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ChemMedAdv 2021 • Paper 1

Standard Enthalpy Change for Solid Decomposition Reaction

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}.

Determine the value of standard enthalpy, ΔH⊖\Delta H^{\ominus} (expressed in kJ mol−1\text{kJ mol}^{-1}) for the given reaction.

Question Diagram 1
Official Numerical Answer166.28

Topics & Concepts

EquilibriumChemical EquilibriumEnthalpy

Step-by-Step Solution

To determine the standard enthalpy change, ΔH⊖\Delta H^{\ominus}, for the reaction:

X(s)⇌Y(s)+Z(g)\mathbf{X}(s) \rightleftharpoons \mathbf{Y}(s) + \mathbf{Z}(g)

we utilize the van 't Hoff equation given in the question:

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 the equilibrium constant K=pZp⊖K = \frac{p_Z}{p^{\ominus}}.

From the graph of ln⁡(pZp⊖)\ln\left(\frac{p_Z}{p^{\ominus}}\right) versus 104T\frac{10^4}{T}, we identify two points on the line:

  1. (104T∣1,ln⁡K1)=(10,−3)\left(\left.\frac{10^4}{T}\right|_1, \ln K_1\right) = (10, -3)
  2. (104T∣2,ln⁡K2)=(12,−7)\left(\left.\frac{10^4}{T}\right|_2, \ln K_2\right) = (12, -7)

The slope of the line with respect to 104T\frac{10^4}{T} is given by:

Slope=d(ln⁡K)d(104T)=−7−(−3)12−10=−42=−2\text{Slope} = \frac{\mathrm{d}(\ln K)}{\mathrm{d}\left(\frac{10^4}{T}\right)} = \frac{-7 - (-3)}{12 - 10} = \frac{-4}{2} = -2

To express this in terms of 1T\frac{1}{T}, we multiply the slope by 10410^4:

d(ln⁡K)d(1T)=104×d(ln⁡K)d(104T)=104×(−2)=−2×104 K\frac{\mathrm{d}(\ln K)}{\mathrm{d}\left(\frac{1}{T}\right)} = 10^4 \times \frac{\mathrm{d}(\ln K)}{\mathrm{d}\left(\frac{10^4}{T}\right)} = 10^4 \times (-2) = -2 \times 10^4 \text{ K}

Now, substituting this value into the van 't Hoff equation:

−ΔH⊖R=−2×104 K-\frac{\Delta H^{\ominus}}{R} = -2 \times 10^4 \text{ K}

ΔH⊖=2×104 K×R\Delta H^{\ominus} = 2 \times 10^4 \text{ K} \times R

Given R=8.314 J K−1 mol−1R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}:

ΔH⊖=2×104×8.314 J mol−1=166280 J mol−1=166.28 kJ mol−1\Delta H^{\ominus} = 2 \times 10^4 \times 8.314 \text{ J mol}^{-1} = 166280 \text{ J mol}^{-1} = 166.28 \text{ kJ mol}^{-1}

Final Answer: The value of standard enthalpy, ΔH⊖\Delta H^{\ominus}, for the given reaction is 166.28166.28 (or 166.28 kJ mol−1166.28\text{ kJ mol}^{-1}).

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