Question

For the reaction $2N{O}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\rightleftharpoons N_2{O}_5\left(g\right)$. If the equilibrium constant is $K_p$, then the equilibrium constant for the reaction $2N_2{O}_5\left(g\right)\rightleftharpoons 4N{O}_2\left(g\right)+O_2\left(g\right)$ would be:

• A. $K_p^2$
• B. $\frac{2}{K_p}$
• C. $\frac{1}{K_p^2}$
• D. $\frac{1}{\sqrt{K_p}}$

Answer 1

Answer: (C)

By writing the equation for the reverse reaction$N_2{O}_5\left(g\right)\to 2N{O}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)$

New pressure constant ${K_p}^{{}^{\prime }}$ for this reaction

${K_p}^{{}^{\prime }}=\frac{1}{K_p}$

$K_p^{{}^{\prime }}=\frac{\left[N{O}_2{]}^2\right[O_2{]}^{\frac{1}{2}}}{\left[N_2{O}_5\right]}$

Thus pressure constant for the reaction,

$2N_2{O}_5\left(g\right)\rightleftharpoons 4N{O}_2\left(g\right)+O_2\left(g\right)$

$K_p^{{}^{\prime \prime }}=\frac{\left[N{O}_2{]}^4\right[O_2]}{[N_2{O}_5{]}^2}$