For the reaction $$2H_2(g) + 2NO(g) \rightarrow N_2(g) + 2H_2O(g)$$ the observed rate expression is, rate $$= k_f[NO]^2[H_2]$$. The rate expression for the reverse reaction is:

We have the forward reaction

$$2\,H_2(g)+2\,NO(g)\;\longrightarrow\;N_2(g)+2\,H_2O(g)$$

and its experimentally observed rate law

$$\text{rate}_{f}=k_f[NO]^2[H_2].$$

To obtain a suitable rate law for the reverse reaction we make use of the fact that, at equilibrium, the forward rate equals the reverse rate. First we write a general form for the backward rate:

$$\text{rate}_{b}=k_b\,[N_2]^{\,x}[H_2O]^{\,y}[H_2]^{\,z}[NO]^{\,w},$$

where the exponents $$x,y,z,w$$ are to be chosen so that (i) equality of the two rates at equilibrium gives a constant ratio $$k_f/k_b$$ and (ii) no term on the right is superfluous. Because only $$N_2$$ and $$H_2O$$ appear as products, we take $$w=0$$; hence

$$\text{rate}_{b}=k_b\,[N_2]^{\,x}[H_2O]^{\,y}[H_2]^{\,z}.$$

At equilibrium we must have

$$k_f[NO]^2[H_2]=k_b\,[N_2]^{\,x}[H_2O]^{\,y}[H_2]^{\,z}. \qquad (1)$$

Meanwhile, the equilibrium‐constant expression for the reaction is

$$K_c=\frac{[N_2][H_2O]^2}{[NO]^2[H_2]^2}. \qquad (2)$$

We now substitute the right side of (2) into (1) so that every concentration on the right of (1) can be replaced by a power of those on the left:

From (2) we have

$$[NO]^2[H_2]^2=\frac{[N_2][H_2O]^2}{K_c},$$

and therefore

$$[NO]^2[H_2]=\frac{[N_2][H_2O]^2}{K_c[H_2]}. \qquad (3)$$

Substituting (3) into (1) gives

$$k_f\left(\frac{[N_2][H_2O]^2}{K_c[H_2]}\right)=k_b\,[N_2]^{\,x}[H_2O]^{\,y}[H_2]^{\,z}.$$

Canceling common factors $$ [N_2] $$ and $$ [H_2O]^2 $$ on both sides yields

$$\frac{k_f}{K_c[H_2]}=k_b\,[N_2]^{\,x-1}[H_2O]^{\,y-2}[H_2]^{\,z}.$$

The left side is a constant (because $$k_f$$ and $$K_c$$ are constants), so the concentration factors on the right must together reduce to unity. This is possible only when

$$x-1=0,\quad y-2=0,\quad z+1=0.$$

Thus

$$x=1,\; y=2,\; z=-1.$$

Substituting these values back into the general form of the reverse rate law gives

$$\text{rate}_{b}=k_b\;\frac{[N_2][H_2O]^2}{[H_2]}\;.$$

Therefore, the observed rate expression for the reverse reaction is

$$\text{rate}_{b}=k_b\,[N_2][H_2O]^2/[H_2].$$

This corresponds to Option D.

Hence, the correct answer is Option D.