Chemical Kinetics: Common Mistakes and Fixes (5)

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Tags Chemical Kinetics

Question

A first-order reaction has a rate constant k=6.93×10−3k = 6.93 \times 10^{-3} s−1^{-1} at 25°C25°C. Find: (a) the half-life, (b) the time required for 90%90\% completion, (c) the fraction remaining after 55 minutes.

Solution — Step by Step

For first-order: t1/2=0.693kt_{1/2} = \tfrac{0.693}{k}.

t1/2=0.6936.93×10−3=100 st_{1/2} = \tfrac{0.693}{6.93 \times 10^{-3}} = 100 \text{ s}

90%90\% completion means [A]=0.1[A]0[A] = 0.1[A]_0. Use ln⁡([A]0/[A])=kt\ln([A]_0/[A]) = kt:

t=1kln⁡[A]0[A]=2.303klog⁡[A]0[A]t = \tfrac{1}{k}\ln\tfrac{[A]_0}{[A]} = \tfrac{2.303}{k}\log\tfrac{[A]_0}{[A]} t=2.3036.93×10−3log⁡(10)=2.3036.93×10−3≈332.3 st = \tfrac{2.303}{6.93 \times 10^{-3}}\log(10) = \tfrac{2.303}{6.93 \times 10^{-3}} \approx 332.3 \text{ s} [A][A]0=e−kt=e−6.93×10−3×300=e−2.079\tfrac{[A]}{[A]_0} = e^{-kt} = e^{-6.93 \times 10^{-3} \times 300} = e^{-2.079} e−2.079≈0.125=18e^{-2.079} \approx 0.125 = \tfrac{1}{8}

So 1/81/8 of the original concentration remains.

300300 s =3×t1/2= 3 \times t_{1/2}. So three half-lives have passed: 1→1/2→1/4→1/81 \to 1/2 \to 1/4 \to 1/8. Matches!

Final answers: t1/2=100 st_{1/2} = \mathbf{100 \text{ s}}, t90%≈332.3 st_{90\%} \approx \mathbf{332.3 \text{ s}}, fraction remaining =1/8= \mathbf{1/8}.

Why This Works

For first-order kinetics, half-life is independent of initial concentration. This is a defining feature — every half-life period leaves half of whatever was there. After nn half-lives, fraction remaining is 1/2n1/2^n.

The general formula ln⁡([A]0/[A])=kt\ln([A]_0/[A]) = kt comes from integrating −d[A]dt=k[A]-\tfrac{d[A]}{dt} = k[A].

Alternative Method

Using natural log directly: [A]=[A]0e−kt[A] = [A]_0 e^{-kt}. Same result, just one form vs. the other. JEE often gives log⁡10\log_{10} tables, so the 2.3032.303 form (using log⁡\log) is more practical in the exam.

Common Mistake

Computing time for 90%90\% completion as t90%=9×t1/2t_{90\%} = 9 \times t_{1/2}. The relationship t90%=3.32×t1/2t_{90\%} = 3.32 \times t_{1/2} comes from ln⁡(10)/ln⁡(2)≈3.32\ln(10)/\ln(2) \approx 3.32, not by simple multiplication. Always go through the formula.

Confusing rate (−d[A]/dt-d[A]/dt, units of mol/L/s) with rate constant (kk, units depend on order). For first-order, kk has units of s−1\text{s}^{-1} — independent of concentration units, a useful diagnostic.

Half-life questions appear in every NEET paper. Memorise: t1/2(1)=0.693/kt_{1/2}^{(1)} = 0.693/k, t1/2(2)=1/(k[A]0)t_{1/2}^{(2)} = 1/(k[A]_0), t1/2(0)=[A]0/(2k)t_{1/2}^{(0)} = [A]_0/(2k). The order is encoded in how t1/2t_{1/2} depends on [A]0[A]_0.

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