Question

For the following equlibrium, $N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$
$K_P=K_C$ is attained when

• A. $T=1.0K$
• B. $T=27.3K$
• C. $T=12.18K$
• D. $T=273K$

Answer 1

The correct option is C $T=12.18K$

Relation between $K_p$ and $K_c$ is given by,
$K_p=K_C(RT{)}^{\Delta n_g}$
where,
$\Delta n_g=$ Number of moles of gaseous product $-$ Number of moles of gaseous reactant.

For given reaction,
$N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$
$\Delta n_g=2-1=1$

$\therefore K_P=K_C\left(RT\right)atmLmo{l}^{-1}$
$K_P=K_C,ifRT=1atmLmo{l}^{-1}$
or $T=\frac{1}{R}=\frac{1atmLmo{l}^{-1}}{0.0821L-atmmo{l}^{-1}{K}^{-1}}=12.18K$