Chemistry - Chemical Equilibrium Question with Solution | TestHub

The equilibrium constant $K$ is the ratio of the forward and reverse rate constants: $K=\frac{k_f}{k_r}=\frac{6.4\times 10^{-4}}{1.6\times 10^{-4}}=4$.

The reaction is $\mathrm{A}(\mathrm{g})+\mathrm{B}(\mathrm{g})\rightleftharpoons 2\mathrm{C}(\mathrm{g})$. Initially, $[\mathrm{A}]=[\mathrm{B}]=\frac{2}{2}=1\mathrm{M}$, and $[\mathrm{C}]=0$. At equilibrium, let $[\mathrm{C}]=2x$. Then $[\mathrm{A}]=1-x$ and $[\mathrm{B}]=1-x$.

$K=\frac{[\mathrm{C}{]}^2}{[\mathrm{A}][\mathrm{B}]}=\frac{(2x{)}^2}{(1-x)(1-x)}=4$. Taking the square root: $\frac{2x}{1-x}=2$, so $2x=2-2x$, $4x=2$, and $x=0.5$. Therefore, $[\mathrm{C}]=2x=1.00\mathrm{M}$.