For the reaction $2 \mathrm{H}_{2}(\mathrm{~g})+2 \mathrm{NO}(\mathrm{g}) \longrightarrow \mathrm{N}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}$, the observed rate expression is, rate $= \mathrm{k}_{\mathrm{f}}[\mathrm{NO}]^{2}\left[\mathrm{H}_{2}\right]$. The rate expression for the reverse reaction is:

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

(A) $\mathrm{K}_{\mathrm{b}}\left[\mathrm{N}_{2}\right]\left[\mathrm{H}_{2} \mathrm{O}\right]$

(B) $\mathrm{k}_{\mathrm{b}}\left[\mathrm{N}_{2}\right]\left[\mathrm{H}_{2} \mathrm{O}\right]^{2} /[\mathrm{NO}]$

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

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Chemical Kinetics

For the reaction

2 H_{2} (g) + 2 NO (g) ⟶ N_{2} (g) + 2 H_{2} O

, the observed rate expression is, rate

= k_{f} [NO]^{2} [H_{2}]

. The rate expression for the reverse reaction is:

Answer: (C)

At equilibrium the forward and reverse rates are equal, and the equilibrium constant is

K = \frac{k _{f}}{k _{b}} = \frac{[ N X _{2} ] [ H X _{2} O ] ^{2}}{[ NO ] ^{2} [ H X _{2} ]}

. Setting the forward rate

k_{f} [NO]^{2} [H X_{2}]

equal to the reverse rate and rearranging shows the reverse rate must be

\frac{k _{b} [ N X _{2} ] [ H X _{2} O ] ^{2}}{[ H X _{2} ]}

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Single correct option