The concentration of a reactant undergoing decomposition was $0.1, 0.08$ and $0.067\,\mathrm{mol\,L^{-1}}$ after $1.0, 2.0$ and $3.0$ hr respectively. The order of the reaction is
(a)0
(b)1
(c)2
(d)3
Answer
Answer: C ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Key was missing. 1/[A] = 10, 12.5, 14.93 L mol⁻¹ rises by a constant ≈2.5 per hour, so the reaction is second order.
Explanation
Test each order with the data (t = 1, 2, 3 h; [A] = 0.100, 0.080, 0.067 mol L⁻¹).
Zero order: Δ[A] = 0.020, 0.013 — not constant. First order: ln([A]₁/[A]₂) = 0.223, ln([A]₂/[A]₃) = 0.177 — not constant.
Second order: 1/[A] = 10.0, 12.5, 14.9 L mol⁻¹; Δ(1/[A]) = 2.5, 2.4 — constant, so k ≈ 2.5 L mol⁻¹ h⁻¹.