Problem 55 - Chemistry An Atoms First Approach

Question

For the reaction \(\mathrm{A} \rightarrow\) products, successive half-lives are observed to be \(10.0,20.0,\) and 40.0 min for an experiment in which \([\mathrm{A}]_{0}=0.10 \mathrm{M} .\) Calculate the concentration of \(\mathrm{A}\) at the following times. a. \(80.0 \mathrm{min}\) b. \(30.0 \mathrm{min}\)

Step-by-Step Solution

Answer

a. At 80.0 min, the concentration of A is \(0.0123 M\). b. At 30.0 min, the concentration of A is \(0.0323 M\).

Step 1: Determine the reaction order

Observe the relationship between the successive half-lives given for the reaction. Compare each half-life with the previous one to see if they increase, decrease, or remain constant. In this case: For 1st half-life: 10.0 min For 2nd half-life: 20.0 min (which is 2 x 10.0 min) For 3rd half-life: 40.0 min (which is 2 x 20.0 min) The half-lives are increasing in a doubling pattern. Hence, the reaction is most likely a second-order reaction.

Step 2: Calculate the rate constant

For a second-order reaction, we know that the half-life (t1/2) is equal to \(\frac{1}{k[A]_0}\). We will use the first half-life and initial concentration value to find the rate constant k. Given t1/2 = 10.0 min and \([A]_0\) = 0.10 M, we set up the equation: 10.0 = \(\frac{1}{k(0.10)}\) Solve for k: k = \(\frac{1}{10.0(0.10)}\) k = 1.00 min\(^{-1}\)M\(^{-1}\)

Step 3: Calculate the concentration of A at 80.0 min

For a second-order reaction, the integrated rate law is given by: \(\frac{1}{[A]} = kt + \frac{1}{[A]_0}\) We want to find the concentration of A at 80.0 min. Plug in the values for k, t, and \([A]_0\): \(\frac{1}{[A]} = (1.00)(80.0) + \frac{1}{0.10}\) Solve for [A]: \(\frac{1}{[A]} = 81\) \([A] = \frac{1}{81}\) \([A] = 0.0123 M\) The concentration of A at 80.0 min is 0.0123 M.

Step 4: Calculate the concentration of A at 30.0 min

Using the second-order integrated rate law again, plug in the values for k, t, and \([A]_0\) to find the concentration of A at 30.0 min: \(\frac{1}{[A]} = (1.00)(30.0) + \frac{1}{0.10}\) Solve for [A]: \(\frac{1}{[A]} = 31\) \([A] = \frac{1}{31}\) \([A] = 0.0323 M\) The concentration of A at 30.0 min is 0.0323 M. To summarize: a. At 80.0 min, the concentration of A is 0.0123 M. b. At 30.0 min, the concentration of A is 0.0323 M.

Key Concepts

Half-Life

The concept of half-life is particularly important when studying chemical reactions. It refers to the time required for half of the reactant to be consumed in a reaction. For second-order reactions, unlike first-order reactions where the half-life remains constant, the half-life changes as the concentration of the reactant changes. In the context of second-order reactions, the half-life is proportional to the inverse of the initial concentration and is expressed as:

• First half-life (\( t_{1/2} \)): \( \frac{1}{k[A]_0} \)
• Second half-life: double the first half-life
• Third half-life: double the second half-life

In our exercise example, the half-lives increase from 10.0 min to 20.0 min to 40.0 min, illustrating the doubling pattern typical of second-order reactions.

Rate Constant

The rate constant \(k\) is a crucial element in the rate equation of a chemical reaction. For second-order reactions, finding \(k\) involves using the half-life and the initial concentration of the reactant. Specifically, for a second-order reaction, the half-life can be calculated using \( \frac{1}{k[A]_0} \), allowing for the determination of \(k\) when the first half-life is known.

In the case provided, the first half-life is 10.0 mins and the initial concentration is 0.10 M. Using the equation:

• \( 10.0 = \frac{1}{k \times 0.10} \)
• Solve for \(k\): \( k = 1.00 \, \text{min}^{-1}\text{M}^{-1} \)

This rate constant is then useful for predicting concentrations at various times, which is calculated using the integrated rate law.

Integrated Rate Law

The integrated rate law is invaluable for determining the reactant concentration at a given time in a second-order reaction. This formula takes the format:
\[ \frac{1}{[A]} = kt + \frac{1}{[A]_0} \]where \([A]\) is the concentration of the reactant at time \(t\), \(k\) is the rate constant, and \([A]_0\) is the initial concentration.

Applying this equation to our specific example:

• For 80.0 min: \( \frac{1}{[A]} = (1.00)(80.0) + \frac{1}{0.10} = 81 \)
• Solve for \([A]\): \([A] = \frac{1}{81} = 0.0123 \, \text{M} \)
• For 30.0 min: \( \frac{1}{[A]} = (1.00)(30.0) + \frac{1}{0.10} = 31 \)
• Solve for \([A]\): \([A] = \frac{1}{31} = 0.0323 \, \text{M} \)

This law helps to foresee changes in concentration over time, by providing a clear mathematical representation you can rely on.