Assuming ideal behaviour, the… - JEE Main 2021 Chemistry PYQ

JEE Main 2021 — Chemistry Physical Chemistry

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Assuming ideal behaviour, the magnitude of $\mathrm{log}K$ for the following reaction at $25^{\circ}C$ is $x\times {10}^{-1}.$ The value of $x$ is __________. (Integer answer) $3\mathrm{HC}\equiv {\mathrm{CH}}_{(g)}\rightleftharpoons {C}_{6}{H}_{6(l)}$

[Given: ${\Delta }_{f}G^{\circ}(\mathrm{HC}\equiv \mathrm{CH})=-2.04\times {10}^{5}J{\mathrm{mol}}^{-1};{\Delta }_{f}G^{\circ}({C}_{6}{H}_{6})=-1.24\times {10}^{5}J{\mathrm{mol}}^{-1};R=8.314J{K}^{-1}{\mathrm{mol}}^{-1}$ ]

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Held on 24 Feb 2021 · Verified 6 Jul 2026.

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Solution

$3\mathrm{HC}\equiv {\mathrm{CH}}_{(g)}\rightarrow {C}_{6}{H}_{6}(\ell ):{\Delta G}^{0}=-\mathrm{RT}\mathrm{ln}K$

${\Delta G}_{f}^{0}-2.04\times {10}^{5}\frac{J}{\mathrm{mol}}-1.24\times {10}^{5}J/\mathrm{mol}$

$\Rightarrow {\Delta G}^{0}=\Sigma {({\Delta G}_{f}^{0})}_{P}-\Sigma {({\Delta G}_{f}^{0})}_{R}$

$\Rightarrow -RT\ell nK=1\times (-1.24\times {10}^{5})-(-3\times 2.04\times {10}^{5})$

$\Rightarrow -2.303\times R\times T\mathrm{log}K=4.88\times {10}^{5}$

$\Rightarrow \mathrm{log}K=-\frac{4.88\times {10}^{5}}{2.303\times R\times T}=-\frac{488000}{5705.848}=-85.52$

$=855\times {10}^{-1}$

$\Rightarrow x=855$

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Solution

$3\mathrm{HC}\equiv {\mathrm{CH}}_{(g)}\rightarrow {C}_{6}{H}_{6}(\ell ):{\Delta G}^{0}=-\mathrm{RT}\mathrm{ln}K$

${\Delta G}_{f}^{0}-2.04\times {10}^{5}\frac{J}{\mathrm{mol}}-1.24\times {10}^{5}J/\mathrm{mol}$

$\Rightarrow {\Delta G}^{0}=\Sigma {({\Delta G}_{f}^{0})}_{P}-\Sigma {({\Delta G}_{f}^{0})}_{R}$

$\Rightarrow -RT\ell nK=1\times (-1.24\times {10}^{5})-(-3\times 2.04\times {10}^{5})$

$\Rightarrow -2.303\times R\times T\mathrm{log}K=4.88\times {10}^{5}$

$\Rightarrow \mathrm{log}K=-\frac{4.88\times {10}^{5}}{2.303\times R\times T}=-\frac{488000}{5705.848}=-85.52$

$=855\times {10}^{-1}$

$\Rightarrow x=855$

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Solution

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Did you get this right?

Solution

Your note

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More JEE Main Physical Chemistry questions

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