Calculate $\Delta S$ for the conversion of: (ii) Vapours to liquid at $35^{\circ} \mathrm{C}$, $\Delta S_{v a p . $\therefore$ The enthalpy of polymerization $\left(\Delta H_{\text {poly }}\right)$ must be negative. (ii) Due to settling of solid $A g C l$ from solution, entropy decreases. $C(\text { graphite })+2 N_{2} O(g) \rightarrow C O_{2}(g)+2 N_{2}(g)$, $\Delta_{R} H^{\circ}=-557.5 \mathrm{kJ} .$ Calculate the heat of the formation of, Q. $\Delta H=8 B_{C-H}+2 B_{C-C}+5 B_{O=O}-6 B_{C=O}-8 B_{O-H}$, $-2050 k J=8 \times 414+2 \times 347+5 B_{O=O}-6 \times 741-8 \times 464$, $-2050 k J=3312 k_{0} J+694 k J+5 B_{O=0}-4446 k J-3712 k_{\circlearrowright} J$, $-2050 k J=4006 k_{0} J+5 B_{O=0}-8158 k_{0} J$, $-2050 \mathrm{kJ}=-4152 \mathrm{kJ}+5 \mathrm{B}_{\mathrm{O}=0}$, $B_{O=O}=\frac{2102}{5}=420.4 \mathrm{kJ} \mathrm{mol}^{-1}$. The carefully crafted questions and answers provide students with a comprehensive understanding of the chapters involved. $\Delta n_{g}=2-(1+3)=-2 m o l, T=298 K$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Q. $0=\Delta H-T \Delta S$ or $\Delta H=T \Delta S$ or $T=\frac{\Delta H}{\Delta S}$ Internal energy change is measure at constant volume. The temperature falls from $296.5 \mathrm{Kto} 286.4 \mathrm{K.Find}$ out the value of $q$ for calorimeter and its contents. Give suitable examples. $C+O_{2} \longrightarrow C O_{2} ; \Delta G=-380 k J$ $H C l(g)+H_{2} O \rightarrow H_{3} O^{+}(a q)+C l^{-}(a q)$ The enthalpy change for the reaction: Also calculate the enthalpy of combustion of octane. 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 $H C l s-95 k J m o l^{-1}$ SHOW SOLUTION All the commercial liquid fuels are derived from natural petroleum (or crude oil). We know Also calculate enthalpy of solution of ammonium nitrate. $A+B \rightarrow C+D$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. You will get here all the important questions with answers for class 11 Chemistry Therodynamics and  chapters. Welcome to 5.1 THERMODYNAMICS. $\Delta_{r} G^{\circ}=\Delta_{r} H^{o}-T \Delta_{r} S^{\circ}$, $\Delta_{r} H^{\circ}=+491.18 k J \mathrm{mol}^{-1}$, $\Delta_{r} S^{o}=197.67 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}, T=298 \mathrm{K}$, $\Delta_{r} G^{\circ}=491.18 k J \mathrm{mol}^{-1}-298 \mathrm{Kx}$, $\left(197.67 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}\right)$, $=491.18 \mathrm{kJ} \mathrm{mol}^{-1}-58.9 \mathrm{kJ} \mathrm{mol}^{-1}=432.28 \mathrm{kJ} \mathrm{mol}^{-1}$. (iii) 1 mol of a liquid X. Now $C_{v}=\frac{\Delta U}{\Delta T}$ and $\Delta T=1^{\circ} \mathrm{C}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $20.0 \mathrm{g}$ of ammonium nitrate $\left(\mathrm{NH}_{4} \mathrm{NO}_{3}\right)$ is dissolved in $125 \mathrm{g}$ of water in a coffee cup calorimeter, the temperature falls from $296.5 \mathrm{Kto} 286.4 \mathrm{K} .$ Find the value of $q$ for the calorimeter. (iii) $\quad \Delta \mathrm{S}=-\mathrm{ve}$, (i) $\quad \mathrm{O}_{2}(g)+2 S O_{2}(g) \rightarrow 2 S O_{3}(g)$, (ii) $\quad \mathrm{CaC}_{2} \mathrm{O}_{4}(\mathrm{s}) \rightarrow \mathrm{CaCO}_{3}(\mathrm{s})+\mathrm{CO}(\mathrm{g})$, (iii) $2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$, (iii) $\quad \Delta \mathrm{S}=-\mathrm{ve}$, Q. $T=\frac{\Delta H}{\Delta S}=\frac{-10000 J m o l^{-1}}{-33.3 J K^{-1} m o l^{-1}}=300.3 \mathrm{K}$ Now, $\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$ SHOW SOLUTION the condensation of diethyl ether is the reverse process, therefore, $\Delta S_{\text {condensation }}=-84.4 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$, Q. $C O_{2}=-393.5 k_{U}$ Therefore, $C_{v}=3 / 2 R$, At $(T+1) K,$ the kinetic energy per mole $\left(E_{k}\right)=3 / 2 R(T+1)$ Therefore, increase in the average kinetic energy of the gas for $1^{\circ} \mathrm{C}(\text { or } 1 \mathrm{K})$ rise in temperature $\Delta E_{k}=3 / 2 R(T+1)-3 / 2 R T=3 / 2 R$, 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., $\Delta U=\Delta E_{K}$, Now $C_{v}=\frac{\Delta U}{\Delta T}$ and $\Delta T=1^{\circ} \mathrm{C}$, Q. $\Delta G=\Delta H-T \Delta S$ (ii) $4 A l+3 O_{2} \rightarrow 2 A l_{2} O_{3} ; \Delta G^{\ominus}=+22500 \mathrm{kJ}$ From 2021 Material and much more... 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