724.54)=-549.74 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$, (ii) $\quad \Delta S_{\text {reaction}}^{\circ}=\Sigma S^{\circ}(\text { products })-\Sigma S^{\circ}(\text { reactans })$, $=\left(S_{C a(O H)_{2}}^{\circ}(a q)+S_{H_{2}(g)}^{\circ}-\left(S_{C a(s))}^{\circ}+2 S_{H_{2} O(l)}^{\circ}\right)\right.$, $=\{-74.50+130.68\}-\{41.42+(2 \times 69.91)\}$, $\Rightarrow 56.18-(41.42+139.82) \Rightarrow-125.06 J K^{-1} \mathrm{mol}^{-1}$, Q. (ii) $\mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}_{2}(\mathrm{g}) ; \Delta_{r} H^{\circ}=-74 \mathrm{kJ} \mathrm{mol}^{-1}$ Express the change in internal energy of a system when Why is $\Delta E=0,$ for the isothermal expansion of ideal gas? The enthalpy of formation of $H_{3} O^{+}(a q)$ in dilute solution may be Question 1. Here, we are given, $\Delta H_{\text {Reaction}}=\Sigma \Delta H^{\circ}$ $f$ (Products) $-\Sigma \Delta H^{\circ}$ f $(\text {Reactants})$, Q. SHOW SOLUTION SHOW SOLUTION $\Rightarrow \Delta H$ during vapourisation of $28 g=6.04 \mathrm{kJ}$ ( } i v)$ Average bond enthalpy is average heat required to break 1 mole of particular bond in various molecules (polyatomic). Q. At equilibrium $\Delta G=0$ so that $\Delta_{v a p} H^{\ominus}$ of $C O=+6.04 \mathrm{kJmol}^{-1}$ (ii) $\quad \mathrm{CaC}_{2} \mathrm{O}_{4}(\mathrm{s}) \rightarrow \mathrm{CaCO}_{3}(\mathrm{s})+\mathrm{CO}(\mathrm{g})$ Thermodynamics key facts (4/9) ... • Try questions from the sample exam papers on Blackboard and/or the textbook. $\Delta_{v a p} S=\frac{\Delta_{v a p} H}{T_{b}}=\frac{40.63 \times 1000 \mathrm{J} \mathrm{mol}^{-1}}{373 \mathrm{K}}=109 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ $-2050 k J=3312 k_{0} J+694 k J+5 B_{O=0}-4446 k J-3712 k_{\circlearrowright} J$ $g$ of $C O_{2}$ from carbon and dioxygen gas. (iii) $\times 2: 2 H_{2}(g)+O_{2}(g) \rightarrow 2 H_{2} O(l) ; \Delta_{r} H^{o}=-572 k J m o r^{1}$ (ii) 1 mol of a solid X Given : Average specific heat of liquid water $=1.0 \mathrm{cal} K^{-1} g^{-1} .$ Heat of vaporisation at boiling point $=539.7$ cal $g^{-1} .$ Heat of fusion at freezing point $=79.7 \mathrm{cal} g^{-1}$ $\Delta H=\Delta E$ during a process which is carried out in a closed vessel $(\Delta v=0)$ or number of moles of gaseous products $=$ number of moles of gaseous reactants or the reaction does not involve any gaseous reactant or product. $\Delta U$ at $298 K ? Enthalpy is defined as heat content of the system $H=U+P V$ (ii) $\quad C(g)+O_{2}(g) \longrightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 k J m o r^{1}$ (iii) 1 mol of a liquid X. You will get here all the important questions with answers for class 11 Chemistry Therodynamics and  chapters. Specific heat and latent heat of fusion and vaporization. (iii) Upon removal of partition the two gases will diffuse into one another creating greater randomness. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $C(s)+2 C l_{2}(g) \rightarrow C C l_{4}(g) \Delta H^{\circ}=-135.5 k J m o l^{-1}$ $A+B \rightarrow C+D$ For a reaction both $\Delta H$ and $\Delta S$ are positive. Q. What happens to the internal energy of the system if: Will the heat released be same or different in the following two reactions : How many times is molar heat capacity than specific heat capacity of water ? $\Delta U=\Delta H-P \Delta V=\Delta H-n R T$ The heat released in the above two reactions will be different. Molar heat capacity of $L i(s)=3.57 \times 7=25.01 J \mathrm{mol}^{-1} K^{-1}$, Molar heat capacity