Question

In a first-order decomposition reaction, \(50.0 \%\) of a compound decomposes in \(10.5 \mathrm{~min}\). (a) What is the rate constant of the reaction? (b) How long does it take for \(75.0 \%\) of the compound to decompose?

Short answer

The rate constant is approximately 0.066 min^{-1}. It takes about 21.0 minutes for 75.0% decomposition.

Step-by-step solution

Understanding First-Order Reactions

First-order reactions follow the rate law: \[ \text{rate} = k[A] \] where \( k \) is the rate constant and \( [A] \) is the concentration of the reactant.

Use the First-Order Integrated Rate Law

The integrated rate law for a first-order reaction is: \[ \text{ln}([A]_t/[A]_0) = -kt \] where \( [A]_0 \) is the initial concentration, \( [A]_t \) is the concentration at time \( t \), and \( k \) is the rate constant.

Calculate the Rate Constant

Given that 50.0% of the compound decomposes in 10.5 minutes, \( \frac{[A]_t}{[A]_0} = 0.50 \). Substitute into the integrated rate law: \[ \text{ln}(0.50) = -k(10.5) \] Solve for \( k \): \[ k = -\frac{\text{ln}(0.50)}{10.5} \] Calculate \( k \): \[ k \approx 0.066 \thinspace \text{min}^{-1} \]

Determine the Time for 75% Decomposition

For 75.0% decomposition, \( \frac{[A]_t}{[A]_0} = 0.25 \). Using the integrated rate law: \[ \text{ln}(0.25) = -0.066t \] Solve for \( t \): \[ t = -\frac{\text{ln}(0.25)}{0.066} \] Calculate \( t \): \[ t \approx 21.0 \thinspace \text{min} \]

Key concepts

rate constant

The rate constant, denoted as 'k', is a crucial factor in chemical kinetics. It indicates how quickly a reaction progresses. For first-order reactions, the rate constant can be determined using the integrated rate law.

In the context of a first-order decomposition reaction, the rate constant is calculated based on the time it takes for a given percentage of the compound to decompose. In our exercise, 50% decomposition occurs in 10.5 minutes.

Using the integrated rate law for a first-order reaction: \ \[\text{ln}( [A]_t / [A]_0 ) = -kt\] \ we start by substituting the known values into the equation:

\[\text{ln}(0.50) = -k(10.5)\]

Solving for 'k', we get:

\[k = - \frac{\text{ln}(0.50)}{10.5} \approx 0.066 \text{ min}^{-1} \]

This tells us the reaction rate is 0.066 per minute.

integrated rate law

The integrated rate law connects the concentration of reactants to time, making it easier to predict how long a reaction will take given specific conditions.

For first-order reactions, the integrated rate law is expressed as:

\[\text{ln}( [A]_t / [A]_0 ) = -kt\]

where:\ * \([A]_0\) is the initial concentrationewline* \([A]_t\) is the concentration at time 't'ewline* 'k' is the rate constant

This law allows us to calculate the reaction time or the remaining concentration at any moment during the reaction. For instance, to find out how long it takes for 75% of a compound to decompose (leaving 25% of the original concentration), we use:

\[\text{ln}(0.25) = -kt\]

Rearranging to solve for time, 't', we get:

\[t = - \frac{\text{ln}(0.25)}{0.066} \approx 21.0 \text{ min} \]

This tells us that it will take about 21 minutes for 75% of the compound to decompose.

reaction kinetics

Reaction kinetics is the study of how fast chemical reactions proceed and the factors that influence these rates. Several core concepts in reaction kinetics include reaction order, rate laws, and reaction mechanisms.

First-order reactions, like the one in our exercise, have a reaction rate that depends linearly on the concentration of a single reactant. The rate law for a first-order reaction is given by:

\[\text{rate} = k[A]\]

To understand how reactants convert to products over time, we use the integrated rate law. This helps us relate the concentration of reactants at any time to the initial concentration, allowing predictions about the progress of the reaction.

Knowing the rate constant 'k' and applying the integrated rate law, we can determine both the speed and extent of the reaction under different conditions. This information is invaluable for predicting how a reaction will behave in practical scenarios, such as in industrial processes or biochemical pathways.