Chemical Kinetics: Exam-Pattern Drill (10)

easy • 2 min read
Tags Chemical Kinetics

Question

A first-order reaction has a half-life of 30 minutes at 300 K. Find the rate constant kk, the time required for 75% of the reactant to be consumed, and the concentration remaining after 90 minutes if the initial concentration was 1.0 M.

Solution — Step by Step

For a first-order reaction, k=0.693t1/2=0.69330=0.0231k = \dfrac{0.693}{t_{1/2}} = \dfrac{0.693}{30} = 0.0231 min−1^{-1}.

If 75% is consumed, 25% remains. So [A]/[A]0=0.25=(1/2)2[A]/[A]_0 = 0.25 = (1/2)^2 — that’s exactly 2 half-lives.

t=2×30=60t = 2 \times 30 = 60 minutes.

Verify with the first-order integrated rate law:

ln⁡[A]0[A]=kt⇒ln⁡4=0.0231⋅t⇒t=1.3860.0231=60 min ✓\ln\frac{[A]_0}{[A]} = kt \Rightarrow \ln 4 = 0.0231 \cdot t \Rightarrow t = \frac{1.386}{0.0231} = 60 \text{ min}\ ✓

90 minutes = 3 half-lives. After 3 half-lives, [A]=[A]0(1/2)3=1.0×0.125=0.125[A] = [A]_0 (1/2)^3 = 1.0 \times 0.125 = 0.125 M.

[A]=[A]0e−kt=1.0⋅e−0.0231×90=e−2.079≈0.125 M ✓[A] = [A]_0 e^{-kt} = 1.0 \cdot e^{-0.0231 \times 90} = e^{-2.079} \approx 0.125 \text{ M}\ ✓

Why This Works

For first-order reactions, the half-life is independent of initial concentration — every half-life consumes another 50% of what’s there. So we can compute remaining concentration just by counting half-lives.

This makes first-order kinetics extremely clean: k=0.693/t1/2k = 0.693/t_{1/2} is the master link, and all other quantities follow from there.

Three speed reflexes for first-order kinetics:

  1. Half-life shortcut. When t/t1/2t/t_{1/2} is an integer nn, fraction remaining is (1/2)n(1/2)^n.
  2. Common percentages: 50% remaining = 1 half-life; 25% = 2 half-lives; 12.5% = 3; 6.25% = 4.
  3. Use natural logs. When the fraction isn’t a power of 1/2, use t=(1/k)ln⁡([A]0/[A])t = (1/k)\ln([A]_0/[A]).

Alternative Method

Direct application of the first-order integrated rate law: ln⁡([A]0/[A])=kt\ln([A]_0/[A]) = kt. For 75% consumed: ln⁡(1/0.25)=ln⁡4=1.386=(0.0231)t⇒t=60\ln(1/0.25) = \ln 4 = 1.386 = (0.0231) t \Rightarrow t = 60 min. Same answer.

Students apply the half-life formula k=0.693/t1/2k = 0.693/t_{1/2} to second-order or zeroth-order reactions. Wrong. Each order has its own formula:

  • Zeroth: t1/2=[A]0/2kt_{1/2} = [A]_0 / 2k
  • First: t1/2=0.693/kt_{1/2} = 0.693/k
  • Second: t1/2=1/(k[A]0)t_{1/2} = 1/(k[A]_0)

Always verify the order before using a half-life formula.

Final answer: k=0.0231k = 0.0231 min−1^{-1}; 75% consumed in 60 min; [A]=0.125[A] = 0.125 M after 90 min.

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