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Chapter 14: Problem 97

For a first order reaction \(\mathrm{A} \longrightarrow \mathrm{B}+\mathrm{C},\) if the half-life of \(\mathrm{A}\) at \(25^{\circ} \mathrm{C}\) is $3.05 \times 10^{4} \mathrm{~s},\( what is the rate constant \)k$ at this temperature? What percentage of A will not have reacted after one day?

Short Answer

Expert verified
The rate constant \(k\) at 25°C is \(2.27 \times 10^{-5} \text{ s}^{-1}\), and after one day, 46.7% of A will not have reacted.

Step by step solution

01

Calculate the Rate Constant k

Using the half-life formula for a first-order reaction, we can solve for the rate constant k: \(t_{1/2} = \frac{0.693}{k}\), Rearranging the equation to solve for k: \(k = \frac{0.693}{t_{1/2}}\), Now, substitute the given half-life value of \(3.05 \times 10^4\) s: \(k = \frac{0.693}{3.05 \times 10^4 \text{ s}}\). Calculating k: \(k = 2.27 \times 10^{-5} \text{ s}^{-1}\).
02

Calculate the Percentage of A Remaining after One Day

To find the remaining percentage of A after one day, we can use the integrated rate law for a first-order reaction: \[A_t = A_0 e^{-kt}\]. We are interested in finding the ratio between the concentration of A after one day to the initial concentration: \(\frac{A_t}{A_0} = e^{-kt}\). First, let's find the value of t in seconds after one day: 1 day = 24 hours × 60 minutes/hour × 60 seconds/minute = 86400 seconds. Now, substitute the value of k and t into the equation: \(\frac{A_t}{A_0} = e^{-(2.27 \times 10^{-5} \text{ s}^{-1})(86400 \text{ s})}\), Calculating the ratio: \(\frac{A_t}{A_0} = 0.467\).
03

Calculate the Remaining Percentage of A

To find the remaining percentage of A, multiply the ratio by 100: Remaining percentage = \(\frac{A_t}{A_0} \times 100\% = 0.467 \times 100\% = 46.7\%\). Therefore, 46.7% of A will not have reacted after one day.

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Most popular questions from this chapter

Ozone in the upper atmosphere can be destroyed by the following two-step mechanism: $$ \begin{aligned} \mathrm{Cl}(g)+\mathrm{O}_{3}(g) & \longrightarrow \mathrm{ClO}(g)+\mathrm{O}_{2}(g) \\ \mathrm{ClO}(g)+\mathrm{O}(g) & \longrightarrow \mathrm{Cl}(g)+\mathrm{O}_{2}(g) \end{aligned} $$ (a) What is the overall equation for this process? (b) What is the catalyst in the reaction? (c) What is the intermediate in the reaction?

The isomerization of methyl isonitrile $\left(\mathrm{CH}_{3} \mathrm{NC}\right)\( to acetonitrile \)\left(\mathrm{CH}_{3} \mathrm{CN}\right)$ was studied in the gas phase at \(215^{\circ} \mathrm{C}\), and the following data were obtained: $$ \begin{array}{rc} \hline \text { Time (s) } & {\left[\mathrm{CH}_{3} \mathrm{NC}\right](M)} \\ \hline 0 & 0.0165 \\ 2000 & 0.0110 \\ 5000 & 0.00591 \\ 8000 & 0.00314 \\ 12,000 & 0.00137 \\ 15,000 & 0.00074 \\ \hline \end{array} $$ (a) Calculate the average rate of reaction, in \(M / s\), for the time interval between each measurement. (b) Calculate the average rate of reaction over the entire time of the data from \(t=0\) to \(t=15,000 \mathrm{~s} .(\mathbf{c})\) Which is greater, the average rate between \(t=2000\) and $t=12,000 \mathrm{~s}\(, or between \)t=8000\( and \)t=15,000 \mathrm{~s} ?(\mathbf{d})$ Graph \(\left[\mathrm{CH}_{3} \mathrm{NC}\right]\) versus time and determine the instantaneous rates in \(M / \mathrm{s}\) at \(t=5000 \mathrm{~s}\) and $t=8000 \mathrm{~s}$.

As shown in Figure 14.23 , the first step in the heterogeneous hydrogenation of ethylene is adsorption of the ethylene molecule on a metal surface. One proposed explanation for the "sticking" of ethylene to a metal surface is the interaction of the electrons in the \(\mathrm{C}-\mathrm{C} \pi\) bond with vacant orbitals on the metal surface. (a) If this notion is correct, would ethane be expected to adsorb to a metal surface, and, if so, how strongly would ethane bind compared to ethylene? (b) Based on its Lewis structure, would you expect ammonia to adsorb to a metal surface using a similar explanation as for ethylene?

(a) The gas-phase decomposition of sulfuryl chloride $\left(\mathrm{SO}_{2} \mathrm{Cl}_{2}\right), \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)\( is first order in \)\mathrm{SO}_{2} \mathrm{Cl}_{2}\(. At \)300^{\circ} \mathrm{C}$ the half-life for this process is two and a half days. What is the rate constant at this temperature? (b) At \(400^{\circ} \mathrm{C}\) the rate constant is \(0.19 \mathrm{~min}^{-1}\). What is the half-life at this temperature?

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