The half life for the decomposition of gaseous compound A is 240 s when the gaseous pressure was 500 — Chemical Kinetics Chemistry Question
Question
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 & Explanation
**Step 1: Relate half-life to reaction order** The half-life formula depends on reaction order: - For order n: t₁/₂ ∝ P₀^(1-n) where P₀ is initial pressure. **Step 2: Set up the ratio of half-lives** $$\frac{t_{1/2}^{(1)}}{t_{1/2}^{(2)}} = \left(\frac{P_0^{(1)}}{P_0^{(2)}}\right)^{1-n}$$ **Step 3: Substitute given values** - t₁/₂⁽¹⁾ = 240 s; P₀⁽¹⁾ = 500 Torr - t₁/₂⁽²⁾ = 4.0 min = 240 s; P₀⁽²⁾ = 250 Torr $$\frac{240}{240} = \left(\frac{500}{250}\right)^{1-n}$$ **Step 4: Solve for n** $$1 = (2)^{1-n}$$ $$2^0 = 2^{1-n}$$ $$0 = 1 - n$$ $$n = 1$$ Therefore, the answer is 1.