Question

At $350$K, $K_p$ for the reaction given below is $3.0times{10}^{10}ba{r}^{-1}$ at equilibrium. What will be the value of $K_c$ at this temperature?
$2N_{2\left(g\right)}+O_{2\left(g\right)}\rightleftharpoons 2N_2{O}_{(}g)$

• A. $7.4\times {10}^{11}Lmo{l}^{-1}$
• B. $8715\times {10}^{10}mo{l}^{-1}$
• C. $0.08Lmo{l}^{-1}$
• D. $8715\times {10}^{11}mo{l}^{-1}$

Answer 1

Answer: (D)

$8715\times {10}^{11}mo{l}^{-1}$

$2N_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2N_{2}O\left(g\right)$

$K_c=\frac{K_p}{(RT{)}^{\Delta n}}$

$\Delta n=2-(2+1)=-1,\therefore K_p<K_c$

$K_c=\frac{3.0\times {10}^{10}ba{r}^{-1}}{(0.083Lbar{K}^{-1}mo{l}^{-1}\times 350K{)}^{-1}}$

$\therefore K_c=(3.0\ast {10}^{10})\times 0.083\times 350=8.715\times {10}^{11}L/mol$