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Chapter 14: Problem 108

The following expression shows the dependence of the half-life of a reaction \(\left(t_{\frac{1}{2}}\right)\) on the initial reactant concentration \([\mathrm{A}]_{0}:\) $$ t_{\frac{1}{2}} \propto \frac{1}{[\mathrm{~A}]_{0}^{n-1}} $$ where \(n\) is the order of the reaction. Verify this dependence for zero-, first-, and second-order reactions.

Short Answer

Expert verified
The given dependence of the half-life of a reaction on the initial reactant concentration holds true for zero-, first-, and second-order reactions.

Step by step solution

01

Understanding the Order of Reactions

Order of reaction could be zero, first or second order. It's defined by how the reaction rate changes with concentration. A zero-order reaction means the rate does not change with concentration, a first order means the rate changes proportionately with concentration, and for second order reaction the rate changes as the square of concentration.
02

Zero Order Reaction

For a zero order reaction (\(n=0\)), the rate is constant and independent of the concentration of reactants. Hence the half-life \(\left(t_{\frac{1}{2}}\right)\) should also be independent of the initial concentration. Substituting this in the given equation, \(t_{\frac{1}{2}} \propto \frac{1}{[\mathrm{A}_{0}]^{0-1}}\) yields \(t_{\frac{1}{2}} \propto [\mathrm{A}_{0}]\), which matches our expectation.
03

First Order Reaction

For a first order reaction (\(n=1\)), the rate depends on the concentration of reactants. Hence the half-life \(\left(t_{\frac{1}{2}}\right)\) should be independent of the initial concentration. Substituting this in the given equation, \(t_{\frac{1}{2}} \propto \frac{1}{[\mathrm{A}_{0}]^{1-1}}\) yields \(t_{\frac{1}{2}} \propto 1\), which matches our expectation.
04

Second Order Reaction

For a second order reaction (\(n=2\)), the rate depends on the square of the concentration of reactants. Hence the half-life \(\left(t_{\frac{1}{2}}\right)\) should be inversely proportional to the initial concentration. Substituting this in the equation, \(t_{\frac{1}{2}} \propto \frac{1}{[\mathrm{A}_{0}]^{2-1}}\) yields \(t_{\frac{1}{2}} \propto \frac{1}{[\mathrm{A}_{0}]} \), which matches our expectation.

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Most popular questions from this chapter

To carry out metabolism, oxygen is taken up by hemoglobin (Hb) to form oxyhemoglobin \(\left(\mathrm{HbO}_{2}\right)\) according to the simplified equation $$ \mathrm{Hb}(a q)+\mathrm{O}_{2}(a q) \stackrel{k}{\longrightarrow} \mathrm{HbO}_{2}(a q) $$ where the second-order rate constant is \(2.1 \times\) \(10^{6} / M \cdot \mathrm{s}\) at \(37^{\circ} \mathrm{C}\). (The reaction is first order in \(\mathrm{Hb}\) and \(\mathrm{O}_{2} .\) ) For an average adult, the concentrations of \(\mathrm{Hb}\) and \(\mathrm{O}_{2}\) in the blood at the lungs are \(8.0 \times 10^{-6} M\) and \(1.5 \times 10^{-6} M,\) respectively. (a) Calculate the rate of formation of \(\mathrm{HbO}_{2}\). (b) Calculate the rate of consumption of \(\mathrm{O}_{2}\). (c) The rate of formation of \(\mathrm{HbO}_{2}\) increases to \(1.4 \times 10^{-4} \mathrm{M} / \mathrm{s}\) during exercise to meet the demand of increased metabolism rate. Assuming the Hb concentration to remain the same, what must be the oxygen concentration to sustain this rate of \(\mathrm{HbO}_{2}\) formation?

What are the units of the rate of a reaction?

“The rate constant for the reaction $$ \mathrm{NO}_{2}(g)+\mathrm{CO}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{CO}_{2}(g) $$ is \(1.64 \times 10^{-6} / M \cdot \mathrm{s} .\)." What is incomplete about this statement?

Explain why most metals used in catalysis are transition metals.

As we know, methane burns readily in oxygen in a highly exothermic reaction. Yet a mixture of methane and oxygen gas can be kept indefinitely without any apparent change. Explain.
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