Question

The reaction \(\mathrm{X} \longrightarrow\) Product follows first-order kinetics, in 40 minutes, the concentration of \(\mathrm{X}\) changes from \(0.1 \mathrm{M}\) to \(0.025 \mathrm{M}\), then the rate of reaction when concentration of \(\mathrm{X}\) is \(0.01 \mathrm{M}\) is? (a) \(3.47 \times 10^{-5} \mathrm{M} / \mathrm{min}\) (b) \(1.73 \times 10^{-4} \mathrm{M} / \mathrm{min}\) (c) \(1.73 \times 10^{-5} \mathrm{M} / \mathrm{min}\) (d) \(3.47 \times 10^{-4} \mathrm{M} / \mathrm{min}\)

Short answer

The rate of reaction is \(3.47 \times 10^{-5} \text{ M/min}\) (option a).

Step-by-step solution

Understanding First-Order Kinetics

For a first-order reaction, the rate of reaction depends on the concentration of a single reactant raised to the first power. The rate law expression for a first-order reaction is given by: \( r = -\frac{d[X]}{dt} = k[X] \) where \( k \) is the rate constant and \( [X] \) is the concentration of the reactant.

Use First-Order Kinetics Equation

The integrated rate law for a first-order reaction is: \( \ln \left(\frac{[X]_0}{[X]t}\right) = kt \). Here, \([X]_0\) is the initial concentration, \([X]t\) is the concentration at time \( t \), and \( k \) is the rate constant.

Solve for the Rate Constant \( k \)

Given \([X]_0 = 0.1 \text{ M} \) and \([X]t = 0.025 \text{ M} \) in 40 minutes:\(\ln \left(\frac{0.1}{0.025}\right) = 40k\). This simplifies to \( \ln(4) = 40k \) or \( k = \frac{\ln 4}{40} \).

Calculate the Rate of Reaction

Now use the rate law: \( r = k[X] \), with \( [X] = 0.01 \text{ M} \).Substitute \( k = \frac{\ln 4}{40} \) into the equation: \( r = \frac{\ln 4}{40} \times 0.01 \). Evaluate \( r \) to get the rate of reaction.

Simplify the Rate Expression

Calculate the rate: \( \ln 4 \approx 1.386 \). Substitute to get \( r = \frac{1.386}{40} \times 0.01 = 0.0003465 \text{ M/min} \) or \( 3.465 \times 10^{-5} \text{ M/min} \).

Key concepts

Rate Constant

The rate constant, often denoted by the symbol \( k \), is a crucial factor in defining the speed of a chemical reaction. For a first-order reaction, the rate constant has units of \( ext{time}^{-1} \), like \( ext{min}^{-1} \). This constant is specific to each reaction and is influenced by factors such as temperature and the presence of a catalyst.
To find the rate constant, you generally need experimental data about the concentration of a reactant over time, as shown in the original exercise. Here, the concentration of \( X \) dropped from \( 0.1 \, \text{M} \) to \( 0.025 \, \text{M} \) in 40 minutes. By using the integrated rate law equation \( \ln \left(\frac{[X]_0}{[X]_t}\right) = kt \), you can solve for \( k \), thus making it possible to quantify the strength or weakness of the reaction's rate in relation to changes in concentration.

Integrated Rate Law

The integrated rate law for first-order reactions connects concentration with time. It's an essential tool because it allows us to calculate the concentration of a reactant at any given time point, knowing its initial concentration and the rate constant. The equation is:\[ \ln \left(\frac{[X]_0}{[X]_t}\right) = kt \]
This equation states that the natural logarithm of the fraction of the initial concentration divided by the concentration at time \( t \) equals the product of the rate constant and the time. It rearranges to also allow for the derivation of the rate constant when multiple time points are known. In practical terms, understanding the integrated rate law helps chemists and students visualize how reactions progress over time, offering insights into reaction mechanisms and the influence of various external conditions.

Rate of Reaction

The rate of a reaction indicates how quickly or slowly a reaction takes place and is frequently calculated using the relationship between the rate constant and the reactant concentration, given by:\[ r = k[X] \]
This formula shows that for first-order reactions, the rate of reaction directly depends on the concentration of the reactant \( X \) and the rate constant \( k \). For example, if the concentration is halved, the rate of reaction will halve as well in a first-order reaction.
In the exercise, by substituting \( k \) calculated earlier and the given concentration \( 0.01 \, \text{M} \), the rate was determined to be approximately \( 3.47 \times 10^{-5} \, \text{M/min} \). This conversion of theoretical calculations into practical rate values assists chemists in planning experiments, optimizing reactions, and predicting reaction behavior under various conditions.