JEE Main 2022 · 29 July · morning · Numerical answer · +4 / −0
The reaction between X and Y is first order with respect to X and zero order with respect to Y.
Examine the data of table and calculate ratio of numerical values of M and L. (Nearest Integer)
| Experiment | [X]/(mol L⁻¹) | [Y]/(mol L⁻¹) | (Initial rate)/(mol L⁻¹ min ⁻¹) |
|---|---|---|---|
| I | 0.1 | 0.1 | 2 × 10⁻³ |
| I | L | 0.2 | 4 × 10⁻³ |
| III | 0.4 | 0.4 | M × 10⁻³ |
| IV | 0.1 | 0.2 | 2 × 10⁻³ |
Answer and solution
Answer 40
Solution
Chemistry ma’am · 5 steps
Y does not affect the rate at all, so the rate depends only on [X]. Find k once and use it everywhere.
- Write the rate law. First order in X and zero order in Y gives rate = k[X]. Experiment IV confirms this: doubling [Y] from experiment I leaves the rate at 2 × 10⁻³.
- Find k. From experiment I, k = (2 × 10⁻³)/0.1 = 0.02 min⁻¹.
- Find L. The second row (experiment II) has rate 4 × 10⁻³, so L = (4 × 10⁻³)/0.02 = 0.2 mol L⁻¹. The [Y] value of 0.2 plays no part.
- Find M. In experiment III, rate = 0.02 × 0.4 = 8 × 10⁻³ mol L⁻¹ min⁻¹. The table writes this as M × 10⁻³, so M = 8.
M/L = 8/0.2 = 40
- So. The ratio of M to L is 40.
The common slip: 0.04. Taking M as the full rate 8 × 10⁻³ instead of the number 8 gives (8 × 10⁻³)/0.2 = 0.04, not 40.
The idea to keep. A zero-order reactant can change as much as it likes without changing the rate. Get k from one clean experiment, then use rate = k[X] for every missing entry.
Answer from the NTA final answer key. Solution by PrepWiser.