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Chapter 12: Problem 57

For the reaction \(\mathrm{A} \rightarrow\) products, successive half-lives are observed to be \(10.0,20.0,\) and 40.0 \(\mathrm{min}\) for an experiment in which \([\mathrm{A}]_{0}=0.10 M .\) Calculate the concentration of \(\mathrm{A}\) at the following times. a. 80.0 \(\mathrm{min}\) b. 30.0 \(\mathrm{min}\)

Short Answer

Expert verified
a. At 80.0 minutes: \([\mathrm{A}] \approx 3.93 \times 10^{-3} M\) b. At 30.0 minutes: \([\mathrm{A}] \approx 0.0126\, M\)

Step by step solution

01

Write the first-order integrated rate law

In a first-order reaction, the integrated rate law is given by: \[ln\frac{[\mathrm{A}]_0}{[\mathrm{A}]} = kt\] Where: - \( [\mathrm{A}] \) is the concentration of A at time t, - \( [\mathrm{A}]_0 \) is the initial concentration of A, - k is the rate constant, and - t is the time.
02

Determine the rate constant (k) using first half-life

We know that the half-life of a first-order reaction is given by: \[t_{1/2} = \frac{0.693}{k}\] So, we can solve for k: \[k = \frac{0.693}{t_{1/2}}\] Given the first half-life, \(t_{1/2} = 10.0\, minutes\), we can plug this value to calculate the rate constant k: \[k = \frac{0.693}{10.0\,minutes} = 0.0693\,min^{-1}\]
03

Calculate the concentration of A at 80.0 minutes

Now that we have the rate constant (k), we can use the integrated rate law to find the concentration of A at 80.0 minutes. We know \( [\mathrm{A}]_0 = 0.10\, M\), \(k=0.0693\,min^{-1}\), and \(t=80.0\, min\). Plugging these values into the integrated rate law equation: \[ln\frac{0.10\, M}{[\mathrm{A}]} = (0.0693\,min^{-1})(80.0\,min)\] Solving for [\mathrm{A}]: \[ [\mathrm{A}] = 0.10\, M \times e^{-(0.0693\,min^{-1})(80.0\,min)}\] \[ [\mathrm{A}] = 0.10\, M \times e^{-5.544}\] \[ [\mathrm{A}] \approx 3.93 \times 10^{-3} M\]
04

Calculate the concentration of A at 30.0 minutes

Similarly, we can use the integrated rate law to find the concentration of A at 30.0 minutes. We know \( [\mathrm{A}]_0 = 0.10\, M\), \(k=0.0693\,min^{-1}\), and \(t=30.0\, min\). Plugging these values into the integrated rate law equation: \[ln\frac{0.10\, M}{[\mathrm{A}]} = (0.0693\,min^{-1})(30.0\,min)\] Solving for [\mathrm{A}]: \[ [\mathrm{A}] = 0.10\, M \times e^{-(0.0693\,min^{-1})(30.0\,min)}\] \[ [\mathrm{A}] = 0.10\, M \times e^{-2.079}\] \[ [\mathrm{A}] \approx 0.0126\, M\] Now we have the concentration of A at the given times: a. At 80.0 minutes: \([\mathrm{A}] \approx 3.93 \times 10^{-3} M\) b. At 30.0 minutes: \([\mathrm{A}] \approx 0.0126\, M\)

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Most popular questions from this chapter

Consider the reaction $$ 3 \mathrm{A}+\mathrm{B}+\mathrm{C} \longrightarrow \mathrm{D}+\mathrm{E} $$ where the rate law is defined as $$ -\frac{\Delta[\mathrm{A}]}{\Delta t}=k[\mathrm{A}]^{2}[\mathrm{B}][\mathrm{C}] $$ An experiment is carried out where $[\mathrm{B}]_{0}=[\mathrm{C}]_{0}=1.00 \mathrm{M}$ and \([\mathrm{A}]_{0}=1.00 \times 10^{-4} \mathrm{M}\) a. If after \(3.00 \min ,[\mathrm{A}]=3.26 \times 10^{-5} M,\) calculate the value of \(k .\) b. Calculate the half-life for this experiment. c. Calculate the concentration of \(B\) and the concentration of A after 10.0 min.

The decomposition of \(\mathrm{NO}_{2}(g)\) occurs by the following bimolecular elementary reaction: $$ 2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ The rate constant at 273 \(\mathrm{K}\) is $2.3 \times 10^{-12} \mathrm{L} / \mathrm{mol} \cdot \mathrm{s}\( , and the activation energy is 111 \)\mathrm{kJ} / \mathrm{mol}$ . How long will it take for the concentration of \(\mathrm{NO}_{2}(g)\) to decrease from an initial partial pressure of 2.5 \(\mathrm{atm}\) to 1.5 \(\mathrm{atm}\) at \(500 . \mathrm{K}\) ? Assume ideal gas behavior.

For the reaction $$ 2 \mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) $$ the following data were collected, where $$ \text {Rate} =-\frac{\Delta\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]}{\Delta t} $$ Calculate \(E_{\mathrm{a}}\) for this reaction.

A first-order reaction has rate constants of $4.6 \times 10^{-2} \mathrm{s}^{-1}\( and \)8.1 \times 10^{-2} \mathrm{s}^{-1}\( at \)0^{\circ} \mathrm{C}\( and \)20 .^{\circ} \mathrm{C},$ respectively. What is the value of the activation energy?

Consider two reaction vessels, one containing \(\mathrm{A}\) and the other containing \(\mathrm{B}\) , with equal concentrations at \(t=0 .\) If both substances decompose by first-order kinetics, where $$ \begin{array}{l}{k_{\mathrm{A}}=4.50 \times 10^{-4} \mathrm{s}^{-1}} \\\ {k_{\mathrm{B}}=3.70 \times 10^{-3} \mathrm{s}^{-1}}\end{array} $$ how much time must pass to reach a condition such that \([\mathrm{A}]=\) 4.00\([\mathrm{B}]\) ?
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