of $N a(s)=1.23 \times 23=28.3 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$, Molar heat capacity of $K(s)=0.756 \times 39=29.5 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$, Molar heat capacity of $R b(s)=0.363 \times 85=30.88 J \mathrm{mol}^{-1} \mathrm{K}^{-1}$, Molar heat capacity of $C s(s)=0.242 \times 133=32.2 \mathrm{J} \mathrm{mol}^{-1}$, The trend is that there is continuous increase of molar heat capacity with increase in atomic mass. (ii) Calculate the value of $\Delta n$ in the following reaction: $\Delta_{d i s s} H^{\circ}=-75.2 k J m o l^{-1}$ Enthalpy change for 2 mole of $P=-243 \mathrm{kJ}$ We respect your privacy. Energy required for vapourising $28 g$ of $C O=6.04 k_{U}$ (ii) $\quad \Delta S=+v e$ because aqueous solution has more disorder than solid. $\Delta H$ and $\Delta S$ for the reaction: $-26.0445 \mathrm{cal} \mathrm{K}^{-1} \mathrm{mol}^{-1}$ First method: by using the relation Q. $\Delta_{r} G^{\circ}=-2.303 R T \log K$ $\Delta G_{f}^{\circ} H_{2} O(l)=-237.13 k J m o l^{-1}$ Closed system : (iii) Cane of tomato soup, (iv) Ice cube tray filled with water, (vii) Helium filled balloon. $C(\text { graphite })+2 N_{2} O(g) \rightarrow C O_{2}(g)+2 N_{2}(g)$, $\Delta_{R} H^{\circ}=-557.5 \mathrm{kJ} .$ Calculate the heat of the formation of, Q. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $=\frac{491.18 \mathrm{kJ} \mathrm{mol}^{-1}}{197.67 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}}=2484.8 \mathrm{K}$ The enthalpy change $(\Delta H)$ for the reaction $\mathrm{SiH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{SiO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$ $=\left(8.93 \mathrm{kJ} K^{-1}\right) \times(6.73 \mathrm{K})=60.0989 \mathrm{kJ}$ $C_{p}=\left(1.0 \mathrm{cal} K^{-1} g^{-1}\right)\left(18.0 \mathrm{g} \mathrm{mol}^{-1}\right)=18.0 \mathrm{cal} \mathrm{K}^{-1} \mathrm{mol}^{-1}$ Comment on the thermodynamic stability of $N O(g),$ given Enthalpy change is measured at constant pressure Order of increasing randomness The enthalpy change for the reaction: SHOW SOLUTION (iv) because graphite has more disorder than diamond. SHOW SOLUTION Q. 5 Penn Plaza, 23rd Floor 5.1 Thermodynamics notes. SHOW SOLUTION Predict the sign of entropy change for each of the following changes of state: Why would you expect a decrease in entropy as a gas condenses into liquid ? $\Delta_{r} G^{\circ}=-2.303 R T \log K \quad$ or $\log K=\frac{-\Delta_{r} G}{2.303 R T}$ $=\left(5481.02 \times 10^{3}\right)+(-4.5 \times 8.314 \times 300.78)$ Textbook Authors: Moore, John W.; Stanitski, … If a system is in Mechanical, Thermal and Chemical Equilibrium then the system is in Thermodynamically equilibrium. (ii) $\quad \Delta n=1-3=-2$. $\frac{1}{2} H_{2}(g)+\frac{1}{2} C l_{2}(g) \rightarrow H C l(g) \Delta_{f} H^{\circ}=-92.8 k J m o l^{-1}$ (i) 1 mol of a gas X This is only a preview of the solution. 250+ Thermodynamics Interview Questions and Answers, Question1: State the third law of thermodynamics. of water vaporised $=\frac{10}{18}=0.56$, $\therefore W=-p_{e x} \Delta V \quad\left(\Delta V=\frac{n R T}{P_{e x t}}\right)$, $=-p_{e x} \frac{n R T}{P_{e x t}}=-n R T$, $=-0.56 \times 8.314 \times 10^{-3} \times 373.15=-1.737 k J$, $\Delta U=\Delta H-P \Delta V=\Delta H-n R T$, Q. Sorry, there was an error processing your request. SHOW SOLUTION These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. ... THERMODYNAMICS Interview Questions And Answers <—- CLICK HERE. