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  • The gas phase reaction <img alt="2 \mathrm{~A}(\mathrm{~g}) \rightleftharpoons \mathrm{A}_{2}(\mathrm{~g})" src="/latex-image/?2%20%5Cmathrm%7B%7EA%7D%28%5Cmathrm%7B%7Eg%7D%29%20%5Crightleftharpoons%20%5Cmathrm%7BA%7D_%7B2%7D%28%5Cmathrm%7B%7Eg%7D%29"

The gas phase reaction 2 \mathrm{~A}(\mathrm{~g}) \rightleftharpoons \mathrm{A}_{2}(\mathrm{~g})  at 400K has \Delta \mathrm{G}^{\mathrm{o}}=+25.2 \mathrm{~kJ} \mathrm{~mol}^{-1}  The equilibrium constant Kc for this reaction is _____ \times 10 ^{-2} (Rounded off to the nearest integer) [Use : R = 8.3 J mol-1K-1 , ln 10 = 2.3 log102 = 0.30, 1 atm = 1 bar] [antilog (-0.3) = 0.501]
 

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best_answer

Using formula

\Delta_{\mathrm{r}} \mathrm{G}^{0}=-\mathrm{R} \mathrm{T} \ln \mathrm{K}_{\mathrm{p}}

25200=-2.3 \times 8.3 \times 400 \log \left(\mathrm{K}_{\mathrm{p}}\right)

\mathrm{K}_{\mathrm{p}}=10^{-3.3}=10^{-3} \times 0.501

\mathrm{K}_{\mathrm{p}}=5.01 \times 10^{-4}\ \mathrm{Bar}^{-1}

\mathrm{K}_{\mathrm{p}}=5.01 \times 10^{-9} \mathrm{~Pa}^{-1}

\mathrm{K}_{\mathrm{p}}=\frac{\mathrm{K}_{\mathrm{C}}}{8.3 \times 400}

\mathrm{K}_{\mathrm{C}}=1.66 \times 10^{-5} \mathrm{~m}^{3} / \mathrm{mole}

\mathrm{K_C}=1.66 \times 10^{-2} \mathrm{~L} / \mathrm{mol}

Ans = 2

Posted by

Kuldeep Maurya

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