JEE Main 2023 · 24 January · evening · Single correct · +4 / −1
A student has studied the decomposition of a gas AB₃ at 25°C. He obtained the following data.
The order of the reaction is
| p (mm Hg) | 50 | 100 | 200 | 400 |
|---|---|---|---|---|
| relative t_(1/2) (s) | 4 | 2 | 1 | 0.5 |
- (a) 2
- (b) 0.5
- (c) 1
- (d) 0 (zero)
Answer and solution
Answer (a) 2
Solution
Chemistry ma’am · 3 steps
Double the pressure and the half-life halves every time, so t_(1/2) is inversely proportional to p.
- Half-life rule. For a reaction of order n, the half-life depends on the starting amount. For a gas that amount is its initial pressure p₀, and t_(1/2) ∝ p₀¹⁻ⁿ.
p (mm Hg) t_(1/2) (s) p × t_(1/2) 50 4 200 100 2 200 200 1 200 400 0.5 200 - Read the pattern. The product p × t_(1/2) stays at 200 for every reading. So t_(1/2) ∝ p⁻¹.
1 - n = -1 ⇒ n = 2
- So. The decomposition of AB₃ is second order, which is option (a).
The idea to keep. Use t_(1/2) ∝ [A]₀¹⁻ⁿ. When the starting amount doubles, a half-life that stays the same means order 1, one that doubles means order 0, and one that halves means order 2.
Answer from the NTA final answer key. Solution by PrepWiser.