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For the reaction, ${PCl}_{5} (g) ⇌ {PCl}_{3} (g) + {Cl}_{2} (g)$ at 121.8K . logK_p - logK_c is equal to (R = 0.0821 lit atm K^-1 mole^-1)

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Detailed Solution

$\begin{array}{l} {Δn}_{g} = 2 - 1 = 1 \\ K_{p} = K_{c} {(RT)}^{1} \\ \frac{K_{p}}{K_{c}} = (0 .0821 \times 121 .8) \\ \frac{K_{p}}{K_{c}} = 10 \\ {logK}_{p} - {logK}_{c} = \log 10 \\ {logK}_{p} - {logK}_{c} = 1 \end{array}$

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