Question

Consider the following reaction,
\(2H_2(g) + 2NO(g) \rightarrow N_2(g) + 2H_2O(g)\)
which follows the mechanism given below:
\(2NO(g)\underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}} N_2O_2(g)\) (fast equilibrium)
\(N_2O_2(g)+H_2(g) \xrightarrow[]{K_2}N_2O(g)+H_2O\)  (slow reaction)
\(N_2O(g)+H_2(g) \xrightarrow[]{K_3}N_2(g)+H_2O\)  (fast reaction)
The order of the reaction is________

Answer 1

Correct Answer: 3

Step 1: Identify the Rate-Determining Step 

The slowest step controls the overall reaction rate. Here, the slow step is:

\[ N_2O_2(g) + H_2(g) \rightarrow N_2O(g) + H_2O(g) \]

Step 2: Express Rate Law in Terms of Intermediate

The rate law for the slow step is:

\[ \text{Rate} = k_2 [N_2O_2] [H_2] \]

Since \( N_2O_2 \) is an intermediate, we express it in terms of reactants using the equilibrium step:

\[ K = \frac{[N_2O_2]}{[NO]^2} \Rightarrow [N_2O_2] = K[NO]^2 \]

Substituting this into the rate equation:

\[ \text{Rate} = k_2 K [NO]^2 [H_2] \]

Step 3: Determine Reaction Order

• The exponent of \([NO]\) is 2.
• The exponent of \([H_2]\) is 1.
• Total order of reaction = 2 + 1 = 3.