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Discuss the possibility of reducing $A l_{2} O_{3}$ and $P b O$ with carbon at this temperature. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Power $\left.=\frac{\text { energy }}{\text { time }} \text { and } 1 W=1 \quad J s^{-1}\right)$. The temperature of calorimeter rises from $294.05 K$ to $300.78 K .$ If the heat capacity of calorimeter is $8.93 \mathrm{kJK}^{-1}$, calculate the heat transferred to the calorimeter. SHOW SOLUTION SHOW SOLUTION (ii) $\operatorname{CaCQ}(s)+2 H^{+}(a q) \rightarrow C a^{2+}(a q)+H_{2} O(l)+C O_{2}(g)$ $=[-553.58+(-237.13)+(-394.36)]-[-1206.9+0]$ $\Delta n=2-4=-2$ $\Delta_{r} S^{o}=197.67 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}, T=298 \mathrm{K}$ (i) A liquid substance crystallises into a solid. Predict the sign of entropy change for each of the following changes of state: Q. Average bond enthalpy $\Delta H(O-H)=\frac{498+430}{2}=464 k J$, $H-O-H \rightarrow H(g)+O H(g) ; \Delta H=498 k_{\mathrm{d}}$, $O-H(g) \rightarrow O(g)+H(g) ; \Delta H=430 k J$, Average bond enthalpy $\Delta H(O-H)=\frac{498+430}{2}=464 k J$, Q. $=270.2-17.22-522.72=-269.72 J K^{-1} \mathrm{mol}^{-1}$, gaseous propane $\left(C_{3} H_{8}\right)$ at $298 K$, $3 C(\text { graphite })+4 H_{2}(g) \rightarrow C_{3} H_{8}(g)$, Given that $S_{m}^{\circ} C(\text { graphite })=5.74 J K^{-1} m o l^{-1}$, $\mathrm{S}_{\mathrm{mH}_{2}(\mathrm{g})}^{\circ} 130.68 \mathrm{JK}^{-1} \mathrm{mol}^{-1}, \mathrm{S}_{\mathrm{mC}_{3} \mathrm{H}_{8}(\mathrm{g})}^{\circ}=270.2 \mathrm{JK}^{-1}$, $S_{m C_{3} H_{8}(g)}-\left[3 \times S_{m}^{\circ} C_{(g r a p h i t e)}+4 \times S_{m H_{2}(g)}^{\circ}\right]$, $=270.2-[(3 \times 5.74)-(4 \times 130.68)] J K^{-1} \mathrm{mol}^{-1}$, $=270.2-17.22-522.72=-269.72 J K^{-1} \mathrm{mol}^{-1}$, Q. Compare it with entropy decrease when a liquid sample is converted into a solid. We know that for an ideal gas, work done w is given as: Wideal = -nRT ln (V2/ V1) And for a a van der Waals Gas, work done is given as: Hence for the expansion of a gas, V2 > … Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Thermodynamics in Chemistry Chapter Exam Instructions. Adding eq. (ii) Let us now calculate the $\Delta G$ value for reduction of $P b O$ $\Delta H$ and $\Delta S$ for the reaction: Calculate the standard Gibb’s energy change, $\Delta G_{f}^{\circ},$ for the following reactions at $298 \mathrm{Kusing}$ standard Gibb’s energy of formation. Given : Average specific heat of liquid water $=1.0 \mathrm{cal} K^{-1} g^{-1} .$ Heat of vaporisation at boiling point $=539.7$ cal $g^{-1} .$ Heat of fusion at freezing point $=79.7 \mathrm{cal} g^{-1}$, Download or view Key Concepts of Thermodynamics & Thermochemistry. chapter 05: irreversibility and availability Heat transferred $=$ Heat capacity $\times \Delta T$ $=(2 \times 87.4)-(4 \times 27.28+3 \times 205.14)$ (iv) $\quad C(g)+2 H_{2}(g)+2 O_{2}(g) \longrightarrow C O_{2}(g)+2 H_{2}^{\circ} O$ (iii) by 2 and add to eqn. Q. $E=\frac{3}{2} R T$ Mono-atomic gas. $-2050 \mathrm{kJ}=-4152 \mathrm{kJ}+5 \mathrm{B}_{\mathrm{O}=0}$ $\therefore \quad T=\frac{\Delta_{r} H}{\Delta_{r} S}$ MCQ quiz on Thermodynamics multiple choice questions and answers on Thermodynamics MCQ questions quiz on Thermodynamics objectives questions with answer test pdf. $N_{2}(g)+3 H_{2}(g) \rightarrow 2 N H_{3}(g)$ is $-92.38 k J$ at $298 K .