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Chapter 14: Problem 5

In the reaction \(A \longrightarrow\) products, 4.40 min after the reac- tion is started, \([\mathrm{A}]=0.588 \mathrm{M}\). The rate of reaction at this point is rate \(=-\Delta[\mathrm{A}] / \Delta t=2.2 \times 10^{-2} \mathrm{M} \mathrm{min}^{-1}.\) Assume that this rate remains constant for a short period of time. (a) What is \([\mathrm{A}] 5.00\) min after the reaction is started? (b) At what time after the reaction is started will \([\mathrm{A}]=0.565 \mathrm{M} ?\)

Short Answer

Expert verified
(a) The concentration of A 5.00 min after the reaction started is approximately 0.575 M. (b) The concentration of A will be 0.565 M approximately 5.45 min after the reaction has started.

Step by step solution

01

Using Given Reaction Rate

The given rate of reaction is \(-\Delta[\mathrm{A}] / \Delta t = 2.2 \times 10^{-2} \mathrm{M} \mathrm{min}^{-1}\). Assuming this rate to be constant, it can be applied to find out the changes in concentration over given periods of time.
02

(a) Find Concentration of A After 5 min

The time difference between 5.00 min and 4.40 min (when the concentration was last known) is 0.60 min. As the rate is given in M min^-1, we multiply this time period by the rate: \( \Delta[\mathrm{A}] = (2.2 \times 10^{-2} \mathrm{M} \mathrm{min}^{-1})(0.60 \mathrm{min}) = 0.0132 \mathrm{M}\). As this is the amount of A used up (hence negative), the concentration after 5.00 min will be \(0.588 \mathrm{M} - 0.0132 \mathrm{M}\).
03

(b) Find Time When Concentration of A is 0.565 M

Now we need to find out how much time it takes for the concentration to decrease from 0.588 M to 0.565 M. The difference in concentration is \( \Delta[\mathrm{A}] = 0.588 \mathrm{M} - 0.565 \mathrm{M} = 0.023 \mathrm{M}\). The time this takes at the given rate can be found with: \( \Delta t = \Delta[\mathrm{A}] / rate = 0.023 \mathrm{M} / 2.2 \times 10^{-2} \mathrm{M} \mathrm{min}^{-1}\). Add this time duration to the initial 4.40 min to find the time after the reaction begins.

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Most popular questions from this chapter

For the reaction \(A \longrightarrow\) products, the following data give \([\mathrm{A}]\) as a function of time \(t=0 \mathrm{s},[\mathrm{A}]=0.600 \mathrm{M};100 \mathrm{s}, 0.497 \mathrm{M} ; 200 \mathrm{s}, 0.413 \mathrm{M} ; 300 \mathrm{s}, 0.344 \mathrm{M} ; 400 \mathrm{s}\) \(0.285 \mathrm{M} ; 600 \mathrm{s}, 0.198 \mathrm{M} ; 1000 \mathrm{s}, 0.094 \mathrm{M}.\) (a) Show that the reaction is first order. (b) What is the value of the rate constant, \(k ?\) (c) What is \([\mathrm{A}]\) at \(t=750 \mathrm{s} ?\)

In the first-order reaction \(A \longrightarrow\) products, it is found that \(99 \%\) of the original amount of reactant \(A\) decomposes in 137 min. What is the half-life, \(t_{1 / 2}\), of this decomposition reaction?

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?

Derive a plausible mechanism for the following reaction in aqueous solution, \(\mathrm{Hg}_{2}^{2+}+\mathrm{Tl}^{3+} \longrightarrow 2 \mathrm{Hg}^{2+}+\mathrm{Tl}^{+}\) for which the observed rate law is: rate \(=k\left[\mathrm{Hg}_{2}^{2+1}\right]\) \(\left.[\mathrm{T}]^{3+}\right] /\left[\mathrm{Hg}^{2+}\right].\)

In the reaction \(A \longrightarrow\) products, at \(t=0\), the \([\mathrm{A}]=0.1565 \mathrm{M} .\) After \(1.00 \mathrm{min},[\mathrm{A}]=0.1498 \mathrm{M},\) and after \(2.00 \mathrm{min},[\mathrm{A}]=0.1433 \mathrm{M}\) (a) Calculate the average rate of the reaction during the first minute and during the second minute. (b) Why are these two rates not equal?
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