The half life for the decomposition of gaseous compound A is 240 s when the gaseous pressure was 500 Torr initially. When the pressure was 250 Torr, the half life was found to be 4.0 min. The order of the reaction is ______________. (Nearest integer)

Solution

Chemistry ma’am · 4 steps

Put both half-lives in seconds first: 4.0 min is 240 s, so halving the pressure did not change the half-life.

1. Same units. Convert the second half-life: 4.0 min = 4.0 × 60 = 240 s. So both runs have t_(1/2) = 240 s.
2. Half-life and order. For a reaction of order n, the half-life depends on the starting pressure as t_(1/2) ∝ 1/p₀ⁿ⁻¹. Write this for both runs and divide one by the other.

   (t_(1/2))₁/(t_(1/2))₂ = ((p₀)₂/(p₀)₁)ⁿ⁻¹ ⇒ 240/240 = (250/500)ⁿ⁻¹

3. Solve for n. This gives 1 = (1/2)ⁿ⁻¹. A power of 1/2 equals 1 only when the exponent is zero, so n - 1 = 0.
4. So. The half-life does not depend on the initial pressure, which is the mark of a first-order reaction. The order is 1.

The idea to keep. Half-life tells you the order: unchanged when the starting amount changes means first order, proportional to it means zero order, inversely proportional means second order. Always match the units first.

Answer from the NTA final answer key. Solution by PrepWiser.