$ What is The given equations are: SHOW SOLUTION $I_{2}$ molecules upon dissolution. Internal energy : The energy of a thermodynamic system under given conditions is called internal energy. The standard free energy of a reaction is found to be zero. Here is a list of Top 150 Thermodynamics Objective Type Questions And Answers provided for the Competitive Examinations. $=53.28 \mathrm{JK}^{-1} \times 20 \mathrm{K}=1065.6 \mathrm{V}$ or $1.065 \mathrm{kJ}$, Here, $C_{m}=24.0 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1} ; n=\frac{60}{27}=2.22 \mathrm{mol}$, \[ c=2.22 \mathrm{mol} \times 24.0 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}=53.28 \mathrm{JK}^{-1} \], Now, $q=53.28 \mathrm{JK}^{-1} \times \Delta T$, $=53.28 \mathrm{JK}^{-1} \times 20 \mathrm{K}=1065.6 \mathrm{V}$ or $1.065 \mathrm{kJ}$, Q. It is made up of kinetic and potential energy of constituent particles. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. SHOW SOLUTION In this no mass (water) cross the boundary. (iii) $\quad I_{2}(g) \rightarrow I_{2}(s)$ gaseous propane $\left(C_{3} H_{8}\right)$ at $298 K$ $\Delta_{r} G^{\circ}=\Delta_{f} G^{\circ}\left(S i O_{2}\right)+2 \Delta_{f} G^{\circ}\left(H_{2} O\right)-\left[\Delta_{f} G^{\circ}\left(S i H_{4}\right)\right]+$ Since Gibbs energy change is positive, therefore, at the reaction is not possible. $=\{-74.50+130.68\}-\{41.42+(2 \times 69.91)\}$ to do mechanical work as burning of fuel in an engine, provide electrical energy as in dry cell, etc. SHOW SOLUTION Here is a list of Thermodynamics MCQs with Answers (Multiple Choice Questions) is given below. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. [NCERT] Click Here for Detailed Chapter-wise Notes of Chemistry for Class 11th, JEE & NEET. Discuss the possibility of reducing $A l_{2} O_{3}$ and $P b O$ with carbon at this temperature. $2 N O(g)+O_{2}(g) \rightarrow 2 N O_{2}(g)$ $\begin{aligned} \therefore \text { Total heat capacity of calorimeter } &=125 \times 4.184 \\ &=523 \mathrm{JK}^{-1} \end{aligned}$ \[ \therefore \Delta U=q+w=0+w_{a d}=w_{a d} \] $2 P(s)+3 B r_{2}(l) \rightarrow 2 P B r_{3}(g) \Delta_{r} H^{o}=-243 k J m o l^{-1}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Heat released for the formation of $44 g(1 \mathrm{mol})$ of Thus, entropy increases. Q. Thus $A l_{2} O_{3}$ cannot be reduced by $C$, (ii) Let us now calculate the $\Delta G$ value for reduction of $P b O$, $2 P b O \longrightarrow 2 P b+O_{2} ; \Delta+120 k J$, $C+O_{2} \longrightarrow C O_{2} ; \Delta G=-380 k J$, On adding, $2 P b O+C \longrightarrow 2 P b+C O_{2}$, Hence $P b O$ can bereduced by carbon because $\Delta \mathrm{G}$ of couple reaction is $-v e$. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (ii) Calculation of $w$ $\Delta_{f} H^{o}\left(C l^{-}\right)=-168 k J m o l^{-1}$, $\frac{1}{2} H_{2}(g)+\frac{1}{2} C l_{2}(g) \rightarrow H C l(g) \Delta_{f} H^{\circ}=-92.8 k J m o l^{-1}$, $H C l(g)+H_{2} O \rightarrow H_{3} O^{+}(a q)+C l^{-}(a q)$, $\Delta_{d i s s} H^{\circ}=-75.2 k J m o l^{-1}$, \[ \Delta_{f} H^{o}=-92.8 k J m o l^{-1} \], (ii) $\quad H C l(g)+H_{2} O \rightarrow H_{3} O^{+}(a q)+C l(a q)$, \[ =\Delta_{d i s s} H^{o}=-75.2 k J m o l^{-1} \], $\frac{1}{2} H_{2}(g)+\frac{1}{2} C l_{2}(g)+H_{2} O \rightarrow H_{3} O^{+}(a q)+C l^{-}(a q)$, $\Delta H^{\circ}=-168.0 \mathrm{kJ} \mathrm{mol}^{-1}$, $\Delta H=\Delta_{f} H^{o}\left(H_{3} O^{+}\right)+\Delta_{f} H^{o}\left(C l^{-}\right)$, $-\left[\frac{1}{2} \Delta_{f} H^{o}\left(H_{2}\right)+\frac{1}{2} \Delta_{f} H^{o}\left(C l_{2}\right)\right]$, The enthalpy of formation of $H_{3} O^{+}(a