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Chemical Kinetics - Half-life of a reaction

Grade 12ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The half-life of a reaction (t1/2t_{1/2}) is defined as the time required for the concentration of a reactant to decrease to exactly one-half of its initial value.

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For a zero-order reaction, the half-life is directly proportional to the initial concentration of the reactants [R]0[R]_0. This means as the initial concentration increases, the half-life also increases.

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For a first-order reaction, the half-life is constant and completely independent of the initial concentration. It depends only on the rate constant kk.

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In general, for a reaction of the nthn^{th} order, the half-life is related to the initial concentration by the proportionality: t1/2∝1[R]0n−1t_{1/2} \propto \frac{1}{[R]_0^{n-1}}.

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Radioactive decay is a natural example of a first-order reaction, where the half-life is used to determine the age of samples (Carbon dating).

📐Formulae

t1/2=[R]02k(For Zero-Order Reaction)t_{1/2} = \frac{[R]_0}{2k} \quad \text{(For Zero-Order Reaction)}

t1/2=0.693k(For First-Order Reaction)t_{1/2} = \frac{0.693}{k} \quad \text{(For First-Order Reaction)}

[R]t=[R]02nwhere n=tt1/2[R]_t = \frac{[R]_0}{2^n} \quad \text{where } n = \frac{t}{t_{1/2}}

k=2.303tlog⁡[R]0[R]t(First-Order Rate Equation)k = \frac{2.303}{t} \log \frac{[R]_0}{[R]_t} \quad \text{(First-Order Rate Equation)}

💡Examples

Problem 1:

A first-order reaction is found to have a rate constant k=5.5×10−14 s−1k = 5.5 \times 10^{-14} \text{ s}^{-1}. Find the half-life of the reaction.

Solution:

Given k=5.5×10−14 s−1k = 5.5 \times 10^{-14} \text{ s}^{-1}. For a first-order reaction, t1/2=0.693kt_{1/2} = \frac{0.693}{k}. Substituting the value: t1/2=0.6935.5×10−14≈1.26×1013 st_{1/2} = \frac{0.693}{5.5 \times 10^{-14}} \approx 1.26 \times 10^{13} \text{ s}.

Explanation:

Since the reaction is first-order, we use the standard half-life formula which is independent of the initial concentration of the reactant.

Problem 2:

Show that for a first-order reaction, the time required for 99.9%99.9\% completion is about 1010 times its half-life (t1/2t_{1/2}).

Solution:

For 99.9%99.9\% completion, [R]t=[R]0−0.999[R]0=0.001[R]0[R]_t = [R]_0 - 0.999[R]_0 = 0.001[R]_0. Using t=2.303klog⁡[R]0[R]tt = \frac{2.303}{k} \log \frac{[R]_0}{[R]_t}: t99.9%=2.303klog⁡[R]00.001[R]0=2.303klog⁡(103)=2.303×3k=6.909kt_{99.9\%} = \frac{2.303}{k} \log \frac{[R]_0}{0.001[R]_0} = \frac{2.303}{k} \log(10^3) = \frac{2.303 \times 3}{k} = \frac{6.909}{k}. Since t1/2=0.693kt_{1/2} = \frac{0.693}{k}, then t99.9%t1/2=6.909/k0.693/k≈10\frac{t_{99.9\%}}{t_{1/2}} = \frac{6.909/k}{0.693/k} \approx 10.

Explanation:

By comparing the integrated rate law for 99.9%99.9\% completion with the half-life formula, we derive a mathematical ratio showing that the time required is tenfold.

Half-life of a reaction - Revision Notes & Key Formulas | ICSE Class 12 Chemistry