Read the Passage carefully and answer the Questions.
Kinetic studies not only help us to determine the rate of a chemical reaction but also describe the conditions by which the reaction rates can be altered. The factors such as concentration, temperature, pressure and catalyst affect the rate of a reaction. The speed of a reaction or the rate of a reaction can be defined as the change in concentration of a reactant or product in unit time.
Rate = \( k [A]^x [B]^y \)
Here k is the rate constant
\( x + y \) gives the overall order of a reaction, whereas x and y represent the order with respect to the reactants A and B, respectively.
Table: Initial rate of formation of \( NO_2 \)
|
S.No. |
Initial \( [NO]/mol\, L^{-1} \) |
Initial \( [O_2]/mol\, L^{-1} \) |
Initial rate of formation of \( NO_2/mol\, L^{-1}\, s^{-1} \) |
|
1 |
0.30 |
0.30 |
0.096 |
|
2 |
0.60 |
0.30 |
0.384 |
|
3 |
0.30 |
0.60 |
0.192 |
|
4 |
0.60 |
0.60 |
0.768 |
Based on the above observations, the rate of formation of \( NO_2 \) can be represented as
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Rate of reaction = \( k[NO]^2 [O_2] \)
The order with respect to each reactant is found by the initial rate method: change the concentration of one reactant while keeping the other constant, and see how the rate changes. If doubling a concentration multiplies the rate by \( 2^{x} \), the order with respect to that reactant is x.
Let \( \text{Rate} = k[\ce{NO}]^{x}[\ce{O2}]^{y} \).
Experiments 1 and 2: [NO] is doubled, \( \ce{[O2]} \) is unchanged.
\( \displaystyle \frac{0.384}{0.096} = 4 = 2^{x} \Rightarrow x = 2 \)
Experiments 1 and 3: \( \ce{[O2]} \) is doubled, [NO] is unchanged.
\( \displaystyle \frac{0.192}{0.096} = 2 = 2^{y} \Rightarrow y = 1 \)
Check with experiment 4, where both are doubled: the rate should become \( 2^{2} \times 2 = 8 \) times, and \( 0.768/0.096 = 8 \).
So the rate law is Rate = k[NO]2[O2], second order in NO, first order in \( \ce{O2} \), and of order 3 in all.
Shortcut: Experiment 4 doubles both concentrations and the rate becomes 8 times, so the sum of the orders is 3. Only Options (1) and (2) are of order 3, and experiments 1 and 2 (rate four times when [NO] is doubled) leave Option (1).
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Rate Expression and Rate Constant
Table 3.2: Initial rate of formation of NO2
| Experiment | Initial [NO]/mol L–1 | Initial [O2]/mol L–1 | Initial rate of formation of NO2/mol L–1s–1 |
|---|---|---|---|
| 1. | 0.30 | 0.30 | 0.096 |
| 2. | 0.60 | 0.30 | 0.384 |
| 3. | 0.30 | 0.60 | 0.192 |
| 4. | 0.60 | 0.60 | 0.768 |
“It is obvious, after looking at the results, that when the concentration of NO is doubled and that of O2 is kept constant then the initial rate increases by a factor of four from 0.096 to 0.384 mol L–1s–1. This indicates that the rate depends upon the square of the concentration of NO. When concentration of NO is kept constant and concentration of O2 is doubled the rate also gets doubled indicating that rate depends on concentration of O2 to the first power. Hence, the rate equation for this reaction will be Rate = k [NO]2[O2]”