At equilibrium $\mathrm{r}_{\mathrm{f}}=\mathrm{r}_{\mathrm{b}}$
$\mathrm{k}_{\mathrm{f}}\left[\mathrm{H}_{2}\right][\mathrm{NO}]^{2}=\mathrm{k}_{\mathrm{b}} \frac{\left[\mathrm{N}_{2}\right]\left[\mathrm{H}_{2} \mathrm{O}\right]^{2}}{\left[\mathrm{H}_{2}\right]}$
Hence, rate expression for reverse reaction.
$=\mathrm{k}_{\mathrm{b}} \frac{\left[\mathrm{N}_{2}\right]\left[\mathrm{H}_{2} \mathrm{O}\right]^{2}}{\left[\mathrm{H}_{2}\right]}$