JEE Main 2021 · 18 March · morning · Numerical answer · +4 / −0
2NO(g) + Cl₂(g) ⇌ 2NOCl(s)
This reaction was studied at -10° and the following data was obtained
[NO]₀ and [Cl₂]₀ are the initial concentrations and r₀ is the initial reaction rate.
The overall order of the reaction is __________. (Round off to the Nearest Integer).
| Run | [NO]₀ | [Cl₂]₀ | r₀ |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 0.18 |
| 2 | 0.10 | 0.20 | 0.35 |
| 3 | 0.20 | 0.20 | 1.40 |
Answer and solution
Answer 3
Solution
Chemistry ma’am · 4 steps
Pick two runs where only one concentration changes, and see what that change does to the rate.
- Set up rate law. Write r₀ = k[NO]ˣ[Cl₂]ʸ. The overall order is x + y, and it comes from the data, not from the coefficients in the equation.
- Order in Cl2. Runs 1 and 2 have the same [NO]₀ while [Cl₂]₀ doubles. The rate goes from 0.18 to 0.35, a ratio of 1.94 ≈ 2. So 2ʸ = 2 and y = 1.
- Order in NO. Runs 2 and 3 have the same [Cl₂]₀ while [NO]₀ doubles. The rate goes from 0.35 to 1.40, a ratio of exactly 4.
r₃/r₂ = (0.20/0.10)ˣ = 1.40/0.35 = 4 ⇒ 2ˣ = 4 ⇒ x = 2
- So. The rate law is r₀ = k[NO]²[Cl₂], and the overall order is 2 + 1 = 3.
The idea to keep. To find the order with respect to one reactant, compare two runs where only that reactant's concentration changes. If the rate is multiplied by 2ⁿ when the concentration doubles, the order is n.
Answer from the NTA final answer key. Solution by PrepWiser.