Question

The decomposition of phosphine, \(\mathrm{PH}_{3}\), proceeds according to the equation $$ \mathrm{PH}_{3}(\mathrm{g}) \rightarrow^{1 / 4} \mathrm{P}_{4}(\mathrm{g})+3 / 2 \mathrm{H}_{2}(\mathrm{g}) $$ It is found that the reaction has the following rate equation: Rate \(=k\left[\mathrm{PH}_{3}\right] .\) The half-life of \(\mathrm{PH}_{3}\) is 37.9 seconds at \(120^{\circ} \mathrm{C}\) (a) How much time is required for three fourths of the \(\mathrm{PH}_{3}\) to decompose? (b) What fraction of the original sample of \(\mathrm{PH}_{3}\) remains after 1.00 minute?

Short answer

(a) 75.86 seconds; (b) 33.3% remains.

Step-by-step solution

Understand the Reaction and Rate Equation

The decomposition of phosphine follows the reaction \[ \mathrm{PH}_{3}(\mathrm{g}) \rightarrow^{1 / 4} \mathrm{P}_{4}(\mathrm{g})+3 / 2 \mathrm{H}_{2}(\mathrm{g}) \].This is a first-order reaction because the rate equation is given by\[\text{Rate} = k [\mathrm{PH}_{3}].\] In first-order reactions, the half-life \( t_{1/2} \) is independent of the concentration and is calculated as \( t_{1/2} = \frac{0.693}{k} \).

Calculate the Rate Constant k

Given the half-life of \( \mathrm{PH}_{3} \) is 37.9 seconds, we can calculate the rate constant \( k \) using the first-order half-life formula:\[t_{1/2} = \frac{0.693}{k}\]Substitute \( t_{1/2} = 37.9 \) seconds to find \( k \):\[37.9 = \frac{0.693}{k}\]Solving for \( k \) gives:\[sk = \frac{0.693}{37.9} \approx 0.0183 \text{ s}^{-1}.\]

Calculate Time for Three Fourths Decomposition

To determine the time required for three fourths of \( \mathrm{PH}_{3} \) to decompose, note that we are left with one fourth of the original concentration. For a first-order reaction, the relation is:\[\ln \left(\frac{[A]_0}{[A]}\right) = kt,\]where \([A]_0\) is the initial concentration and \([A] = \frac{1}{4}[A]_0\). Thus,\[\ln \left(\frac{[A]_0}{[A]}\right) = \ln(4) = kt.\]Solving for \( t \):\[\ln(4) = 0.0183 \times t.\]\[t = \frac{\ln(4)}{0.0183} \approx 75.86 \text{ seconds}.\]

Calculate Fraction Remaining After 1 Minute

One minute is 60 seconds. For a first-order reaction, the concentration remaining is given by:\[[A] = [A]_0 e^{-kt}.\]Substitute \( t = 60 \) seconds and \( k = 0.0183 \ \text{s}^{-1} \) into the equation:\[[A] = [A]_0 e^{-0.0183 \times 60} = [A]_0 e^{-1.098}.\]The fraction of the original sample remaining is:\[e^{-1.098} \approx 0.333.\]

Conclusion

We have calculated that it takes approximately 75.86 seconds for three fourths of \( \mathrm{PH}_{3} \) to decompose. After 1 minute, approximately 33.3% of the original phosphine remains.

Key concepts

Decomposition Reaction

In chemistry, a decomposition reaction involves the breakdown of a single compound into two or more simpler substances. Specifically, in the decomposition of phosphine (\( \mathrm{PH}_{3} \)), the chemical equation is expressed as \( \mathrm{PH}_{3}(\mathrm{g}) \rightarrow^{1/4} \mathrm{P}_{4}(\mathrm{g}) + 3/2 \mathrm{H}_{2}(\mathrm{g}) \). This implies that every molecule of phosphine breaks down to form phosphorus gas and hydrogen gas.
The decomposition of \( \mathrm{PH}_{3} \) is also noted to follow a first-order reaction. This means the reaction rate is directly proportional to the concentration of phosphine at any given time. As such, understanding the nature of decomposition reactions is crucial in predicting and calculating how long these processes take under given conditions.
For this particular reaction, knowing it is a first-order process helps simplify the calculations of time needed for decomposition and remaining concentration after a certain period.

Rate Constant Calculation

For first-order reactions, the rate constant (\( k \)) is a key parameter that helps in understanding the speed of a reaction. It is determined by using the half-life formula for first-order reactions, given by \( t_{1/2} = \frac{0.693}{k} \).
In the given exercise, the half-life of phosphine is 37.9 seconds. By substituting this value into the half-life equation, we find that:
\[ t_{1/2} = 37.9 \text{ seconds} \quad \Rightarrow \quad k = \frac{0.693}{37.9} \approx 0.0183 \text{ s}^{-1} \]
Thus, the rate constant for the decomposition reaction of \( \mathrm{PH}_{3} \) is approximately \( 0.0183 \text{ s}^{-1} \). This constant is crucial for further calculations to predict how the concentration changes over time and how quickly the reaction proceeds.

Half-life of Reactions

The concept of half-life is essential in understanding how reactions progress over time. Half-life, denoted as \( t_{1/2} \), is the time required for half of the original concentration of a reactant to be consumed in a reaction. For first-order reactions, the half-life is independent of the initial concentration, making these calculations more straightforward.
In our example, with a half-life of 37.9 seconds, we can determine how much phosphine remains at any given time, or how much time is required for a certain fraction to decompose. For instance, calculating how much time it takes for three-fourths of phosphine to decompose involves finding when only one-fourth remains. Using the formula \( \ln \left( \frac{[A]_0}{[A]} \right) = kt \), where \([A]\) is a portion of the original concentration, you can find that it takes approximately 75.86 seconds for this scenario.
This predictable pattern of concentration decay is valuable in both laboratory and industrial settings where reaction times need to be managed accurately.