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In which of the following equilibria are $K_c$ and $K_p$ not equal?
A
$2C(s) + O_2(g) \rightleftharpoons 2CO_2(g)$
B
$2NO(g) \rightleftharpoons N_2(g) + O_2(g)$
C
$SO_2(g) + NO_2(g) \rightleftharpoons SO_3(g) + NO(g)$
D
$H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$
Detailed Solution
$K_p = K_c(RT)^{\Delta n_g}$, where $\Delta n_g$ = moles of gaseous products − moles of gaseous reactants. $K_p = K_c$ only when $\Delta n_g = 0$.
$2NO(g) \rightleftharpoons N_2(g) + O_2(g)$: $\Delta n_g = 2 - 2 = 0$
$SO_2(g) + NO_2(g) \rightleftharpoons SO_3(g) + NO(g)$: $\Delta n_g = 2 - 2 = 0$
$H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$: $\Delta n_g = 2 - 2 = 0$
$2C(s) + O_2(g) \rightleftharpoons 2CO_2(g)$: carbon is a solid and is not counted, so $\Delta n_g = 2 - 1 = 1$ and $K_p = K_c(RT)$
Hence $K_c$ and $K_p$ are not equal for $2C(s) + O_2(g) \rightleftharpoons 2CO_2(g)$.
Note: the worked line in the source PDF treats carbon as a gas and gets $\Delta n_g = 0$, which contradicts its own answer; the equation as printed in the question, with C(s), is used here.
$2NO(g) \rightleftharpoons N_2(g) + O_2(g)$: $\Delta n_g = 2 - 2 = 0$
$SO_2(g) + NO_2(g) \rightleftharpoons SO_3(g) + NO(g)$: $\Delta n_g = 2 - 2 = 0$
$H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$: $\Delta n_g = 2 - 2 = 0$
$2C(s) + O_2(g) \rightleftharpoons 2CO_2(g)$: carbon is a solid and is not counted, so $\Delta n_g = 2 - 1 = 1$ and $K_p = K_c(RT)$
Hence $K_c$ and $K_p$ are not equal for $2C(s) + O_2(g) \rightleftharpoons 2CO_2(g)$.
Note: the worked line in the source PDF treats carbon as a gas and gets $\Delta n_g = 0$, which contradicts its own answer; the equation as printed in the question, with C(s), is used here.