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In which of the following equilibrium $\mathrm{K}_C$ and $\mathrm{K}_{\mathrm{P}}$ are not equal ?
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$2 \mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{CO}_2(\mathrm{~g})$
Key Idea The reaction for which the number of moles of gaseous products $\left(n_{\mathrm{p}}\right.$ ) is not equal to the number of moles of gaseous reactants $\left(\mathbf{n}_{\mathrm{R}}\right)$, has different value of $\mathrm{K}_{\mathrm{c}}$ and $\mathrm{K}_{\mathrm{p}}$.
(a) $\boldsymbol{n}_{\mathrm{P}}=\mathbf{n}_{\mathrm{R}}=2$, thus, $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}$
(b) $\mathrm{n}_{\mathrm{P}}=\mathrm{n}_{\mathrm{R}}=2$, thus, $K_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}$
(c) $\mathrm{n}_{\mathrm{P}}=\mathrm{n}_{\mathrm{R}}=2$, thus, $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}$
(d) $n_p=2, n_R=1$, thus, $K_p \neq K_c$
(a) $\boldsymbol{n}_{\mathrm{P}}=\mathbf{n}_{\mathrm{R}}=2$, thus, $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}$
(b) $\mathrm{n}_{\mathrm{P}}=\mathrm{n}_{\mathrm{R}}=2$, thus, $K_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}$
(c) $\mathrm{n}_{\mathrm{P}}=\mathrm{n}_{\mathrm{R}}=2$, thus, $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}$
(d) $n_p=2, n_R=1$, thus, $K_p \neq K_c$
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