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Chemical Kinetics - Half-Life of a Reaction

Grade 12CBSEChemistry

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 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, t1/2t_{1/2} is directly proportional to the initial concentration of reactants [R]0[R]_0. The expression is t1/2=[R]02kt_{1/2} = \frac{[R]_0}{2k}.

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For a first-order reaction, t1/2t_{1/2} is independent of the initial concentration. It is constant and depends only on the rate constant kk. The expression is t1/2=0.693kt_{1/2} = \frac{0.693}{k}.

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General relationship: For a reaction of nthn^{th} order, t1/2∝1[R]0n−1t_{1/2} \propto \frac{1}{[R]_0^{n-1}}.

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In a first-order reaction, the time required to complete a certain fraction of the reaction is independent of the initial concentration.

📐Formulae

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

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

k=2.303tlog⁡[R]0[R] (Integrated Rate Law for First Order)k = \frac{2.303}{t} \log \frac{[R]_0}{[R]} \text{ (Integrated Rate Law for First Order)}

t1/2∝[R]01−n (General proportionality for order n)t_{1/2} \propto [R]_0^{1-n} \text{ (General proportionality for order } n \text{)}

💡Examples

Problem 1:

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

Solution:

t1/2=0.693k=0.6935.5×10−14 s−1=1.26×1013 st_{1/2} = \frac{0.693}{k} = \frac{0.693}{5.5 \times 10^{-14} \text{ s}^{-1}} = 1.26 \times 10^{13} \text{ s}

Explanation:

Since the reaction is first-order, we use the formula t1/2=0.693kt_{1/2} = \frac{0.693}{k}. Substituting the given value of kk yields the half-life in seconds.

Problem 2:

Show that for a first-order reaction, the time required for 99.9%99.9\% completion is roughly 1010 times the half-life of the reaction.

Solution:

For 99.9%99.9\% completion, the remaining concentration [R][R] is: 100.0−99.90.1\begin{array}{r} 100.0 \\ - 99.9 \\ \hline 0.1 \end{array} So, [R]=0.1%[R] = 0.1\% of [R]0=0.001[R]0[R]_0 = 0.001 [R]_0. Using t=2.303klog⁡[R]0[R]t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}: 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} Comparing with t1/2=0.693kt_{1/2} = \frac{0.693}{k}: 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:

We calculate the time for 99.9%99.9\% completion using the integrated rate law for first-order reactions and divide it by the expression for half-life to find the ratio.