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Chapter 14: Problem 20

In the first-order reaction \(A \longrightarrow\) products, \([\mathrm{A}]=0.816 \mathrm{M}\) initially and \(0.632 \mathrm{M}\) after \(16.0 \mathrm{min}.\) (a) What is the value of the rate constant, \(k ?\) (b) What is the half-life of this reaction? (c) At what time will \([\mathrm{A}]=0.235 \mathrm{M} ?\) (d) What will [A] be after 2.5 h?

Short Answer

Expert verified
The calculated rate constant \(k\) is \( \frac{1}{16} \cdot ln(\frac{0.816}{0.632}) \approx 0.0204 min^{-1}\). The half-life of the reaction is \( \frac{ln(2)}{k} \approx 34.0 min\). The time it takes for the reactant concentration to reach 0.235 M is \( \frac{1}{k} \cdot ln(\frac{0.816}{0.235}) \approx 52.5 min\). The concentration of A after 2.5 hours or 150 minutes is \(A_0 \cdot e^{-kt} = 0.816 M \cdot e^{-0.0204 min^{-1} \cdot 150 min} \approx 0.093 M\).

Step by step solution

01

Calculate Rate Constant (k)

The rate constant \(k\) describes how fast a reaction proceeds. It can be calculated using the formula \(k = \frac{1}{t} \cdot ln(\frac{A_0}{A(t)})\), where \(A_0\) is the initial amount, \(A(t)\) is the amount after time \(t\). Insert the given values \(A_0 = 0.816 M\), \(A(t) = 0.632 M\), and \(t = 16.0 min\) into the formula to calculate \(k\).
02

Calculate Half-Life (t_{1/2})

The half-life \(t_{1/2}\) is the time it takes for half of the substance to react. It can be calculated by the formula \(t_{1/2} = \frac{ln(2)}{k}\). Use the calculated value for \(k\) from Step 1 to calculate \(t_{1/2}\).
03

Calculate the Time to Reach 0.235 M

To find the time it takes for the reactant concentration to reach a certain value, the formula \(t = \frac{1}{k} \cdot ln(\frac{A_0}{A(t)})\) can be used. Insert the given values \(A_0 = 0.816 M\), \(A(t) = 0.235 M\), and the previously calculated \(k\) into the formula to calculate the time.
04

Calculate the Concentration After 2.5 Hours

The equation \(A(t) = A_0 \cdot e^{-kt}\) can be used to calculate the amount at a specific time. Convert 2.5 hours to minutes to match the units of the \(k\), and then use this time, along with \(A_0\) and the previously calculated \(k\), to find the concentration of A after 2.5 hours.

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Most popular questions from this chapter

One of the following statements is true and the other is false regarding the first-order reaction $2 \overrightarrow{\mathrm{A}} \longrightarrow \mathrm{B}+\mathrm{C} .$ Identify the true statement and the false one, and explain your reasoning. (a) A graph of [A] versus time is a straight line. (b) The rate of the reaction is one-half the rate of disappearance of A.

If the plot of the reactant concentration versus time is nonlinear, but the concentration drops by \(50 \%\) every 10 seconds, then the order of the reaction is (a) zero order; (b) first order; (c) second order; (d) third order.

The first-order reaction \(A \longrightarrow\) products has \(t_{1 / 2}=180 \mathrm{s}\) (a) What percent of a sample of A remains unreacted \(900 \mathrm{s}\) after a reaction has been started? (b) What is the rate of reaction when \([\mathrm{A}]=0.50 \mathrm{M} ?\)

In the reaction \(A(g) \longrightarrow 2 B(g)+C(g),\) the total pressure increases while the partial pressure of \(\mathrm{A}(\mathrm{g})\) decreases. If the initial pressure of \(\mathrm{A}(\mathrm{g})\) in a vessel of constant volume is \(1.000 \times 10^{3} \mathrm{mmHg}\) (a) What will be the total pressure when the reaction has gone to completion? (b) What will be the total gas pressure when the partial pressure of \(\mathrm{A}(\mathrm{g})\) has fallen to \(8.00 \times 10^{2} \mathrm{mmHg} ?\)

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, \(k ?\)
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