U5.3 · Concentration Changes Over Time (Integrated Rate Laws, Half-Life)Practice · 浓度随时间的变化（积分速率定律、半衰期）

$t_{1/2}$ = 0.693 ÷ 0.0462 = 15.0 min. 1/8 = (1/2)³, so 3 half-lives: 3 × 15.0 = 45.0 min. A first-order half-life doesn’t depend on concentration, so each one is the same length.

First order: ln[A] = ln[A]₀ − kt, so [A] = [A]₀$e^{-kt}$ = 0.500 × $e^{-0.0100 \times 60.0}$ = 0.500 × $e^{-0.600}$ = 0.274 M.

Second order: $\dfrac{1}{[A]} = \dfrac{1}{[A]_0} + kt$ = 1/0.400 + 0.150 × 50.0 = 2.50 + 7.50 = 10.0, so [A] = 0.100 M.

A zero-order reaction loses [A] at a constant rate: $t_{1/2} = \dfrac{[A]_0}{2k}$, proportional to [A]₀, so halving [A]₀ halves the half-life. Compare: first order, $\dfrac{\ln 2}{k}$, stays the same; second order, $\dfrac{1}{k[A]_0}$, doubles.

Compute ln[A]: −1.61, −2.31, −3.01, −3.71, −4.41. It drops by about 0.70 every 20 min, so ln[A] vs. t is linear: first order (slope = −k = −0.035 min⁻¹). The half-life, about 2.0 × 10¹ min, stays constant too.

Reading the curve: at t = 0, 20, 40, and 60 min, [A] is about 0.80, 0.40, 0.20, and 0.10 M. The concentration halves every 20 min, and a half-life that doesn’t depend on concentration is the mark of a first-order reaction (t½ = ln 2/k). The curve flattens, so the slope isn’t constant (not zero order). The amounts lost, 0.40, 0.20, and 0.10 M, are not equal; it’s the fraction lost that stays the same. A second-order reaction would have half-lives that keep getting longer.

A straight line for ln[A] against t means first order: ln[A] = ln[A]₀ − kt, so the slope is −k. Using the end points (0, -0.69) and (80, -2.54), the slope is (-2.54 − (-0.69)) ÷ 80 = -0.0231 s⁻¹, so k = 0.0231 s⁻¹ and t½ = 0.693 ÷ 0.0231 ≈ 30 s. Choosing 43 s calculates 1/k instead of ln 2/k.