Problem 190 - A Complete Resource Book in Chemistry for JEE Main

Question

The rate of a first order reaction at \(20 \mathrm{~min}\) is \(0.55 \mathrm{~mol}\) \(\mathrm{L}^{-1} \min ^{-1}\) and \(0.055 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}\) at \(40 \mathrm{~min}\) after initi- ation. Find half life of the reaction in minutes.

Step-by-Step Solution

Answer

The half-life of the reaction is approximately 6.02 minutes.

Step 1: Understand First Order Reaction

In a first-order reaction, the rate of reaction is directly proportional to the concentration of one reactant. The rate law is given by \( r = k[A] \), where \( r \) is the rate of reaction, \( k \) is the rate constant, and \( [A] \) is the concentration of the reactant.

Step 2: Use the Integrated Rate Law

For a first-order reaction, the integrated rate law is \( [A] = [A]_0 e^{-kt} \), where \( [A]_0 \) is the initial concentration, \( t \) is time, and \( k \) is the rate constant.

Step 3: Relate Rate and Concentration

The rate can also be expressed as \( r = -\frac{d[A]}{dt} = k[A] \). Using this, we can say that \( [A] = \frac{r}{k} \). Let's write this equation for two given times: at 20 minutes and 40 minutes.

Step 4: Calculate the Rate Constant \(k\)

Now, we use the rate expression for both times: \( r_1 = k[A]_1 \) and \( r_2 = k[A]_2 \). This gives us \( [A]_1 = \frac{0.55}{k} \) and \( [A]_2 = \frac{0.055}{k} \). Then, apply the integrated rate law: \( \frac{[A]_2}{[A]_1} = e^{-k(t_2-t_1)} \). Substitute \( \frac{0.055}{0.55} = e^{-k(40-20)} \) to find \( k \).

Step 5: Solve for \( k \)

\( \frac{0.055}{0.55} = e^{-20k} \) simplifies to \( 0.1 = e^{-20k} \). Take the natural logarithm of both sides: \( \ln 0.1 = -20k \). Solve for \( k \): \( k = -\frac{\ln 0.1}{20} \approx 0.1151 \text{ min}^{-1} \).

Step 6: Calculate Half-Life

The half-life \( t_{1/2} \) for a first-order reaction is given by \( t_{1/2} = \frac{0.693}{k} \). Substitute \( k = 0.1151 \text{ min}^{-1} \) into the formula to find \( t_{1/2} \).

Step 7: Solve for Half-Life

\( t_{1/2} = \frac{0.693}{0.1151} \approx 6.02 \text{ minutes} \). Thus, the half-life of the reaction is approximately 6.02 minutes.

Key Concepts

Rate of Reaction

The rate of a reaction refers to how quickly reactants are used up or products are formed in a chemical reaction. In the case of a first-order reaction, the rate is directly proportional to the concentration of one reactant. Imagine you are cooking and you keep adding more ingredients; the dish becomes ready faster, similar to how increasing a reactant affects the reaction rate. Mathematically, this relationship is expressed as

• \( r = k[A] \)

where \( r \) represents the rate of reaction and \( [A] \) is the concentration of the reactant. The rate constant \( k \) is a unique value for each reaction and influences how swiftly the reaction proceeds.

Rate Constant

The rate constant, denoted as \( k \), is a measure of how fast a reaction occurs. It's specific to each reaction and depends on factors like temperature and the presence of a catalyst. For a first-order reaction, the rate constant can be determined by analyzing how the rate changes with the concentration of the reactant. Using the reaction example from the exercise:

• At 20 minutes, \( r_1 = 0.55 \, \mathrm{mol} \, \mathrm{L}^{-1} \, \mathrm{min}^{-1} \)
• At 40 minutes, \( r_2 = 0.055 \, \mathrm{mol} \, \mathrm{L}^{-1} \, \mathrm{min}^{-1} \)

We can calculate the rate constant \( k \) via the ratio of rates at two different times, using the integrated rate law. After simplification, this leads to solving the equation \( e^{-20k} = 0.1 \), resulting in a rate constant \( k \) of approximately \( 0.1151 \, \mathrm{min}^{-1} \). This value helps determine other aspects like half-life without repeating long calculations.

Half-Life

Half-life is the time required for half of the reactant to be consumed in a reaction. For first-order reactions, half-life is a constant and doesn't depend on the initial concentration of the reactant. This peculiar feature makes first-order processes predictable over time. The half-life \( t_{1/2} \) can be calculated using the formula:

• \( t_{1/2} = \frac{0.693}{k} \)

In our exercise, with a rate constant \( k \) of \( 0.1151 \, \mathrm{min}^{-1} \), the half-life calculates to approximately 6.02 minutes. This means every 6.02 minutes, the concentration of the reactant reduces by half, illustrating a steady decline as the reaction progresses.

Integrated Rate Law

The integrated rate law shows how the concentration of reactants changes over time in a chemical reaction. Particularly for first-order reactions, it bridges the concentration at different times through the equation:

• \( [A] = [A]_0 e^{-kt} \)

This equation allows you to calculate the concentration of the reactant \([A]\) at any time \( t \) if the rate constant \( k \) and the initial concentration \([A]_0\) are known. In the context of our example, the integrated rate law helps express the concentration of the substance at 20 and 40 minutes in terms of their respective rates. By rearranging and solving this equation, we found critical parameters like the rate constant and established the time evolution of the reactant concentration. The integrated rate law is crucial in deducing how quickly reactions are happening and planning how to tweak conditions for desired outcomes.