Problem 67 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

In a first-order reaction, the activity of reactant drops from \(800 \mathrm{~mol} / \mathrm{dm}^{3}\) to \(50 \mathrm{~mol} / \mathrm{dm}^{3}\) in \(2 \times 10^{4} \mathrm{~s}\). The rate constant of the reaction, in \(\mathrm{s}^{-1}\), is (a) \(1.386 \times 10^{-4}\) (b) \(1.386 \times 10^{-3}\) (c) \(1.386 \times 10^{-5}\) (d) \(5.0 \times 10^{3}\)

Step-by-Step Solution

Answer

The rate constant k is (c) \(1.386 \times 10^{-5}\) s\(^{-1}\).

Step 1: Understand First-Order Reaction Kinetics

For a first-order reaction, the rate of the reaction is directly proportional to the concentration of the reactant. The integrated rate law for a first-order reaction is given by the equation: ln([A]_t/[A]_0) = -kt, where [A]_t is the concentration of the reactant at time t, [A]_0 is the initial concentration of the reactant, k is the rate constant, and t is the time.

Step 2: Set Up the Integrated Rate Law Equation

Substitute the given values into the integrated rate law equation for a first-order reaction: ln([A]_t/[A]_0) = -kt. Here, [A]_t = 50 mol/dm^3, [A]_0 = 800 mol/dm^3, and t = 2 x 10^4 s. We need to find the value of the rate constant k.

Step 3: Calculate the Rate Constant (k)

Plugging the values into the equation, we get ln(50/800) = -k(2 x 10^4 s). To find k, first calculate the natural logarithm, then isolate k by dividing both sides of the equation by -2 x 10^4 s.

Step 4: Solve for k

Perform the calculation: k = -ln(50/800) / (2 x 10^4 s).

Step 5: Evaluate the Result

Find the value of ln(50/800) using a calculator, then divide by 20000 to obtain the value of k. Match the result to the closest answer in the provided options.

Key Concepts

Chemical Kinetics

Chemical kinetics is the study of rates of chemical processes. It addresses how different variables such as temperature, pressure, concentration, and catalysts impact the speed of chemical reactions. For students tackling chemistry problems, understanding kinetics is crucial for predicting how long a reaction will take and what conditions are necessary for it to occur.

In first-order reactions, like the one in the exercise, the rate at which the reaction occurs is directly related to the concentration of one of the reactants. This type of reaction is common and has a straightforward kinetic analysis. Kinetic studies also help scientists understand reaction mechanisms, which are the step-by-step sequences of events at the molecular level that lead to the overall reaction.

Rate Law

The rate law is an equation that relates the rate of a chemical reaction to the concentration of its reactants. For a first-order reaction, the rate law can be expressed as Rate = k[A], where 'Rate' is the reaction rate, 'k' is the rate constant, 'A' is the concentration of the reactant, and the exponent indicates the order of the reaction—in this case, 1.

In first-order reactions, the change in the concentration of the reactant over time is linear when plotted on a logarithmic scale. Hence, the rate constant 'k' becomes crucial as it represents the proportionality between the rate and reactant concentration, and it reflects the efficiency of the reaction under given conditions.

Reactant Concentration

Reactant concentration is a measure of the amount of a given reactant present in a unit volume of solution at any given time. In chemical kinetics, understanding how the concentration of reactants affects the progress of a reaction is fundamental. As you have seen in the exercise, the initial concentration \( [A]_0 \) of a reactant can significantly affect the rate at which a reaction proceeds.

For first-order reactions, the rate is directly proportional to the reactant's concentration. This implies that a higher concentration of reactant will result in a faster reaction rate. Developing a grasp on how concentration affects reaction rate can help in predicting the behavior of chemical reactions under various conditions.

Integrated Rate Equation

The integrated rate equation for a first-order reaction is \[ \ln(\frac{[A]_t}{[A]_0}) = -kt \. \] It shows the relationship between the concentration of the reactant [A] at any time \( t \) and the original concentration \( [A]_0 \). The \( -k \) value in the equation is the rate constant, which is negative because the reactant concentration decreases over time.

When you solve the integrated rate equation for the rate constant \( k \), you'll typically arrange the equation to isolate k on one side. By doing so, you obtain \[ k = -\frac{1}{t}\ln(\frac{[A]_t}{[A]_0}) \.\]This equation is pivotal in kinetics because it allows you to calculate k when given enough information about the reactant concentrations and time, just as demonstrated in the exercise.