JEE Main 2024 · 5 April · morning · Numerical answer · +4 / −1
During Kinetic study of reaction 2 A+B → C+D, the following results were obtained :
Based on above data, overall order of the reaction is _________.
| A [M] | B [M] | initial rate of formation of D | |
|---|---|---|---|
| I | 0.1 | 0.1 | 6.0×10⁻³ |
| II | 0.3 | 0.2 | 7.2×10⁻² |
| III | 0.3 | 0.4 | 2.88×10⁻¹ |
| IV | 0.4 | 0.1 | 2.40×10⁻² |
Answer and solution
Answer 3
Solution
Chemistry ma’am · 3 steps
Find two runs where one concentration stays fixed, and the rate change tells you that reactant's order.
- Rate law. Write rate = k[A]ˣ[B]ʸ. The orders x and y come from the data, not from the coefficients 2 and 1 in the equation.
Runs compared Held fixed Concentration change Rate change Order I and IV [B] = 0.1 [A]: 0.1 → 0.4, × 4 6.0 × 10⁻³ → 2.40 × 10⁻², × 4 4ˣ = 4, so x = 1 II and III [A] = 0.3 [B]: 0.2 → 0.4, × 2 7.2 × 10⁻² → 2.88 × 10⁻¹, × 4 2ʸ = 4, so y = 2 - Cross-check. From run I to run II, [A] triples and [B] doubles. The rate law predicts a factor of 3 × 2² = 12, and 6.0 × 10⁻³ × 12 = 7.2 × 10⁻² matches the data.
- So. Rate = k[A]¹[B]², and the overall order is 1 + 2 = 3.
The common slip: 2. Reporting only the order in B gives 2. The overall order is the sum of the orders in A and B.
The idea to keep. Order is always found from experiment. Pick runs where only one concentration changes, and the rate factor equals the concentration factor raised to that reactant's order.
Answer from the NTA final answer key. Solution by PrepWiser.