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Chapter 14: Problem 18

The reaction \(A \longrightarrow\) products is first order in A. Initially, \([\mathrm{A}]=0.800 \mathrm{M}\) and after 54min, \([\mathrm{A}]=0.100 \mathrm{M}.\) (a) At what time is \([\mathrm{A}]=0.025 \mathrm{M} ?\) (b) What is the rate of reaction when \([\mathrm{A}]=0.025 \mathrm{M} ?\)

Short Answer

Expert verified
The concentration \([A] = 0.025M\) occurs at the calculated time from Step 2. The rate of reaction when \([A] = 0.025M\) is obtained from Step 3.

Step by step solution

01

Use the Integrated First-Order Rate Law

We will use the integrated rate law in the form: \(\log[A]_t = \log[A]_0 - kt\), where \([A]_t\) and \([A]_0\) are the final and initial concentrations, \(k\) is the rate constant, and \(t\) is the time. Plugging in the given values: \(\log(0.100) = \log(0.800) - k(54min)\). Rearrange this equation and solve for the rate constant \(k\).
02

Apply Found Rate Constant to Solve for Time

Substitute the value of \(k\) from step 1 and the given concentration of 0.025M into the rate law and rearrange for \(t\). Thus we obtain: \(t = (\log(0.800) - \log(0.025)) / k\). Calculate the number of minutes.
03

Determine the Rate of Reaction

The rate of a first-order reaction is defined by the Law of Mass Action as \(Rate = k[A]\). Substitute the given concentration of 0.025M and rate constant \(k\) into the equation to obtain the rate of reaction.

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Most popular questions from this chapter

The reaction \(2 \mathrm{NO}+\mathrm{Cl}_{2} \longrightarrow 2 \mathrm{NOCl}\) has the rate law: rate of reaction \(=k[\mathrm{NO}]^{2}\left[\mathrm{Cl}_{2}\right] .\) Propose a twostep mechanism for this reaction consisting of a fast reversible first step, followed by a slow step.

The decomposition of \(\mathrm{HI}(\mathrm{g})\) at \(700 \mathrm{K}\) is followed for \(400 \mathrm{s},\) yielding the following data: at \(t=0,[\mathrm{HI}]=\) \(1.00 \mathrm{M} ;\) at \(t=100 \mathrm{s},[\mathrm{HI}]=0.90 \mathrm{M} ;\) at \(t=200 \mathrm{s}, [\mathrm{HI}]=0.81 \mathrm{M} ; t=300 \mathrm{s},[\mathrm{HI}]=0.74 \mathrm{M} ;\) at \(t=400 \mathrm{s}, [\mathrm{HI}]=0.68 \mathrm{M} .\) What are the reaction order and the rate constant for the reaction: $$\mathrm{HI}(\mathrm{g}) \longrightarrow \frac{1}{2} \mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{I}_{2}(\mathrm{g}) ?$$ Write the rate law for the reaction at 700 K.

The following first-order reaction occurs in \(\mathrm{CCl}_{4}(1)\) at \(45^{\circ} \mathrm{C}: \mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) .\) The rate constant is \(k=6.2 \times 10^{-4} \mathrm{s}^{-1} .\) An \(80.0 \mathrm{g}\) sample of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in \(\mathrm{CCl}_{4}(\mathrm{l})\) is allowed to decompose at \(45^{\circ} \mathrm{C}.\) (a) How long does it take for the quantity of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to be reduced to \(2.5 \mathrm{g} ?\) (b) How many liters of \(\mathrm{O}_{2},\) measured at \(745 \mathrm{mmHg}\) and \(45^{\circ} \mathrm{C},\) are produced up to this point?

In your own words, define or explain the following terms or symbols: (a) \([\mathrm{A}]_{0} ;\) (b) \(\dot{k} ;\) (c) \(t_{1 / 2} ;\) (d) zeroorder reaction; (e) catalyst.

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?
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