For the reaction PCl_5 , (g) rightleftharpoon | vedclass

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Question

(A) The relationship between $K_p$ and $K_c$ is given by the formula $K_p = K_c(RT)^{\Delta n_g}$.For the reaction $PCl_5 \, (g) \rightleftharpoons PC

Vedclass · Accepted Answer

$\log \frac{K_p}{K_c} - \log RT = 0$

Vedclass · Answer

(A) The relationship between $K_p$ and $K_c$ is given by the formula $K_p = K_c(RT)^{\Delta n_g}$.For the reaction $PCl_5 \, (g) \rightleftharpoons PCl_3 \, (g) + Cl_2 \, (g)$,the change in the number of moles of gaseous species is $\Delta n_g = (1 + 1) - 1 = 1$.Substituting this into the formula,we get $K_p = K_c(RT)^1$,which simplifies to $K_p = K_c RT$.Rearranging this,we get $K_c = K_p(RT)^{-1}$ or $\frac{K_p}{K_c} = RT$.Taking the logarithm on both sides,$\log \frac{K_p}{K_c} = \log RT$,which implies $\log \frac{K_p}{K_c} - \log RT = 0$.Thus,option $A$ is correct.

For the reaction $PCl_5 \, (g) \rightleftharpoons PCl_3 \, (g) + Cl_2 \, (g)$,the two reaction constants $K_p$ and $K_c$ are related to each other by the expression

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When $2 \, mol$ of $PCl_5$ is heated in a closed vessel of $2 \, L$ capacity,$40\%$ of $PCl_5$ dissociates into $PCl_3$ and $Cl_2$ at equilibrium. The value of the equilibrium constant $K_c$ will be:

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$A_{2(g)} + B_{2(g)} \overset {K_1} \leftrightarrows 2AB_{(g)} ......(1)$
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