For the reaction, X(s) leftharpoons Y(s) + Z(g), the plot of \ln pz \over p^\theta versus 10^4 \over T is given below (in solid line), where pz is the pressure (in bar) of the gas Z at temperature T and p^\theta = 1 bar. (Given, d(\ln K) \over d( 1 \over T ) = - \Delta H^\theta \over R , where the equilibrium constant, K = pz \over p^\theta and the gas constant, R = 8.314 J K - 1 mol - 1) The value of \Delta S \theta (in J K - 1 mol - 1) for the given reaction, at 1000 K is \underline\hspace2cm .

Thermodynamics PYQ — JEE Advanced 2021 · Shift 1 · Q8 with Solution

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Correct answer: 141.34

Equilibrium constant for the reaction

For $X(s) \rightleftharpoons Y(s) + Z(g)$ only the गैस contributes to the equilibrium constant, so $K = \frac{p_Z}{p^\theta}$ Hence the graph of $\ln\left(\frac{p_Z}{p^\theta}\right)$ vs $\frac{10^4}{T}$ is effectively the graph of $\ln K$ vs $\frac{10^4}{T}$ .

Use van’t Hoff equation

Given, $\frac{d(\ln K)}{d(1/T)} = -\frac{\Delta H^\theta}{R}$

Let $x = \frac{10^4}{T}$ Then, $\frac{1}{T} = \frac{x}{10^4}$ So,

= \left(-\frac{\Delta H^\theta}{R}\right)\left(\frac{1}{10^4}\right)$$ Thus slope $m$ of the graph is $$m = -\frac{\Delta H^\theta}{R\,10^4}$$ Therefore, $$\Delta H^\theta = -mR\times 10^4$$ --- 3. **Read two points from the straight line** From the graph (solid line), the line passes through points approximately: - at $\frac{10^4}{T}=10$, $\ln K \approx -3$ - at $\frac{10^4}{T}=14$, $\ln K \approx -11$ So slope is $$m = \frac{-11-(-3)}{14-10} = \frac{-8}{4} = -2$$ Hence, $$\Delta H^\theta = -(-2)(8.314)(10^4) = 2\times 8.314\times 10^4 = 1.6628\times 10^5\ \text{J mol}^{-1}$$ So, $$\Delta H^\theta = 166.28\ \text{kJ mol}^{-1}$$ --- 4. **Find $\Delta G^\theta$ at $1000\,\text{K}$** At $T=1000\,\text{K}$, $$\frac{10^4}{T} = \frac{10^4}{1000}=10$$ From graph, at this point $$\ln K = -3$$ Now, $$\Delta G^\theta = -RT\ln K$$ So, $$\Delta G^\theta = -(8.314)(1000)(-3) = 24942\ \text{J mol}^{-1}$$ --- 5. **Use relation $\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta$** Rearranging, $$\Delta S^\theta = \frac{\Delta H^\theta - \Delta G^\theta}{T}$$ Substitute values: $$\Delta S^\theta = \frac{166280 - 24942}{1000}$$ $$\Delta S^\theta = \frac{141338}{1000} = 141.338\ \text{J K}^{-1}\text{mol}^{-1}$$ Therefore, $$\boxed{\Delta S^\theta \approx 141.34\ \text{J K}^{-1}\text{mol}^{-1}}$$ --- 6. **Comparison with stored answer** Stored correct answer = $141.34$ Our derived answer matches exactly.

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View written solutionFree

Correct answer: 141.34

Equilibrium constant for the reaction

Use van’t Hoff equation

Given, $\frac{d(\ln K)}{d(1/T)} = -\frac{\Delta H^\theta}{R}$

Let $x = \frac{10^4}{T}$ Then, $\frac{1}{T} = \frac{x}{10^4}$ So,

= \left(-\frac{\Delta H^\theta}{R}\right)\left(\frac{1}{10^4}\right)$$ Thus slope $m$ of the graph is $$m = -\frac{\Delta H^\theta}{R\,10^4}$$ Therefore, $$\Delta H^\theta = -mR\times 10^4$$ --- 3. **Read two points from the straight line** From the graph (solid line), the line passes through points approximately: - at $\frac{10^4}{T}=10$, $\ln K \approx -3$ - at $\frac{10^4}{T}=14$, $\ln K \approx -11$ So slope is $$m = \frac{-11-(-3)}{14-10} = \frac{-8}{4} = -2$$ Hence, $$\Delta H^\theta = -(-2)(8.314)(10^4) = 2\times 8.314\times 10^4 = 1.6628\times 10^5\ \text{J mol}^{-1}$$ So, $$\Delta H^\theta = 166.28\ \text{kJ mol}^{-1}$$ --- 4. **Find $\Delta G^\theta$ at $1000\,\text{K}$** At $T=1000\,\text{K}$, $$\frac{10^4}{T} = \frac{10^4}{1000}=10$$ From graph, at this point $$\ln K = -3$$ Now, $$\Delta G^\theta = -RT\ln K$$ So, $$\Delta G^\theta = -(8.314)(1000)(-3) = 24942\ \text{J mol}^{-1}$$ --- 5. **Use relation $\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta$** Rearranging, $$\Delta S^\theta = \frac{\Delta H^\theta - \Delta G^\theta}{T}$$ Substitute values: $$\Delta S^\theta = \frac{166280 - 24942}{1000}$$ $$\Delta S^\theta = \frac{141338}{1000} = 141.338\ \text{J K}^{-1}\text{mol}^{-1}$$ Therefore, $$\boxed{\Delta S^\theta \approx 141.34\ \text{J K}^{-1}\text{mol}^{-1}}$$ --- 6. **Comparison with stored answer** Stored correct answer = $141.34$ Our derived answer matches exactly.

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