For the reaction

2 mathrm{H}_{2}(mathrm{g} | vedclass

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Question

$\mathrm{K}_{\mathrm{eq}}=\frac{\mathrm{k}_{\mathrm{f}}}{\mathrm{k}_{\mathrm{b}}}=\frac{\left[\mathrm{N}_{2}\right]\left[\mathrm{H}_{2} \mathrm{O}\rig

Vedclass · Accepted Answer

$\mathrm{k}_{\mathrm{b}}\left[\mathrm{N}_{2}\right]\left[\mathrm{H}_{2} \mathrm{O}\right]^{2} /\left[\mathrm{H}_{2}\right]$

Vedclass · Answer

$\mathrm{K}_{\mathrm{eq}}=\frac{\mathrm{k}_{\mathrm{f}}}{\mathrm{k}_{\mathrm{b}}}=\frac{\left[\mathrm{N}_{2}\right]\left[\mathrm{H}_{2} \mathrm{O}\right]^{2}}{\left[\mathrm{H}_{2}\right]^{2}[\mathrm{NO}]^{2}}$

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.

$=\mathrm{k}_{\mathrm{b}} \frac{\left[\mathrm{N}_{2}\right]\left[\mathrm{H}_{2} \mathrm{O}\right]^{2}}{\left[\mathrm{H}_{2}\right]}$

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