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Chapter 14: Problem 7

For the reaction \(A+2 B \longrightarrow 2 C\), the rate of reaction is \(1.76 \times 10^{-5} \mathrm{M} \mathrm{s}^{-1}\) at the time when \([\mathrm{A}]=0.3580 \mathrm{M}.\) (a) What is the rate of formation of \(\mathrm{C}\) ? (b) What will \([\mathrm{A}]\) be 1.00 min later? (c) Assume the rate remains at \(1.76 \times 10^{-5} \mathrm{M} \mathrm{s}^{-1}\) How long would it take for \([\mathrm{A}]\) to change from 0.3580 to \(0.3500 \mathrm{M} ?\)

Short Answer

Expert verified
The rate of formation of C is \(3.52 \times 10^{-5} \mathrm{M} \mathrm{s}^{-1}\). [A] is 0.357 M after 1.00 minute. It would take approximately 7.58 minutes for [A] to change from 0.3580 M to 0.3500 M.

Step by step solution

01

Calculation of the rate of formation of C

The stoichiometry of the reaction shows that 2 moles of C are formed for every mole of A that reacts. Thus, the rate of formation of C is twice the rate of reaction. This is calculated by multiplying the given rate of reaction (\(1.76 \times 10^{-5} \mathrm{M} \mathrm{s}^{-1}\)) by 2.
02

Calculation of [A] after 1.00 minute

The rate of reaction is the rate of decrease of [A], therefore the change in [A] after 1.00 minute can be calculated by multiplying the rate by the time in seconds (1.00 min = 60 s). This value should then be subtracted from the initial concentration [A] to get [A] after 1.00 minute.
03

Find the time for [A] to change from 0.3580 M to 0.3500 M

This time can be calculated by dividing the change in concentration (0.008 M) by the rate of the reaction. The result gives the time in seconds, which then should be converted into minutes by dividing by 60.

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Most popular questions from this chapter

Ammonia decomposes on the surface of a hot tungsten wire. Following are the half-lives that were obtained at \(1100^{\circ} \mathrm{C}\) for different initial concentrations of \(\mathrm{NH}_{3}:\left[\mathrm{NH}_{3}\right]_{0}=0.0031 \mathrm{M}, t_{1 / 2}=7.6 \mathrm{min} ; 0.0015 \mathrm{M}\) \(3.7 \mathrm{min} ; 0.00068 \mathrm{M}, 1.7 \mathrm{min.}\) For this decomposition reaction, what is (a) the order of the reaction; (b) the rate constant, \(k ?\)

A reaction is \(50 \%\) complete in 30.0 min. How long after its start will the reaction be \(75 \%\) complete if it is (a) first order; (b) zero order?

A kinetic study of the reaction \(A \longrightarrow\) products yields the data: \(t=0 \mathrm{s},[\mathrm{A}]=2.00 \mathrm{M} ; 500 \mathrm{s}, 1.00 \mathrm{M}; 1500 \mathrm{s}, 0.50 \mathrm{M} ; 3500 \mathrm{s}, 0.25 \mathrm{M} .\) Without performing detailed calculations, determine the order of this reaction and indicate your method of reasoning.

The reaction \(A+B \longrightarrow\) products is first order in \(A\) first order in \(\mathrm{B},\) and second order overall. Consider that the starting concentrations of the reactants are \([\mathrm{A}]_{0}\) and [ \(\mathrm{B}]_{0},\) and that \(x\) represents the decrease in these concentrations at the time \(t .\) That is, \([\mathrm{A}]_{t}=[\mathrm{A}]_{0}-x\) and \([\mathrm{B}]_{t}=[\mathrm{B}]_{0}-x .\) Show that the integrated rate law for this reaction can be expressed as shown below. $$\ln \frac{[\mathrm{A}]_{0} \times[\mathrm{B}]_{t}}{[\mathrm{B}]_{0} \times[\mathrm{A}]_{t}}=\left([\mathrm{B}]_{0}-[\mathrm{A}]_{0}\right) \times k t$$

The following rates of reaction were obtained in three experiments with the reaction \(2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NOCl}(\mathrm{g}).\) $$\begin{array}{llll} \hline & \text { Initial } & \text { Initial } & \text { Initial Rate of } \\ \text { Expt } & \text { [NO], M } & \text { [Cl }_{2} \text { ], M } & \text { Reaction, } \mathrm{M} \mathrm{s}^{-1} \\ \hline 1 & 0.0125 & 0.0255 & 2.27 \times 10^{-5} \\ 2 & 0.0125 & 0.0510 & 4.55 \times 10^{-5} \\ 3 & 0.0250 & 0.0255 & 9.08 \times 10^{-5} \\ \hline \end{array}$$ What is the rate law for this reaction?
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