JEE Main 2022 · 27 July · morning · Numerical answer · +4 / −0
2 NO+2 H₂ → N₂+2 H₂ O
The above reaction has been studied at 800° C. The related data are given in the table below
The order of the reaction with respect to NO is ___________.
| Reaction serial number | Initial Pressure of H₂/kPa | Initial Pressure of NO/kPa | Initial rate ( (- dp)/dt )/(kPa/s) |
|---|---|---|---|
| 1 | 65.6 | 40.0 | 0.135 |
| 2 | 65.6 | 20.1 | 0.033 |
| 3 | 38.6 | 65.6 | 0.214 |
| 4 | 19.2 | 65.6 | 0.106 |
Answer and solution
Answer 2
Solution
Chemistry ma’am · 3 steps
Pick the two runs where only NO changes. Then ask how many times the rate grew when NO doubled.
- Choose the runs. Write the rate law as r = k p_(H₂)ᵐ p_(NO)ⁿ. In runs 1 and 2, p_(H₂) stays at 65.6 kPa and only p_(NO) changes. So the H₂ term cancels when you divide these two rates.
r₁/r₂ = (40.0/20.1)ⁿ ⇒ 0.135/0.033 = (1.99)ⁿ
- Compare the ratios. The pressure of NO roughly doubles (1.99 ≈ 2). The rate goes up about four times (4.09 ≈ 4). So 2ⁿ = 4.
- So. n = 2, which means the reaction is second order in NO. The exact working gives n = (ln 4.09)/(ln 1.99) ≈ 2.05, which also rounds to 2.
The common slip: 1. Runs 3 and 4 keep NO fixed and change H₂. Using them gives the order in H₂, which is 1, not the order in NO.
The idea to keep. To find the order in one reactant, pick two runs where every other concentration stays the same. Then the rate ratio equals the concentration ratio raised to the order.
Answer from the NTA final answer key. Solution by PrepWiser.