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Chapter 11: Problem 39

The decomposition of dimethyl ether \(\left(\mathrm{CH}_{3} \mathrm{OCH}_{3}\right)\) to methane, carbon monoxide, and hydrogen gases is found to be first-order. At \(500^{\circ} \mathrm{C}\), a \(150.0\) -mg 35 sample of dimethyl ether is reduced to \(43.2 \mathrm{mg}\) after three quarters of an hour. Calculate (a) the rate constant. (b) the half-life at \(500^{\circ} \mathrm{C}\). (c) how long it will take to decompose \(95 \%\) of the dimethyl ether.

Short Answer

Expert verified
Question: Determine the rate constant, half-life, and the time it takes to decompose 95% of dimethyl ether (CH3OCH3) at 500掳C, given that the initial mass is 150 mg and the final mass after 0.75 hours is 43.2 mg. Answer: The rate constant (k) is 2.78 h鈦宦�, the half-life is 0.249 hours, and the time to decompose 95% of dimethyl ether is 1.16 hours.

Step by step solution

01

Determine the initial and final concentrations

First, convert the given masses of dimethyl ether to moles using its molar mass (46.07 g/mol). Then, determine the initial and final concentrations. Initial mass = 150.0 mg = 150.0 g / 1000 = 0.150 g Final mass = 43.2 mg = 43.2 g / 1000 = 0.0432 g Molar mass of CH3OCH3 = 46.07 g/mol Initial moles = 0.150 g / 46.07 g/mol = 0.00325 mol Final moles = 0.0432 g / 46.07 g/mol = 0.000937 mol Now, assume a volume of 1 L for calculation purposes. Therefore, we have: Initial concentration, [A]鈧� = 0.00325 mol/L Final concentration, [A] = 0.000937 mol/L
02

Calculate the rate constant (k)

Using the first-order rate law and the given reaction time (0.75 h), we can calculate the rate constant (k). \(ln\frac{[A]}{[A]_0} = -kt\) \(ln\frac{0.000937}{0.00325} = -k(0.75 h)\) Now, solve for k: \(k = -\frac{ln\frac{0.000937}{0.00325}}{0.75\;h} = 2.78\;h^{-1}\) The rate constant, k, is 2.78 h鈦宦�.
03

Calculate the half-life

We can determine the half-life of the reaction using the following formula: \(t_{1/2} = \frac{ln 2}{k}\) \(t_{1/2} = \frac{ln 2}{2.78 \;h^{-1}} = 0.249 \;h\) The half-life at 500掳C is 0.249 h.
04

Calculate the time to decompose 95% dimethyl ether

To find the time it takes for 95% of dimethyl ether to decompose, we need to determine the concentration when only 5% is left. 95% decomposed means 5% remaining: Remaining concentration, [A] = 0.05 x [A]鈧� = 0.05 x 0.00325 mol/L = 0.000163 mol/L Applying the first-order rate law again: \(ln\frac{[A]}{[A]_0} = -kt\) \(ln\frac{0.000163}{0.00325} = -2.78\;h^{-1}t\) Finally, solve for t: \(t = -\frac{ln\frac{0.000163}{0.00325}}{2.78\;h^{-1}} = 1.16\;h\) It will take 1.16 hours to decompose 95% of the dimethyl ether.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First-Order Reactions
Understanding "first-order" reactions is key to mastering the concept of how dimethyl ether decomposes. In a first-order reaction, the rate at which a substance is consumed is directly proportional to its current concentration. This means the reaction rate depends on the amount of one reactant only. In mathematical terms, this relationship is expressed as:
\[\text{Rate} = k [A]\]where:
  • \(k\) is the rate constant.
  • \([A]\) is the concentration of reactant A at any time.

Dimethyl ether decomposes in this manner, which simplifies the calculations because the change in concentration over time is straightforward to relate to the decomposition process.
These equations help predict how quickly the decomposition occurs under specific conditions, such as temperature.
Rate Constant
The rate constant \(k\) is an essential part of chemical kinetics, playing a pivotal role in first-order reactions like the decomposition of dimethyl ether. It provides information on how fast a reaction progresses. The larger the rate constant, the quicker the reaction.
For first-order reactions, the rate constant can be derived from the equation:
\[ln\frac{[A]}{[A]_0} = -kt\]where:
  • \([A]\) is the final concentration.
  • \([A]_0\) is the initial concentration.
  • \(t\) is the time elapsed.

