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Chapter 14: Problem 67

The decomposition of gaseous dimethyl ether at ordinary pressures is first- order. Its half-life is 25.0 minutes at \(500^{\circ} \mathrm{C}\) $$ \mathrm{CH}_{3} \mathrm{OCH}_{3}(\mathrm{g}) \rightarrow \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) $$ (a) Starting with 8.00 g of dimethyl ether, what mass remains (in grams) after 125 minutes and after 145 minutes? (b) Calculate the time in minutes required to decrease 7.60 ng (nanograms) to 2.25 ng. (c) What fraction of the original dimethyl ether remains after 150 minutes?

Short Answer

Expert verified
(a) 0.250 g at 125 min, 0.144 g at 145 min. (b) 47.89 min. (c) 0.0156.

Step by step solution

01

Understanding First-Order Reactions

For a first-order reaction, the rate depends on the concentration of the single reactant. The equation is given by:\[[A]_t = [A]_0 e^{-kt}\]where \([A]_t\) is the concentration at time \(t\), \([A]_0\) is the initial concentration, and \(k\) is the rate constant.
02

Calculate Rate Constant (k)

The half-life \(t_{1/2}\) for a first-order reaction is related to the rate constant by:\[t_{1/2} = \frac{0.693}{k}\]Given that the half-life \(t_{1/2} = 25.0\) minutes, we solve for \(k\):\[k = \frac{0.693}{25.0} = 0.02772 \, \text{min}^{-1}\]
03

Determine Mass Remaining After 125 Minutes

Starting mass is 8.00 g. The fraction remaining after a time \(t\) is given by:\[\text{Remaining fraction} = e^{-kt}\]For 125 minutes:\[\text{Remaining fraction} = e^{-0.02772 \times 125} = e^{-3.465} \approx 0.0312\]Thus, the remaining mass is:\[0.0312 \times 8.00 = 0.2496 \, \text{g}\]
04

Determine Mass Remaining After 145 Minutes

For 145 minutes:\[\text{Remaining fraction} = e^{-0.02772 \times 145} = e^{-4.019} \approx 0.0180\]Thus, the remaining mass is:\[0.0180 \times 8.00 = 0.144 \, \text{g}\]
05

Calculate Time to Reduce 7.60 ng to 2.25 ng

Use the equation \([A]_t = [A]_0 e^{-kt}\), solving for \(t\):\[\frac{2.25}{7.60} = e^{-0.02772t}\]\[de^{-0.02772t} = 0.2961\]Taking natural logarithms:\[-0.02772t = \ln(0.2961)\]Solving for \(t\):\[t = \frac{\ln(0.2961)}{-0.02772} = 47.89 \, \text{minutes}\]
06

Fraction of Original Ether Remaining After 150 Minutes

The remaining fraction after 150 minutes can be calculated as:\[\text{Remaining fraction} = e^{-0.02772 \times 150} = e^{-4.158} \approx 0.0156\]

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Key Concepts

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

Understanding Half-Life in First-Order Reactions
Half-life is an important concept when studying reactions, especially those of first-order. It refers to the time required for a concentration of a reactant to decrease by half. This is a key feature of first-order reactions because the half-life is constant regardless of the starting concentration. It's determined by the formula \( t_{1/2} = \frac{0.693}{k} \), where \( k \) is the rate constant.
Let's say we know the half-life is 25 minutes at a certain temperature. This means that every 25 minutes, the concentration of the reactant will be halved. Even if you start with a large or small amount, as long as the reaction is first-order, every half-life period will reduce the concentration by 50%. This predictable pattern helps in calculating how much of a reactant is left at any time during the reaction.
Rate Constant: The Speed Governor of Reactions
The rate constant \( k \) is a crucial factor in determining the speed of a first-order reaction. It doesn't depend on the concentration but rather on the conditions like temperature. You can find \( k \) by rearranging the half-life formula: \( k = \frac{0.693}{t_{1/2}} \).
In our example, with a half-life of 25 minutes, the rate constant \( k \) is calculated as 0.02772 \( \, \text{min}^{-1} \). The value of \( k \) provides insight into how fast or slow a reaction happens. A higher \( k \) means a faster reaction, while a lower \( k \) indicates a slower process. It's like the throttle setting on a car engine, controlling how quickly the decomposition of a substance like dimethyl ether occurs.
Dimethyl Ether Decomposition: A First-Order Reaction
The decomposition of dimethyl ether into methane, carbon monoxide, and hydrogen is a fascinating example of a first-order reaction. This reaction can be described by the simple equation:
  • \( \text{Dimethyl Ether} \rightarrow \text{Methane} + \text{CO} + \text{H}_2 \)
The reaction model is important because it tells us that the rate is directly proportional to the concentration of dimethyl ether. As the reaction proceeds, the amount of dimethyl ether decreases in a pattern defined by its half-life and rate constant.
For students looking to understand this concept, it's key to grasp that the concentration decrease is exponential due to the first-order nature. Whether you're calculating remaining mass after a period or determining how long it will take for a specific mass to reach a certain level, the principles remain the same. By applying the formula \( [A]_t = [A]_0 e^{-kt} \), you can predict how much of the reactant is consumed over time.

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Most popular questions from this chapter

Draw a reaction coordinate diagram for an exothermic reaction that occurs in a single step. Identify the activation energy and the net energy change for the reaction on this diagram. Draw a second diagram that represents the same reaction in the presence of a catalyst, assuming a single-step reaction is involved here also. Identify the activation energy of this reaction and the energy change. Is the activation energy in the two drawings different? Does the energy evolved in the two reactions differ?

Describe each of the following statements as true or false. If false, rewrite the sentence to make it correct. (a) The rate-determining elementary step in a reaction is the slowest step in a mechanism. (b) It is possible to change the rate constant by changing the temperature. (c) As a reaction proceeds at constant temperature, the rate remains constant. (d) A reaction that is third-order overall must involve more than one step.

When heated, cyclopropane is converted to propene (Example \(14.5) .\) Rate constants for this reaction at \(470^{\circ} \mathrm{C}\) and \(510^{\circ} \mathrm{C}\) are \(k=\) \(1.10 \times 10^{-4} \mathrm{s}^{-1}\) and \(k=1.02 \times 10^{-3} \mathrm{s}^{-1},\) respec- tively. Determine the activation energy, \(E_{a}\), from these data.

To determine the concentration dependence of the rate of the reaction $$ \mathrm{H}_{2} \mathrm{PO}_{3}^{-}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{HPO}_{3}^{2-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) $$ you might measure \(\left[\mathrm{OH}^{-}\right]\) as a function of time using a pH meter. (To do so, you would set up conditions under which \(\left[\mathrm{H}_{2} \mathrm{PO}_{3}\right]\) remains constant by using a large excess of this reactant.) How would you prove a second-order rate dependence for \(\left[\mathrm{OH}^{-}\right] ?\)

Calculate the activation energy, \(E_{a}\) for the reaction $$ 2 \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \rightarrow 4 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) $$ from the observed rate constants: \(k\) at \(25^{\circ} \mathrm{C}=\) \(3.46 \times 10^{-5} \mathrm{s}^{-1}\) and \(k\) at \(55^{\circ} \mathrm{C}=1.5 \times 10^{-3} \mathrm{s}^{-1}\).
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