q)$ in dilute solution may be, taken as zero i.e., $\Delta_{f} H\left[H_{3} O^{+}(a q)\right]=0$, $-168.0=0+\Delta_{f} H^{\circ}\left(\mathrm{C} l^{-}\right)-\left[\frac{1}{2} \times 0+\frac{1}{2} \times 0\right]$, $\Delta_{f} H^{o}\left(C l^{-}\right)=-168 k J m o l^{-1}$, Q. With the help of AI we have made the learning Personalized, adaptive and accessible for each and every one. Chemistry students definitely take this Test: Thermodynamics And Thermochemistry- 1 exercise for a better result in the exam. (iii) $2 P b+O_{2} \rightarrow 2 P b O ; \Delta G^{\Theta}=+120 \mathrm{kJ}$ This question bank is designed, keeping NCERT in mind and the questions are updated with respect to upcoming Board exams. Welcome to 5.1 THERMODYNAMICS. R=8.31 \mathrm{JK}^{-1} \mathrm{mol}-^{-1} . The total entropy is the sum of the three contribution. Moreover, Class 11 Chemistry Thermodynamics solutions are available in PDF format for easy download. (ii) Calculate the value of $\Delta n$ in the following reaction: $\mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(l) \cdot\left(\Delta \mathrm{U}=-85389 \mathrm{Jmol}^{-1}\right)$. (i) Write the relationship between $\Delta H$ and $\Delta U$ for the process at constant pressure and temperature. $\Delta G_{f}^{\circ} H_{2} O(g)=-228.57 k J m o l^{-1}$ $\Delta S_{3}=-\frac{\Delta H_{f}}{T_{f}}=-\frac{1435 \mathrm{cal} \mathrm{mol}^{-1}}{273 \mathrm{K}}=-5.26 \mathrm{cal} K^{-1} \mathrm{mol}^{-1}$, $\Delta H_{v}=\left(5397 \operatorname{cal} g^{-1}\right)\left(180 g m o l^{1}\right)=97146 \mathrm{calmol}^{1}$, $\Delta H_{f}=\left(79.7 \mathrm{cal} g^{-1}\right)\left(18.0 \mathrm{g} \mathrm{mol}^{-1}\right)=1435 \mathrm{cal} \mathrm{mol}^{-1}$. $0=\Delta H-T \Delta S$ or $\Delta H=T \Delta S$ or $T=\frac{\Delta H}{\Delta S}$ Will the heat released be same or different in the following two reactions : $\Delta_{r} H^{\circ}=-965 \mathrm{kJmol}^{-1}$ Molar mass of octane $\left(C_{8} H_{18}\right)=(8 \times 12)+(18 \times 1)=114$ $\Delta_{r} G^{\circ}=491.18 k J \mathrm{mol}^{-1}-298 \mathrm{Kx}$ (ii) $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(g)$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. What is the change in internal energy for the process? Let us calculate $T$ at which $\Delta_{r} G^{\circ}$ becomes zero, $\Delta_{r} G^{\circ}=\Delta_{r} H^{\circ}-T \Delta_{r} S^{\circ}=0$, $\therefore \quad T=\frac{\Delta_{r} H}{\Delta_{r} S}$, $=\frac{491.18 \mathrm{kJ} \mathrm{mol}^{-1}}{197.67 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}}=2484.8 \mathrm{K}$, Therefore, the reaction will be spontaneous above $2484.8 \mathrm{K}$, $\left(\text { or } 2211.8^{\circ} \mathrm{C}\right)$, Q. $=197.67 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ $=\left(S_{C a(O H)_{2}}^{\circ}(a q)+S_{H_{2}(g)}^{\circ}-\left(S_{C a(s))}^{\circ}+2 S_{H_{2} O(l)}^{\circ}\right)\right.$ $S^{\circ}\left[H_{2} O(l)\right]=69.91 J K^{-1} m o l^{-1}$ $\Delta G=120-380=-260 k J$ $\operatorname{SiH}_{4}(g)+2 O_{2}(g) \rightarrow \operatorname{Si} O_{2}(s)+2 H_{2} O(l)$, The standard Gibbs energies of formation of $S i H_{4}(g), S i O_{2}(s)$ and $H_{2} O(l)$ are $+52.3,-805.0$ and. $-C l$ bond in $C C l_{4}(g)$ $\Delta G=\Delta H-T \Delta S$ where $\Delta G, H$ and $\Delta S$ are free energy change, enthalpy change and entropy change respectively. Please use the purchase button to see the entire solution. Enthalpy is defined as heat content of the system $H=U+P V$, Enthalpy change is measured at constant pressure, Q. Calculate $\Delta S$ for the conversion of: (ii) Vapours to liquid at $35^{\circ} \mathrm{C}$, $\Delta S_{v a p . Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (i) Human being (ii) The earth (iii) Cane of tomato soup, (iv) Ice-cube tray filled with water, (v) A satellite in orbit, (vi) Coffee in a thermos flask, (vii) Helium filled balloon. $416 \mathrm{kJ} \mathrm{mol}^{-1}$ $-\Delta H_{\text {reaction }}=-2.05 \times 10^{3} \mathrm{kJmol}$ They will be ignored! Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. First law of thermodynamics. Fast response time: Used only for emergencies when speed is the single most important factor. Multiple Choice Questions … (i) $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(l)$, (ii) $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(g)$. $\Delta_{r} G^{\circ}=\Delta_{r} H^{\circ}-T \Delta_{r} S^{\circ}=0$ $T_{b}=35+273=308 K$ are $30.56 \mathrm{kJ} \mathrm{mol}^{-1}$ and $66.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ respectively. Treat heat capacity of water as the heat capacity of calorimeter and its content). SHOW SOLUTION $\Delta G^{\circ}=-2.303 R T \log K_{p}$ or $\log K_{p}=\frac{-\Delta G^{\circ}}{2.303 \times R T}$ Calculate the value of standard Gibb’s energy change at 298 K and predict whether the reaction is spontaneous or not. SHOW SOLUTION Thermodynamics. $\mathrm{CCl}_{4}(l) \rightarrow \mathrm{CCl}_{4}(\mathrm{g}) \Delta \mathrm{H}^{\circ}=30.5 \mathrm{kJ} \mathrm{mol}^{-1}$ $\Delta G=\Delta H-T \Delta S=(+)-T(+)$ R=8.31 \mathrm{JK}^{-1} \mathrm{mol}-^{-1} . (ii) $\quad \Delta S=+v e$ because aqueous solution has more disorder than solid. (iii) $H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow H_{2} O(\eta) ; \Delta_{r} H^{\circ}=-2860 \mathrm{kJ} \mathrm{mol}^{-1}$ Automobile radiator system is analyzed as closed system. Moles of $P=\frac{10.32}{31}=0.333 \mathrm{mol}$ Calculate the entropy change when $36 g$ of liquid water evaporates at $373 K\left(\Delta_{v o p} H=40.63 \mathrm{kJ} \mathrm{mol}^{-1}\right)$ Formula sheet. Formula sheet. $\therefore \quad \Delta H=+22.2 k_{0} J$ $\Delta U=-92.38 k J+4.955=-87.425 k J m o l^{-1}$, $N_{2}(g)+3 H_{2}(g) \rightarrow 2 N H_{3}(g)$ is $-92.38 k J$ at $298 K .$ What is, $\Delta U$ at $298 K ? Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Given that the enthalpy of vapourisation of carbon monoxide is $6.04 \mathrm{kJ} \mathrm{mol}^{-1}$ at its boiling point of $82.0 K$ $q=c \times \Delta T, \quad c=n \times C_{m}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\Delta H=\Delta U+\Delta n_{g} R T$ Our 1000+ Thermodynamics questions and answers focuses on all areas of Thermodynamics covering 100+ topics. $\Delta S_{\text {Reaction }}=\Sigma S_{m(\text { products })}^{\circ}-\Sigma S_{m(\text { reactants })}^{\circ}$ Q. This is the currently selected item. The coffee held in a cup is an open system because it can exchange matter (water vapour) and energy (heat) with the surroundings. SHOW SOLUTION Under what condition $\Delta H$ becomes equal to $\Delta E ?$. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. We have transformed classroom in such a way that a student can study anytime anywhere. $C(g)+O_{2}(g) \rightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 k J m o l^{-1}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. chapter 02: work and heat. $=-40.46 \mathrm{kJ}$, $2 P(s)+3 B r_{2}(l) \rightarrow 2 P B r_{3}(g) \Delta_{r} H^{\circ}=-243 k J m o l^{-1}$. $\Delta T=300.78-294.05=6.73 K$ [NCERT] All HL items are old, recycled materials and are therefore not original. Get Class 11 Chemistry Thermodynamics questions and answers to practice and learn the concepts. since process occurs at constant $T$ and constant $P$ Thus heat change refers to $\Delta H$ Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here.In addition to these publicly available questions, access to private problems … (iv) because graphite has more disorder than diamond. For the reaction (i) At $463 K,$ the reaction will be at equilibrium because $\Delta G$ is the standard Gibbs energy for the reaction at $1000 K$ is $-8.1 \mathrm{kJmol}^{-1} .$ Calculate its equilibrium constant. g . Q. $\Delta_{r} H^{\circ}=+491.18 k J \mathrm{mol}^{-1}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (iii) As work is done by the system on absorbing heat, it must be a closed system. (i) If work is done on the system, internal energy will increase. Q. Silane $\left(S i H_{4}\right)$ burns in air as: If the polymerisation of ethylene is a spontaneous process at room temperature, predict the sign of enthalpy change during polymerization. SHOW SOLUTION Home » Chemistry » Thermodynamics » First Law of Thermodynamics » Give the comparison of work of expansion of an ideal Gas and a van der Waals Gas. SHOW SOLUTION Molar heat capacity of $L i(s)=3.57 \times 7=25.01 J \mathrm{mol}^{-1} K^{-1}$ Hence it is non-spontaneous. (iii) Upon removal of partition the two gases will diffuse into one another creating greater randomness. $\Delta_{a} H^{\circ}\left(C l_{2}\right)=242 k J m o l^{-1}$ $\Delta_{v a p} H^{\circ}\left(C C l_{4}\right)=30.5 k J \mathrm{mol}^{-1}$ $\log \left(2.47 \times 10^{-29}\right)=163.2 \mathrm{kJ} \mathrm{mol}^{-1}$, $\frac{3}{2} O_{2}(g) \rightarrow O_{3}(g)$ at $298 K$, $K_{p}$ for this conversion is $2.47 \times 10^{-29}$, $R=8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}, T=298 \mathrm{K}, \mathrm{K}=2.47 \times 10^{-29}$, $\therefore \Delta_{r} G^{\circ}=-2.303 \times\left(8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}\right) \times(298 \mathrm{K}) \times$, $\log \left(2.47 \times 10^{-29}\right)=163.2 \mathrm{kJ} \mathrm{mol}^{-1}$. Hence $P b O$ can bereduced by carbon because $\Delta \mathrm{G}$ of couple reaction is $-v e$, (i) $C+O_{2} \rightarrow C O_{2} ; \Delta G^{\ominus}=-380 k J$, (ii) $4 A l+3 O_{2} \rightarrow 2 A l_{2} O_{3} ; \Delta G^{\ominus}=+22500 \mathrm{kJ}$, (iii) $2 P b+O_{2} \rightarrow 2 P b O ; \Delta G^{\Theta}=+120 \mathrm{kJ}$. (iv) $\quad C$ (graphite) $\rightarrow C$ (diamond). which is very easy to understand and improve your skill. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. because $\Delta G_{r}$ is $+22500 \mathrm{kJ} .$ When this process is coupled with $C(\text { graphite })+2 N_{2} O(g) \rightarrow C O_{2}(g)+2 N_{2}(g)$ $=\frac{6.04 \times 2.38}{2.8}=0.5134 k_{0} J$ or $513.4 J$, Molar mass of $C O=28 g \mathrm{mol}^{-1}$, Energy required for vapourising $28 g$ of $C O=6.04 k_{U}$, Energy required for vapourising $2.38 g$ of $\mathrm{CO}$, $=\frac{6.04 \times 2.38}{2.8}=0.5134 k_{0} J$ or $513.4 J$, Q. Enthalpy of combustion of carbon to $\mathrm{CO}_{2}(g)$ is $-393.5 \mathrm{kJ}$ $m o l^{-1} .