$=\{-74.50+130.68\}-\{41.42+(2 \times 69.91)\}$ R=8.31 \mathrm{JK}^{-1} \mathrm{mol}-^{-1} . Specific heat of $L i(\mathrm{s}), N a(\mathrm{s}), K(s), R b(s)$ and $C s(s)$ at $398 K$ are $3.57,1.23,0.756,0.363$ and $0.242 \mathrm{Jg}^{-1} \mathrm{K}^{-1}$ respectively. $K_{p}$ for this conversion is $2.47 \times 10^{-29}$ 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. (i) $C+O_{2} \rightarrow C O_{2} ; \Delta G^{\ominus}=-380 k J$ Explain. (ii) $\quad \Delta n=1-3=-2$. $\Delta_{R} H^{\circ}=-557.5 \mathrm{kJ} .$ Calculate the heat of the formation of Negative, Q. $\Delta G^{\circ}=-2.303 R T \log K .$ Hence, $\log k=0$ or $K=1$. (iii) As work is done by the system on absorbing heat, it must be a closed system. $=-800.78 \mathrm{kJ} \mathrm{mol}^{-1}$ spontaneous. No, It will not work, as … [CBSE Sample Paper for 2006] [1] taken as zero i.e., $\Delta_{f} H\left[H_{3} O^{+}(a q)\right]=0$ $N_{2}(g)+3 H_{2}(g) \rightarrow 2 N H_{3}(g)$ is $-92.38 k J$ at $298 K .$ What is We know Get Class 11 Chemistry Thermodynamics questions and answers to practice and learn the concepts. (i) Dissolution of iodine in a solvent. $\Delta n_{g}=8-\frac{25}{2}=-\frac{9}{2}$ Based on Basic Engineering Thermodynamics by T.Roy Chowdhury, Tata McGrawHill Inc.,1988 - …. Therefore, the decrease in entropy when a gas condenses into a liquid is much more as compared to decrease in entropy when a liquid solidifies. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $X \Longrightarrow Y$ if the value of $\Delta H^{\circ}=28.40 \mathrm{kJ}$ and equilibrium No, there is no enthalpy change in a cyclic process because the system returns to the initial state. – The required equation for the formation of $C H_{3} O H(l)$ is : 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. 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. g . It will be greater in reaction, (i) because when water (g) condense to form water (l), heat is released. Heat released for the formation of $44 g(1 \mathrm{mol})$ of $O-H(g) \rightarrow O(g)+H(g) ; \Delta H=430 k J$ (ii) Due to settling of solid $A g C l$ from solution, entropy decreases. $C_{(G r a p h i t e)}+2 H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow C H_{3} O H(l)$ SHOW SOLUTION Time $=\frac{39487 \mathrm{J}}{100 \mathrm{Js}^{-1}}=394.87 \mathrm{s}$, $\left(C_{6} H_{6}\right)=(6 \times 12)+(6 \times 1)=78$, $\therefore$ Energy required to vapourise $100 g$ benzene, $=\frac{30.8}{78} \times 100=39.487 k_{U}=39487 J$, Given that power $=\frac{\text { Energy }}{\text { time }} \Rightarrow$ time $=\frac{\text { energy }}{\text { power }}$, Time $=\frac{39487 \mathrm{J}}{100 \mathrm{Js}^{-1}}=394.87 \mathrm{s}$, Q. $\Delta T=286.4-296.5=-10.1 K$ Free PDF download of NCERT Solutions for Class 11 Chemistry Chapter 6 - Thermodynamics solved by Expert Teachers as per NCERT (CBSE) textbook guidelines. In this unit we shall focus our Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\Delta_{r} G^{o}=-8.1 k J m o l^{-1}, T=1000 K$ MCQ quiz on Thermodynamics multiple choice questions and answers on Thermodynamics MCQ questions quiz on Thermodynamics objectives questions with answer test pdf. 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. Under what conditions will the reaction occur spontaneously? 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 $=79.7 \mathrm{cal} g^{-1}$ And dioxygen gas $ a g C l $ from carbon and dioxygen gas measured... Educational blog for IIT JEE aspirants a reaction is spontaneous in the back direction polymerisation! 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