Question

What is the rate expression for the reverse reaction?

$2{\mathrm{H}}_2\left(\mathrm{g}\right)+2\mathrm{NO}\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{N}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

The observed rate expression is $\mathrm{rate}={\mathrm{K}}_{\mathrm{f}}{\left[\mathrm{NO}\right]}^2\left[{\mathrm{H}}_2\right]$.

• A. ${\mathrm{K}}_{\mathrm{b}}\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\mathrm{O}\right]}^2$
• B. ${\mathrm{K}}_{\mathrm{b}}\left[{\mathrm{N}}_2\right]\left[{\mathrm{H}}_2\mathrm{O}\right]$
• C. ${\mathrm{K}}_{\mathrm{b}}\frac{\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\mathrm{O}\right]}^2}{\left[{\mathrm{H}}_2\right]}$
• D. ${\mathrm{K}}_{\mathrm{b}}\frac{\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\mathrm{O}\right]}^2}{\left[\mathrm{NO}\right]}$

Answer 1

The correct option is C

${\mathrm{K}}_{\mathrm{b}}\frac{\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\mathrm{O}\right]}^2}{\left[{\mathrm{H}}_2\right]}$

The explanation for the correct option: C

1. At equilibrium, the rate of forwarding reaction is equal to the rate of backward reaction.
   $2{\mathrm{H}}_2\left(\mathrm{g}\right)+2\mathrm{NO}\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{N}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$
2. The equilibrium constant for the reaction is $\mathrm{K}=\frac{{\mathrm{K}}_{\mathrm{f}}}{{\mathrm{K}}_{\mathrm{b}}}$
3. $\frac{{\mathrm{K}}_{\mathrm{f}}}{{\mathrm{K}}_{\mathrm{b}}}=\frac{\left[N_2\right]{\left[H_{2}O\right]}^2}{{\left[NO\right]}^2{\left[H_2\right]}^2}$
4. ${\mathrm{K}}_{\mathrm{f}}{\left[\mathrm{NO}\right]}^2{\left[{\mathrm{H}}_2\right]}^2={\mathrm{K}}_{\mathrm{b}}\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\mathrm{O}\right]}^2$
5. ${\mathrm{K}}_{\mathrm{f}}\left[{\mathrm{H}}_2\right]{\left[\mathrm{NO}\right]}^2={\mathrm{K}}_{\mathrm{b}}\frac{\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\mathrm{O}\right]}^2}{\left[{\mathrm{H}}_2\right]}$
6. Hence, option C is correct.