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Chapter 14: Problem 19

The reaction \(A \longrightarrow\) products is first order in A. (a) If \(1.60 \mathrm{g} \mathrm{A}\) is allowed to decompose for 38 min, the mass of A remaining undecomposed is found to be 0.40 g. What is the half-life, \(t_{1 / 2}\), of this reaction? (b) Starting with \(1.60 \mathrm{g} \mathrm{A},\) what is the mass of \(\mathrm{A}\) remaining undecomposed after \(1.00 \mathrm{h} ?\)

Short Answer

Expert verified
The half-life of the reaction is 38 min, and the remaining mass of A after 1 hour is 0.20 g.

Step by step solution

01

Calculate the rate constant

In part (a), 1.60 g of A has decomposed to 0.40 g after 38 min, indicating that half of the reactant A has decomposed. The half-life of the reaction is therefore 38 min. Using the equation for the rate constant of a first order reaction, \(k = 0.693 / t_{1/2}\), the rate constant k is calculated by substituting \(t_{1/2}\) with 38 min: \(k = 0.693 / 38 = 0.018 min^{-1}\)
02

Determine the remaining amount of the reactant

In part (b), the remaining mass of A after 1.00 hour (or 60 min) is obtained using the integrated rate law. Rearrange the equation to solve for [A]_t: \(ln[A]_t = -kt + ln[A]_0\), where [A]_0 is the initial concentration (1.60 g), and [A]_t is the concentration at time t. Substituting the known values: \(ln[A]_t = -(0.018 min^{-1} * 60 min) + ln(1.60 g)\). The remaining concentration [A]_t is found by taking the exponential of both sides, since the natural logarithm is involved: \([A]_t = e^{ln[A]_t} = e^{-(0.018 min^{-1} * 60 min) + ln(1.60 g)} = 0.20 g\)
03

Summary

The half-life of the reaction is thus 38 min and the remaining mass of reactant A after 1.00 h is 0.20 g.

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Most popular questions from this chapter

The rate constant for the reaction \(\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \longrightarrow\) \(2 \mathrm{HI}(\mathrm{g})\) has been determined at the following temperatures: \(599 \mathrm{K}, k=5.4 \times 10^{-4} \mathrm{M}^{-1} \mathrm{s}^{-1} ; 683 \mathrm{K}, k=2.8 \times 10^{-2} \mathrm{M}^{-1} \mathrm{s}^{-1} .\) Calculate the activation energy for the reaction.

One example of a zero-order reaction is the decomposition of ammonia on a hot platinum wire, \(2 \mathrm{NH}_{3}(\mathrm{g}) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) .\) If the concentration of ammonia is doubled, the rate of the reaction will (a) be zero; (b) double; (c) remain the same; (d) exponentially increase.

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).

The following substrate concentration [S] versus time data were obtained during an enzyme-catalyzed reaction: \(t=0 \min ,[\mathrm{S}]=1.00 \mathrm{M} ; 20 \mathrm{min}, 0.90 \mathrm{M}; 60 \min , 0.70 \mathrm{M} ; 100 \mathrm{min}, 0.50 \mathrm{M} ; 160 \mathrm{min}, 0.20 \mathrm{M}.\) What is the order of this reaction with respect to \(\mathrm{S}\) in the concentration range studied?

For the reaction \(A \longrightarrow\) products, the following data were obtained: \(t=0 \mathrm{s},[\mathrm{A}]=0.715 \mathrm{M} ; 22 \mathrm{s}, 0.605 \mathrm{M}\) 74 s, 0.345 M; 132 s, 0.055 M. (a) What is the order of this reaction? (b) What is the half-life of the reaction?
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