Question

What is the equilibrium constant for the reaction $P_{4\left(s\right)}+5O_{2\left(g\right)}\rightleftharpoons P_4{O}_{10\left(s\right)}$?

• A. $K_c=\frac{1}{[O_2{]}^5}$
• B. $K_c=\frac{\left[P_4{O}_{10}\right]}{\left[P_4\right][O_2{]}^5}$
• C. $K_c={\left[O_2\right]}^5$
• D. $K_c=\frac{\left[P_4{O}_{10}\right]}{5\left[P_4\right][O_2{]}^5}$

Answer 1

The correct option is A $K_c=\frac{1}{[O_2{]}^5}$

The equilibrium reaction is $P_4\left(s\right)+5O_2\left(g\right)\rightleftharpoons P_4{O}_{10}\left(s\right)$

The expression for the equilibrium reaction is $K_c=\frac{\left[P_4{O}_{10}\right]}{\left[P_4\right][O_2{]}^5}=\frac{1}{1\times [O_2{]}^5}$.

Here, $\left[P_4{O}_{10}\right]=\left[P_4\right]=1$ as the partial pressures of the solid substances are considered as unity.