Question

The thermal decomposition of phosphine \(\left(\mathrm{PH}_{3}\right)\) into phosphorus and molecular hydrogen is a first-order reaction: $$ 4 \mathrm{PH}_{3}(g) \longrightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g) $$ The half-life of the reaction is \(35.0 \mathrm{~s}\) at \(680^{\circ} \mathrm{C}\). Calculate (a) the first-order rate constant for the reaction and (b) the time required for 95 percent of the phosphine to decompose.

Short answer

(a) The rate constant is \(0.0198 \, \text{s}^{-1}\). (b) Time for 95% decomposition is 151.3 seconds.

Step-by-step solution

Understand the Half-Life Formula for First-Order Reactions

In a first-order reaction, the half-life \( t_{1/2} \) is related to the rate constant \( k \) by the formula: \[ t_{1/2} = \frac{0.693}{k} \] Given that the half-life \( t_{1/2} \) is 35.0 seconds, we can use this formula to find the rate constant \( k \).

Calculate the Rate Constant \( k \)

Rearrange the half-life formula to solve for the rate constant: \[ k = \frac{0.693}{t_{1/2}} \] Substitute the given half-life \( t_{1/2} = 35.0 \)s: \[ k = \frac{0.693}{35.0} \approx 0.0198 \, \text{s}^{-1} \] So, the rate constant \( k \) is approximately \( 0.0198 \, \text{s}^{-1} \).

Understand the First-Order Decomposition Formula

The amount of reactant remaining in a first-order reaction after time \( t \) is given by: \[ [A] = [A]_0 e^{-kt} \] To find the time required for 95% decomposition, we need 5% of the original concentration \( [A]_0 \) to remain, or \( [A] = 0.05[A]_0 \).

Set Up the Equation for 95% Decomposition

Starting with: \[ 0.05 [A]_0 = [A]_0 e^{-kt} \] Cancel \( [A]_0 \) from both sides: \[ 0.05 = e^{-kt} \]

Solve for Time \( t \)

Take the natural logarithm of both sides to solve for \( t \): \[ \ln(0.05) = -kt \] Substitute \( k = 0.0198 \, \text{s}^{-1} \): \[ t = \frac{\ln(0.05)}{-0.0198} \] Calculate \( t \): \[ t \approx \frac{-2.9957}{-0.0198} \approx 151.3 \, \text{s} \]

Conclusion: Final Results

The rate constant \( k \) for the reaction is approximately \( 0.0198 \, \text{s}^{-1} \). The time required for 95% of the phosphine to decompose is approximately 151.3 seconds.

Key concepts

Phosphine Decomposition

Phosphine, known chemically as \( \text{PH}_3 \), is a compound that decomposes under heat into phosphorus and molecular hydrogen. This reaction can be illustrated by the chemical equation: \[ 4\text{PH}_3(g) \longrightarrow \text{P}_4(g) + 6\text{H}_2(g) \] This is a first-order reaction. A first-order reaction means that the rate of the reaction depends linearly on the concentration of one reactant. As the phosphine decomposes, the breakdown into new molecules occurs over time, following particular rate laws to describe its kinetics. Understanding how phosphine decomposes is essential to grasp how certain reactions are modeled chemically and mathematically. This foundation helps you appreciate the step-by-step breakdown dynamics, which can be calculated with simple mathematical methods. It leverages concepts such as the rate constant and half-life to describe and predict the reaction's behavior.

Rate Constant Calculation

In order to understand how quickly a reaction proceeds, we need to calculate the rate constant \( k \). The rate constant is a crucial component of the first-order reaction equations and gives insight into the speed of the reaction. For first-order reactions, the relationship between the half-life \( t_{1/2} \) and the rate constant \( k \) is expressed by the equation: \[ t_{1/2} = \frac{0.693}{k} \] To find \( k \), you simply rearrange the formula: \[ k = \frac{0.693}{t_{1/2}} \] Given that the phosphine decomposition has a half-life of \( 35.0 \) seconds at \( 680^{\circ} \text{C} \), substituting this value into the formula gives the rate constant: \[ k = \frac{0.693}{35.0} \approx 0.0198 \, \text{s}^{-1} \] Calculating the rate constant allows you to predict and understand the decomposition process's characteristics. With \( k \), you can figure out not only how long a certain portion of the reaction takes but also model how changing conditions might impact it.

Half-life of Reactions

The concept of half-life is a valuable tool for understanding how chemical reactions proceed over time. In first-order reactions, like the decomposition of phosphine, the half-life remains constant. This means the time it takes for half of the reactant to be converted into product is the same, no matter the initial concentration. The formula for half-life in a first-order reaction is: \[ t_{1/2} = \frac{0.693}{k} \] Where:

• \( t_{1/2}\) represents the half-life
• \( k \) is the rate constant

Knowing the half-life is particularly useful when predicting how long it will take for a significant amount of the reactant to decompose. For example, to determine the time required for 95% decomposition of phosphine, only 5% remains, which sets a different scenario than simply cutting the concentration in half. By setting up the equation \( 0.05 = e^{-kt} \), you'd solve for \( t \) through natural logarithms, giving you a wider understanding of time-dependent reaction progress: \[ t = \frac{\ln(0.05)}{-0.0198} \approx 151.3 \, \text{s} \] This demonstrates how measurable kinetics provide valuable decades of reaction time towards managing and controlling chemical processes effectively.