At 345 K, the half life for the decomposition of a sample of a gaseous compound initially at 55.5 kP — Chemical Kinetics Chemistry Question

💡 Solution & Explanation

**Step 1: Recall the relationship between half-life and reaction order.** For different reaction orders, half-life depends on initial concentration differently: - Order 0: t₁/₂ ∝ [A]₀ - Order 1: t₁/₂ is independent of [A]₀ - Order 2: t₁/₂ ∝ 1/[A]₀ **Step 2: Set up the ratio of half-lives.** Given data: - Initial pressure P₁ = 55.5 kPa, t₁/₂ = 340 s - Initial pressure P₂ = 27.8 kPa, t₁/₂ = 170 s **Step 3: Calculate the ratio.** $$\frac{t_{1/2,1}}{t_{1/2,2}} = \frac{340}{170} = 2$$ $$\frac{P_1}{P_2} = \frac{55.5}{27.8} = 2$$ **Step 4: Test against known relationships.** For zero-order reactions: t₁/₂ ∝ [A]₀ (or P₀) $$\frac{t_{1/2,1}}{t_{1/2,2}} = \frac{P_1}{P_2}$$ $$2 = 2$$ ✓ This matches perfectly! **Step 5: Verify other orders don't fit.** - If order 1: t₁/₂ should be constant (340 ≠ 170) ✗ - If order 2: ratio should be 1/2 (2 ≠ 0.5) ✗ Therefore, the answer is **0**.