At 345 K, the half life for the decomposition of a sample of a gaseous compound initially at 55.5 kPa was 340 s. When the pressure was 27.8 kPa, the half life was found to be 170 s. The order of the reaction is ____________. [integer answer]

Solution

Chemistry ma’am · 4 steps

Watch how the half-life moves with starting pressure. Here they halve together, and that pins the order.

1. Half-life rule. For a reaction of order n, the half-life depends on the starting amount as t_(1/2) ∝ [A]₀¹⁻ⁿ. For a gas, the initial pressure P₀ stands in for [A]₀.

   t_(1/2)(1)/t_(1/2)(2) = (P₀(1)/P₀(2))¹⁻ⁿ

2. Put in numbers. The half-life ratio is 340/170 = 2. The pressure ratio is 55.5/27.8 = 1.996 ≈ 2. So 2 = 2¹⁻ⁿ, which means 1 - n = 1.
3. Check the meaning. Halving the starting pressure halves the half-life, so t_(1/2) ∝ P₀. That is the zero-order signature, t_(1/2) = [A]₀/2k.
4. So. n = 0. The reaction is of order 0.

The common slip: 2. Writing the rule as t_(1/2) ∝ P₀ⁿ⁻¹ flips the sign of the exponent and gives order 2, not 0.

The idea to keep. Remember t_(1/2) ∝ [A]₀¹⁻ⁿ. If the half-life follows the starting amount, the order is zero. If it does not change, the order is first. If it moves the opposite way, the order is second.

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