(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. (i) If work is done on the system, internal energy will increase. SHOW SOLUTION $\Delta_{v a p} H=40.63 \mathrm{kJ} \mathrm{mol}^{-1}, T_{b}=373 \mathrm{K}$ $I_{2}$ molecules upon dissolution. (ii) 1 mol of a solid X $2 P(s)+3 B r_{2}(l) \rightarrow 2 P B r_{3}(g) \Delta_{r} H^{o}=-243 k J m o l^{-1}$ Calculate the enthalpy change when $2.38 g$ of $C O$ vapourise at its boiling point. (iii) $2 P b+O_{2} \rightarrow 2 P b O ; \Delta G^{\Theta}=+120 \mathrm{kJ}$ $\Delta H_{f}=\left(79.7 \mathrm{cal} g^{-1}\right)\left(18.0 \mathrm{g} \mathrm{mol}^{-1}\right)=1435 \mathrm{cal} \mathrm{mol}^{-1}$ [NCERT] $\mathrm{SiH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{SiO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$ $\Delta G_{f}^{\circ} H_{2} O(l)=-237.13 k J m o l^{-1}$ $\Delta_{v a p} H^{\circ}\left(C C l_{4}\right)=30.5 k J \mathrm{mol}^{-1}$ $0=\Delta H-T \Delta S$ or $\Delta H=T \Delta S$ or $T=\frac{\Delta H}{\Delta S}$ Q. Click Here for Detailed Notes of any chapter. Closed system : (iii) Cane of tomato soup, (iv) Ice cube tray filled with water, (vii) Helium filled balloon. The standard Gibbs energy of reaction (at $1000 K)$ is $-8.1$ $k J m o l^{-1} .$ Calculate its equilibrium constant. This is the currently selected item. (ii) $\quad \Delta S_{\text {reaction}}^{\circ}=\Sigma S^{\circ}(\text { products })-\Sigma S^{\circ}(\text { reactans })$ $=\left(5481.02 \times 10^{3}\right)-11253.08 \mathrm{J} \mathrm{mol}^{-1}$ $\Delta_{r} G^{\circ}=-2.303 R T \log K \quad$ or $\log K=\frac{-\Delta_{r} G}{2.303 R T}$, $\Delta_{r} G^{o}=-8.1 k J m o l^{-1}, T=1000 K$, $R=8.314 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}$, $\log K=-\frac{-8.1}{2.303 \times 8.314 \times 10^{-3} \times 1000}$ or $K=2.64$, Q. $\Delta_{r} G^{\circ}=\Delta_{f} G^{\circ}\left(S i O_{2}\right)+2 \Delta_{f} G^{\circ}\left(H_{2} O\right)-\left[\Delta_{f} G^{\circ}\left(S i H_{4}\right)\right]+$ SHOW SOLUTION What will be the direction of the reaction at this temperature and below this temperature and why? [NCERT] Thermodynamics Multiple choice Questions and answers pdf,mcqs,objective type questions,lab viva manual technical basic interview questions download Which of the following process are accompanied by an increase of entropy: Under what conditions will the reaction occur spontaneously? Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. SHOW SOLUTION So, molar heat capacity of these elements can be obtained by multiplying specific heat capacity by atomic mass. $q=125 g \times 4.18 J / g \times(286.4-296.5)$ [NCERT] Silane $\left(S i H_{4}\right)$ burns in air as: Calculate $\Delta_{r} G^{\circ}$ for conversion of oxygen to ozone: Calculate the standard entropy change for the reaction, Calculate standard molar entropy change of the formation of, The standard Gibb’s energy of reactions at $1773 \mathrm{K}$ are given $\mathbf{a} \mathbf{S}$. $=-800.78 \mathrm{kJ} \mathrm{mol}^{-1}$ g . Give suitable examples. SHOW SOLUTION (iii) 1 mol of a liquid X. Given that power $=\frac{\text { Energy }}{\text { time }} \Rightarrow$ time $=\frac{\text { energy }}{\text { power }}$ $\therefore \Delta U=q+w=0+w_{a d}=w_{a d}$. Jump to Page . (i) Work is done on the system, $=-33.84 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$, $X \Longrightarrow Y$ if the value of $\Delta H^{\circ}=28.40 \mathrm{kJ}$ and equilibrium, constant is $1.8 \times 10^{-7}$ at $298 \mathrm{K} ?$, $=-2.303 \times 8.314 \times 298 \times \log \left(1.8 \times 10^{-7}\right)=38484.4$, Now, $\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$, $\therefore \Delta S^{\circ}=\frac{\Delta H^{\circ}-\Delta G^{\circ}}{T}=\frac{28400-38484.4}{298}$, $=-33.84 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$, Q. (i) $\left.\quad \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\right]$ Molar mass of benzene $\therefore \quad \Delta S_{\text {total}}=-26.0445-5.62-5.26$, $\Rightarrow \quad-36.9245 \mathrm{cal} K^{-1} \mathrm{mol}^{-1}$, Download India's Leading JEE | NEET | Class 9,10 Exam preparation app. $\Delta_{v a p} S=\frac{\Delta_{v a p} H}{T_{b}}=\frac{40.63 \times 1000 \mathrm{J} \mathrm{mol}^{-1}}{373 \mathrm{K}}=109 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ So, molar heat capacity of these elements can be obtained by multiplying specific heat capacity by atomic mass. 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 ? 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