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# chemistry thermodynamics questions and answers pdf

The given equations are: $\quad-\left[\Delta G_{f}^{\circ} \mathrm{CaCO}_{3}(\mathrm{s})+2 \Delta G_{f}^{\circ} \mathrm{H}^{+}(a q)\right]$ Q. SHOW SOLUTION 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. $=5492273.082 \mathrm{Jmol}^{-1}=5492.27 \mathrm{kJ} \mathrm{mol}^{-1}$ April 15th, 2018 - Thermodynamics Multiple Choice Questions Has 100 MCQs Thermodynamics Quiz Questions And Answers Pdf MCQs On Applied Thermodynamics First Law Of Thermodynamics MCQs With Answers Second Law Of Thermodynamics Reversible And Irreversible Processes And Working Fluid MCQs And Quizzes To Practice Exam Prep Tests' Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\therefore$ The enthalpy of polymerization $\left(\Delta H_{\text {poly }}\right)$ must be negative. (i) At what temperature the reaction will occur spontaneously from left to right? R=8.31 \mathrm{JK}^{-1} \mathrm{mol}-^{-1} . The temperature falls from $296.5 \mathrm{Kto} 286.4 \mathrm{K.Find}$ out the value of $q$ for calorimeter and its contents. $\mathrm{CCl}_{4}(g) \rightarrow \mathrm{C}(g)+4 \mathrm{Cl}(g)$ (ii) Calculate the value of $\Delta n$ in the following reaction: The carefully crafted questions and answers provide students with a comprehensive understanding of the chapters involved. $-\left[\Delta G_{f}^{\circ} \mathrm{CaCO}_{3}(s)+2 \Delta G_{f}^{\circ} H^{+}(a q)\right]$ Macroscopic properties like pressure and temperature do not change with time for a system in equilibrium state. A piston exerting a pressure of 1 atm rests on the surface of water at $100^{\circ} \mathrm{C} .$ The pressure is reduced to smaller extent when $10 g$ of water evaporates and $22.2 \mathrm{kJ}$ of heat is absorbed. $\Delta_{R} H^{\circ}=-557.5 \mathrm{kJ} .$ Calculate the heat of the formation of According to Gibbs Helmholtz equation, Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Based on Basic Engineering Thermodynamics by T.Roy Chowdhury, Tata McGrawHill Inc.,1988 - вЂ¦ Class 11 Important Questions for Chemistry вЂ“ Thermodynamics NCERT Exemplar Class 11 Chemistry is very important resource for students preparing for XI Board Examination. $\Delta_{d i s s} H^{\circ}=-75.2 k J m o l^{-1}$ Calculate the bond enthalpy of $H C l .$ Given that the bond enthalpies of $H_{2}$ and $C l_{2}$ are $430 \mathrm{kJ} \mathrm{mol}^{-1}$ and $242 \mathrm{kJ} \mathrm{mol}^{-1}$ respectively and $\Delta \mathrm{H}_{f}^{\circ}$ for $H C l s-95 k J m o l^{-1}$, Calculate $\Delta S$ when 1 mole of steam at $100^{\circ} \mathrm{C}$ is converted into ice at $0^{\circ} \mathrm{C}$. (iv) $\quad C$ (graphite) $\rightarrow C$ (diamond). Thus $A l_{2} O_{3}$ cannot be reduced by $C$ You will get here all the important questions with answers for class 11 Chemistry Therodynamics and  chapters. SHOW SOLUTION $C(s)+H_{2} O(g) \rightleftarrows C O(g)+H_{2}(g)$ SHOW SOLUTION What will be sign of for backward reaction? SHOW SOLUTION Compare it with entropy decrease when a liquid sample is converted into a solid. $\left(\text { or } 2211.8^{\circ} \mathrm{C}\right)$, $M g O(s)+C(s) \rightarrow M g(s)+C O(g)$, $\Delta_{r} H^{\circ}=+491.18 k J \mathrm{mol}^{-1}$ and $\Delta_{r} S^{\circ}$, $=197.67 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$. SHOW SOLUTION This will be so if, Q. Molar heat capacity of $K(s)=0.756 \times 39=29.5 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$ (i) $\left.\quad \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\right]$ $-2050 k J=8 \times 414+2 \times 347+5 B_{O=O}-6 \times 741-8 \times 464$ SHOW SOLUTION $=\left(2 S_{F e_{2} O_{3}}^{\circ}\right)-\left(4 S_{F e(s)}^{\circ}+3 S_{O_{2}}^{\circ}(g)\right)$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Formula sheet. For an ideal gas, from kinetic theory of gases, the average kinetic energy per mole $\left(E_{k}\right)$ of the gas at any temperature $T K,$ is given by $E_{k}=3 / 2 R T$ (i) $\quad 4 F e(s)+3 O_{2}(g) \rightarrow 2 F e_{2} O_{3}(s)$ SHOW SOLUTION This will be so if, $\Delta H=-10,000 J \mathrm{mol}^{-1}, \Delta S=-33.3 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$. (i) \$\quad \Delta S