JEE Main 2018 — Chemistry Physical Chemistry
ΔfG∘\Delta_{\mathrm{f}} \mathrm{G}^{\circ}ΔfG∘ at 500 K500 \mathrm{~K}500 K for substance ' S\mathrm{S}S ' in liquid state and gaseous state are +100.7kcalmol−1+100.7 \mathrm{kcal} \mathrm{mol}^{-1}+100.7kcalmol−1 and +103+103+103 kcalmol−1\mathrm{kcal} \mathrm{mol}^{-1}kcalmol−1, respectively. Vapour pressure of liquid ' S\mathrm{S}S ' at 500 K500 \mathrm{~K}500 K is approximately equal to: (R=2calK−1 mol−1). \left(\mathrm{R}=2 \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right) \text {. }(R=2calK−1 mol−1).
Held on 15 Apr 2018 · Verified 6 Jul 2026.
100 atm100 \mathrm{~atm}100 atm
1 atm1 \mathrm{~atm}1 atm
10 atm10 \mathrm{~atm}10 atm
0.1 atm0.1 \mathrm{~atm}0.1 atm
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ΔGrxno=ΔfG∘ (vapour) −ΔfG∘ (liquid) ΔGrxno=103−100.7=2.3kcal/mol=2300cal/molΔGrxno=−RTlnK2300cal/mol=−2cal/molK×500 K×lnKlnK=2.3 K=10 atm= Vapour pressure of liquid ’ S ’ \begin{aligned} &\Delta \mathrm{G}_{\mathrm{rxn}}^{\mathrm{o}}=\Delta_{\mathrm{f}} \mathrm{G}^{\circ} \text { (vapour) }-\Delta_{\mathrm{f}} \mathrm{G}^{\circ} \text { (liquid) } \\ &\Delta \mathrm{G}_{\mathrm{rxn}}^{\mathrm{o}}=103-100.7=2.3 \mathrm{kcal} / \mathrm{mol} \\ &=2300 \mathrm{cal} / \mathrm{mol} \\ &\Delta \mathrm{G}_{\mathrm{rxn}}^{\mathrm{o}}=-\mathrm{RT} \ln \mathrm{K} \\ &2300 \mathrm{cal} / \mathrm{mol}=-2 \mathrm{cal} / \mathrm{mol} \mathrm{K} \times 500 \mathrm{~K} \times \ln \mathrm{K} \\ &\quad \ln \mathrm{K}=2.3 \\ &\mathrm{~K}=10 \mathrm{~atm}=\text { Vapour pressure of liquid ' } \mathrm{S} \text { ' } \end{aligned}ΔGrxno=ΔfG∘ (vapour) −ΔfG∘ (liquid) ΔGrxno=103−100.7=2.3kcal/mol=2300cal/molΔGrxno=−RTlnK2300cal/mol=−2cal/molK×500 K×lnKlnK=2.3 K=10 atm= Vapour pressure of liquid ’ S ’ ∴\therefore \quad∴ Vapour pressure of liquid ' S\mathrm{S}S ' at 500 K500 \mathrm{~K}500 K is approximately equal to 10 atm10 \mathrm{~atm}10 atm.
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