JEE Challenger
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ChemHardAdv 2021 • Paper 1

Standard Entropy Change at Specified Temperature

Comprehension Passage

For the chemical reaction X(s)⇌Y(s)+Z(g)\mathbf{X}(s) \rightleftharpoons \mathbf{Y}(s) + \mathbf{Z}(g), the plot representing ln⁡(pZp⊖)\ln \left(\frac{p_Z}{p^{\ominus}}\right) against 104T\frac{10^4}{T} is provided in the graph (solid line), in which pZp_Z denotes the pressure (in bar) of gas Z\mathbf{Z} at temperature TT, and p⊖=1 barp^{\ominus} = 1\text{ bar}.

Given: d(ln⁡K)d(1T)=−ΔH⊖R\frac{\mathrm{d}(\ln K)}{\mathrm{d}\left(\frac{1}{T}\right)} = -\frac{\Delta H^{\ominus}}{R}, where equilibrium constant K=pZp⊖K = \frac{p_Z}{p^{\ominus}} and gas constant R=8.314 J K−1 mol−1R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}.

Calculate the value of standard entropy, ΔS⊖\Delta S^{\ominus} (in J K−1 mol−1\text{J K}^{-1}\text{ mol}^{-1}) for the given reaction at a temperature of 1000 K1000\text{ K}.

Question Diagram 1
Official Numerical Answer141.34

Topics & Concepts

EquilibriumChemical EquilibriumEntropy and Second Law

Step-by-Step Solution

To find the standard entropy change, ΔS⊖\Delta S^{\ominus}, for the given reaction at T=1000 KT = 1000\text{ K}, we utilize the thermodynamic relationships connecting the equilibrium constant (KK), standard enthalpy change (ΔH⊖\Delta H^{\ominus}), and standard entropy change (ΔS⊖\Delta S^{\ominus}).

Step 1: Express ln⁡K\ln K in terms of temperature TT

The standard Gibbs free energy change is given by: ΔG⊖=ΔH⊖−TΔS⊖\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}

Also, the relationship between ΔG⊖\Delta G^{\ominus} and the equilibrium constant KK is: ΔG⊖=−RTln⁡K\Delta G^{\ominus} = -RT \ln K

Equating the two expressions for ΔG⊖\Delta G^{\ominus}: −RTln⁡K=ΔH⊖−TΔS⊖-RT \ln K = \Delta H^{\ominus} - T\Delta S^{\ominus}

Dividing both sides by −RT-RT: ln⁡K=−ΔH⊖R(1T)+ΔS⊖R\ln K = -\frac{\Delta H^{\ominus}}{R} \left(\frac{1}{T}\right) + \frac{\Delta S^{\ominus}}{R}

Let x=104Tx = \frac{10^4}{T}, so 1T=10−4x\frac{1}{T} = 10^{-4}x. Substituting this gives: ln⁡K=(−ΔH⊖104R)x+ΔS⊖R\ln K = \left(-\frac{\Delta H^{\ominus}}{10^4 R}\right) x + \frac{\Delta S^{\ominus}}{R}

This represents a linear equation of the form ln⁡K=mx+c\ln K = m x + c, where:

  • Slope, m=−ΔH⊖104Rm = -\frac{\Delta H^{\ominus}}{10^4 R}
  • Intercept, c=ΔS⊖Rc = \frac{\Delta S^{\ominus}}{R}

Step 2: Determine the line parameters from the graph

From the given plot of ln⁡K\ln K against x=104Tx = \frac{10^4}{T}:

  • At x1=10x_1 = 10, ln⁡K1=−3\ln K_1 = -3
  • At x2=12x_2 = 12, ln⁡K2=−7\ln K_2 = -7

The slope mm of the line is calculated as: m=ln⁡K2−ln⁡K1x2−x1=−7−(−3)12−10=−42=−2m = \frac{\ln K_2 - \ln K_1}{x_2 - x_1} = \frac{-7 - (-3)}{12 - 10} = \frac{-4}{2} = -2

Using the point-slope form with (x1,ln⁡K1)=(10,−3)(x_1, \ln K_1) = (10, -3): ln⁡K−(−3)=−2(x−10)\ln K - (-3) = -2 (x - 10) ln⁡K+3=−2x+20\ln K + 3 = -2x + 20 ln⁡K=−2x+17\ln K = -2x + 17

Step 3: Calculate the Standard Entropy Change ΔS⊖\Delta S^{\ominus}

Comparing the equation ln⁡K=−2x+17\ln K = -2x + 17 with ln⁡K=(−ΔH⊖104R)x+ΔS⊖R\ln K = \left(-\frac{\Delta H^{\ominus}}{10^4 R}\right) x + \frac{\Delta S^{\ominus}}{R}, we find the intercept cc: ΔS⊖R=17\frac{\Delta S^{\ominus}}{R} = 17

Therefore: ΔS⊖=17×R\Delta S^{\ominus} = 17 \times R

Given R=8.314 J K−1 mol−1R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}: ΔS⊖=17×8.314=141.338 J K−1 mol−1\Delta S^{\ominus} = 17 \times 8.314 = 141.338\text{ J K}^{-1}\text{ mol}^{-1}

Rounding to two decimal places: ΔS⊖≈141.34 J K−1 mol−1\Delta S^{\ominus} \approx 141.34\text{ J K}^{-1}\text{ mol}^{-1}

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