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Chapter 14: Problem 41

In three different experiments, the following results were obtained for the reaction \(A \longrightarrow\) products: \([\mathrm{A}]_{0}=1.00 \mathrm{M}, t_{1 / 2}=50 \mathrm{min} ;[\mathrm{A}]_{0}=200 \mathrm{M}, t_{1 / 2}=\) \(25 \min ;[\mathrm{A}]_{0}=0.50 \mathrm{M}, t_{1 / 2}=100 \mathrm{min} .\) Write the rate equation for this reaction, and indicate the value of \(k.\)

Short Answer

Expert verified
The rate equation for this reaction is \(rate = k [A]\), and the rate constant (\(k\)) is \(0.0139 min^{-1}\).

Step by step solution

01

Determine the Order of Reaction

The half-life of a first-order reaction is constant and it's independent of the initial concentration of the reactant. In our case, when the concentration of reactant \(A\) is doubled, the half-life halves, indicating a first-order reaction.
02

Write the Rate Equation

With the first-order reaction identified, the rate equation can be written as: \(rate = k [A]^{1}\). The exponent '1' indicates that the reaction is indeed first-order.
03

Calculate rate constant (\(k\))

The half-life of a first-order reaction is related to the rate constant through the formula: \(t_{1/2} = 0.693/k\). Rearranging for \(k\), gives: \(k = 0.693 / t_{1/2}\). Plugging in the value from one of the experiments (for example: \(t_{1 / 2}=50 \mathrm{min}\)): \(k = 0.693 / 50 min = 0.0139 min^{-1}\).

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Most popular questions from this chapter

We have seen that the unit of \(k\) depends on the overall order of a reaction. Derive a general expression for the units of \(k\) for a reaction of any overall order, based on the order of the reaction (o) and the units of concentration (M) and time (s).

A first-order reaction, \(\mathrm{A} \longrightarrow\) products, has a halflife of \(75 \mathrm{s},\) from which we can draw two conclusions. Which of the following are those two (a) the reaction goes to completion in 150 s; (b) the quantity of \(A\) remaining after 150 s is half of what remains after 75 s; (c) the same quantity of A is consumed for every 75 s of the reaction; (d) one- quarter of the original quantity of A is consumed in the first 37.5 s of the reaction; (e) twice as much A is consumed in 75 s when the initial amount of \(\mathrm{A}\) is doubled; (f) the amount of \(\mathrm{A}\) consumed in 150 s is twice as much as is consumed in 75 s.

You want to test the following proposed mechanism for the oxidation of HBr. $$\begin{array}{c} \mathrm{HBr}+\mathrm{O}_{2} \stackrel{k_{1}}{\longrightarrow} \mathrm{HOOBr} \\\ \mathrm{HOOBr}+\mathrm{HBr} \stackrel{k_{2}}{\longrightarrow} 2 \mathrm{HOBr} \\\ \mathrm{HOBr}+\mathrm{HBr} \stackrel{k_{3}}{\longrightarrow} \mathrm{H}_{2} \mathrm{O}+\mathrm{Br}_{2} \end{array}$$ You find that the rate is first order with respect to HBr and to \(\mathrm{O}_{2}\). You cannot detect HOBr among the products. (a) If the proposed mechanism is correct, which must be the rate-determining step? (b) Can you prove the mechanism from these observations? (c) Can you disprove the mechanism from these observations?

In the reaction \(2 \mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+3 \mathrm{D},\) reactant \(\mathrm{A}\) is found to disappear at the rate of \(6.2 \times 10^{-4} \mathrm{M} \mathrm{s}^{-1}.\) (a) What is the rate of reaction at this point? (b) What is the rate of disappearance of \(\mathrm{B}\) ? (c) What is the rate of formation of D?

Briefly describe each of the following ideas, phenomena, or methods: (a) the method of initial rates; (b) activated complex; (c) reaction mechanism; (d) heterogeneous catalysis; (e) rate-determining step.
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