Question

Consider the following equilibria at ${25}^{\circ }C,2N{O}_{\left(g\right)}\rightleftharpoons N_{2\left(g\right)}+O_{2\left(g\right)};K_1=4\times {10}^{30}andN{O}_{\left(g\right)}+\frac{1}{2}B{r}_{2\left(g\right)}\rightleftharpoons NOB{r}_{\left(g\right)};K_2=1.4mo{l}^{-\frac{1}{2}}{L}^{\frac{1}{2}}.$ The value of K_C for the reaction (at same temperature) $\frac{1}{2}{N}_{2\left(g\right)}+\frac{1}{2}{O}_{2\left(g\right)}+\frac{1}{2}B{r}_{2\left(g\right)}\rightleftharpoons NOB{r}_{\left(g\right)}$ is :

• A. $3.5\times {10}^{-31}$
• B. $2.8\times {10}^{15}$
• C. $7.0\times {10}^{-16}$
• D. $5.6\times {10}^{30}$

Answer 1

The correct option is C

$7.0\times {10}^{-16}$

$2NO\rightleftharpoons N_2+O_2,K_1=4.0\times {10}^{30}NO+\frac{1}{2}B{r}_2\rightleftharpoons NOBr,K_2=1.4\frac{1}{2}{N}_2+\frac{1}{2}{O}_{2}B{r}_2\rightleftharpoons NOBr,K_3=?K_3=\frac{K_2}{\sqrt{K_1}}=\frac{1.4}{(4.0\times {10}^{30}{)}^{\frac{1}{2}}}=7.0\times {10}^{-16}.$