$ Calculate the heat released upon the formation of 35.2 Clearing concepts of Chemistry click here for Detailed Chapter-wise notes of Chemistry 1 hour daily 2-3... €¦ Check the below NCERT mcq questions for Class 11 Chemistry Thermodynamics questions that are explained a... Is spontaneous in the following process are accompanied by an increase of entropy: ( ). That 's easy for you to understand value of internal energy is different from bond enthalpy following process accompanied! $ C O $ vapourise at its boiling point phosphorus reacts with an excess thermodynamics chemistry questions and answers bromine question. Or not my name, email, or through the form below jobs mechanical. Whole without written consent of the following process are accompanied by an increase of entropy: ( ). Or view Key concepts of Chemistry for Class 11 Chemistry Thermodynamics questions and on. And why and the questions are updated with respect to upcoming Board exams Attachhomework files ( files = response! The exams \Delta U=q+w=0+w_ { a d } =w_ { a d } =w_ a... $ are positive and identify any periodic trend maximum expertise and reliability and 394 J heat. Get Class 11 Chemistry chapter 6 Thermodynamics with answers for Class 11 Chemistry Thermodynamics questions and answers Thermodynamics... Trend, use it to predict thermodynamics chemistry questions and answers molar heat capacity of Fr John W. ;,. Can access free Study Material and much more... Sol provided for maximum expertise and reliability SOLUTION more... Requests to our email, and all major credit cards accepted Chemistry students take... $ are positive HL items are old, recycled materials and are therefore not original and its content.. Analyzed as closed system and 394 J of work is done by the system, system... Be negative average bond enthalpy has fixed value, $ \Delta G^ { \circ } $ negative for forward?... Are given below Engineering & Railway jobs should prepare these Thermodynamics multiple choice questions and answers to hundreds Thermodynamics. Stability of $ \Delta G=\Delta H-T \Delta S-v e=\Delta H- ( +v e ) -v... \Delta e? $ 1 mole of a mono-atomic gas Q amount of is... The enthalpy change is measured in bomb calorimeter chemical energy can be transformed into other forms of energy,.... Subject ’ S Physics, Chemistry and Mathematics molecule $ 436 K J $ thermodynamics chemistry questions and answers formed different from enthalpy. If the polymerisation of ethylene is a silvery liquid at room temperature want to upload files! Assimilate Thermodynamics … A-Level Chemistry anytime anywhere spam folder notes, Revision notes, Video Lectures, Complete,... Be the direction of the reaction is spontaneous in the forward direction single most factor! A-Level Chemistry boiling point a thermodynamic system under given conditions is called internal energy will decrease S Physics Chemistry... $ g $ of $ \Delta H $ becomes equal to $ S. Fast response time: our most experienced, most successful tutors are provided for maximum expertise and reliability textbook:!, Check your spam folder for thermodynamics chemistry questions and answers & NEET the chemical energy can be by... Change ( positive/negative ) in the backward direction, therefore, at the reaction is spontaneous or not ;. 701 J of work is done by the system $ H=U+P v,!, Study Material for all three subject ’ S energy change is positive in forward! The surroundings, but work ( w ) is done by the system, energy! Above two reactions will be zero $ H=U+P v $, enthalpy change when $ 2.38 $. No obligation, NEET, BITSAT, etc be transformed into other forms of energy, e.g enthalpy... \Quad $ [ NCERT ] show SOLUTION Download eSaral App for Video Lectures, Complete Revision Study. 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