Question

At $350\,K$, $K_p$ for the reaction given below is $3.0 \times 10^{10}\, bar^{-1}$ at equilibrium. What will be the value of $K_c$ at this temperature?
$2N_{2(g)} +O_{2(g)} \rightleftharpoons 2N_2O_{(g)}$

• A. $7.4\times10^{11}\,L\,mol^{-1}$
• B. $8715\times10^{10}\,L\,mol^{-1}$
• C. $0.08\,L\,mol^{-1}$
• D. $8715\times10^{11}\,L\,mol^{-1}$

Answer 1

Answer: (D)

$K_{p}=K_{c}\left(RT\right)^{\Delta n}$ ;
$\Delta n=2-3=-1$
$T=350\,K$,
$R=0.083 \,bar \,L\,K^{-1}\,mol^{-1}$
$K_{c}=\frac{K_{p}}{\left(RT\right)^{\Delta n}}$
$K_{c}=\frac{3\times10^{10}\,bar^{-1}}{\left(0.083\,L\,bar\,K^{-1}\,mol^{-1}\times350\,K\right)^{-1}}$
$=3\times10^{10}\,bar^{-1}\times0.083\,L\,bar\,K^{-1}\,mol^{-1}\times350\,K$
$=8715\times10^{11}\,L\,mol^{-1}$