Question

The following data were obtained for the reaction: \[ 2NO(g) + O_2(g) \rightarrow 2N_2O(g) \] at different concentrations: 

 The rate law of this reaction is:

• A. Rate = k[NO][O₂]²
• B. Rate = k[NO][O₂]
• C. Rate = k[NO]²[O₂]²
• D. Rate = k[NO]²[O₂]
• E. Rate = k[NO]²[O₂]³

Hint:

The order of reaction with respect to each reactant is determined by comparing experiments where the concentration of one reactant is changed while the others are held constant. The rate is proportional to the concentration raised to the power of the order.

Answer 1

The Correct Option is D

To determine the rate law, we use the data from the experiments to analyze how the rate depends on the concentrations of \( \text{NO} \) and \( \text{O}_2 \).
1. Experiment 1 and Experiment 2:
The concentration of \( \text{O}_2 \) remains constant, while the concentration of \( \text{NO} \) doubles (from 0.30 to 0.60 mol L\(^{-1}\)), and the rate increases by a factor of 4 (from 0.096 to 0.384 mol L\(^{-1}\) min\(^{-1}\)).
This suggests that the rate is proportional to \( [NO]^2 \), i.e., the order with respect to \( \text{NO} \) is 2.
2. Experiment 1 and Experiment 3:
The concentration of \( \text{NO} \) remains constant, while the concentration of \( \text{O}_2 \) doubles (from 0.30 to 0.60 mol L\(^{-1}\)), and the rate increases by a factor of 2 (from 0.096 to 0.192 mol L\(^{-1}\) min\(^{-1}\)).
This suggests that the rate is proportional to \( [O_2] \), i.e., the order with respect to \( \text{O}_2 \) is 1.
Thus, the rate law is Rate = \[ k[NO]^2[O_2] \], corresponding to option (D).