$2N_2O_{4(g)} \rightleftharpoons 4NO_{2(g)}$ has $K_p = 0.15 \, atm$ at $298 \, K$. Calculate $K_p$ in $torr$ and $K_c$ in $mol \, L^{-1}$. ($1 \, atm = 760 \, torr$; $R = 0.082 \, L \, atm \, mol^{-1} \, K^{-1}$).
(A) For the reaction $2N_2O_{4(g)} \rightleftharpoons 4NO_{2(g)}$,the change in moles of gas is $\Delta n_g = 4 - 2 = 2$.
To calculate $K_p$ in $torr$: $K_p = 0.15 \, atm \times 760 \, torr \, atm^{-1} = 114 \, torr$.
To calculate $K_c$,use the relation $K_p = K_c(RT)^{\Delta n_g}$.
$0.15 = K_c(0.082 \times 298)^2$.
$0.15 = K_c(24.436)^2$.
$0.15 = K_c(597.117)$.
$K_c = 0.15 / 597.117 \approx 0.000251 \, mol \, L^{-1}$.
To calculate $K_p$ in $torr$: $K_p = 0.15 \, atm \times 760 \, torr \, atm^{-1} = 114 \, torr$.
To calculate $K_c$,use the relation $K_p = K_c(RT)^{\Delta n_g}$.
$0.15 = K_c(0.082 \times 298)^2$.
$0.15 = K_c(24.436)^2$.
$0.15 = K_c(597.117)$.
$K_c = 0.15 / 597.117 \approx 0.000251 \, mol \, L^{-1}$.
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