Solving this for the dimethyl ether decomposition, using initial and final amounts and the time provided (0.75 hours), we calculated \(k\) to be 2.78 h-1. This means for each hour, the reaction proceeds at this rate, decreasing the concentration significantly.
Half-Life
The half-life of a reaction is the time it takes for half of the reactant to be consumed. This concept is particularly useful for first-order reactions, as the half-life remains constant regardless of initial concentration.
The formula for half-life \(t_{1/2}\) in a first-order reaction is:
\[t_{1/2} = \frac{ln 2}{k}\]
Using this relation for the dimethyl ether, and knowing \(k\) is 2.78 h-1, the half-life at 500掳C is calculated to be approximately 0.249 hours. It means dimethyl ether reduces to half its original amount in about 15 minutes. Understanding this helps us predict the behavior of the reaction over time.
Dimethyl Ether Decomposition
Dimethyl ether (\(\text{CH}_3\text{OCH}_3\)) decomposition involves breaking down the molecule into methane, carbon monoxide, and hydrogen gases. At 500掳C, this reaction is effectively modeled as a first-order process.
In practical terms, knowing the rate constant and half-life allows us to predict how long it will take for a given percentage of ether to decompose. For instance, to find when 95% of dimethyl ether decomposes, we calculate the time required for the concentration to fall to 5% of its initial value.
Using the rate constant, we've calculated this time to be about 1.16 hours. Thus, you can estimate how long it would take to nearly complete the decomposition under given temperature conditions. Such calculations are critical in diverse fields, from industrial processes to environmental science, illustrating how deeply understanding first-order reactions helps us manipulate and optimize chemical reactions.

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Most popular questions from this chapter

The decomposition of \(\mathrm{A}\) at \(85^{\circ} \mathrm{C}\) is a zero-order reaction. It takes 35 minutes to decompose \(37 \%\) of an initial mass of \(282 \mathrm{mg}\). (a) What is \(k\) at \(85^{\circ} \mathrm{C}\) ? (b) What is the half-life of \(282 \mathrm{mg}\) at \(85^{\circ} \mathrm{C}\) ? (c) What is the rate of decomposition for \(282 \mathrm{mg}\) at \(85^{\circ} \mathrm{C} ?\) (d) If one starts with \(464 \mathrm{mg}\), what is the rate of its decomposition at \(85^{\circ} \mathrm{C} ?\)

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Nitrosyl chloride (NOCl) decomposes to nitrogen oxide and chlorine gases. (a) Write a balanced equation using smallest whole-number coefficients for the decomposition. (b) Write an expression for the reaction rate in terms of \(\Delta[\mathrm{NOCl}] .\) (c) The concentration of NOCl drops from \(0.580 M\) to \(0.238 M\) in \(8.00 \mathrm{~min}\). Calculate the average rate of reaction over this time interval.

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The equation for the iodination of acetone in acidic solution is $$\mathrm{CH}_{3} \mathrm{COCH}_{3}(a q)+\mathrm{I}_{2}(a q) \longrightarrow \mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{I}(a q)+\mathrm{H}^{+}(a q)+\mathrm{I}^{-}(a q)$$ The rate of the reaction is found to be dependent not only on the concentration of the reactants but also on the hydrogen ion concentration. Hence the rate expression of this reaction is $$\text { rate }=k\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]^{m}\left[\mathrm{I}_{2}\right]^{n}\left[\mathrm{H}^{+}\right]^{p}$$ The rate is obtained by following the disappearance of iodine using starch as an indicator. The following data are obtained: $$ \begin{array}{cccc} \hline\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right] & \left.\mathrm{[H}^{+}\right] & {\left[\mathrm{I}_{2}\right]} & \text { Initial Rate }(\mathrm{mol} / \mathrm{L} \cdot \mathrm{s}) \\ \hline 0.80 & 0.20 & 0.001 & 4.2 \times 10^{-6} \\ 1.6 & 0.20 & 0.001 & 8.2 \times 10^{-6} \\ 0.80 & 0.40 & 0.001 & 8.7 \times 10^{-6} \\ 0.80 & 0.20 & 0.0005 & 4.3 \times 10^{-6} \\ \hline\end{array}$$ (a) What is the order of the reaction with respect to each reactant? (b) Write the rate expression for the reaction. (c) Calculate \(k\). (d) What is the rate of the reaction when \(\left[\mathrm{H}^{+}\right]=0.933 M\) and \(\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]=3\left[\mathrm{H}^{+}\right]=10\left[\mathrm{I}^{-}\right] ?\)
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