Chapter 10: Problem 190
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.
Short Answer
Expert verified
The half-life of the reaction is approximately 6.02 minutes.
Step by step solution
01
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.
02
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.
03
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.
04
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 \( k \).
05
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 \( k \): \( k = -\frac{\ln 0.1}{20} \approx 0.1151 \text{ min}^{-1} \).
06
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 \( t_{1/2} \).
07
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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
- \( r = k[A] \)
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, \( r_2 = 0.055 \, \mathrm{mol} \, \mathrm{L}^{-1} \, \mathrm{min}^{-1} \)
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:
- \( t_{1/2} = \frac{0.693}{k} \)
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:
- \( [A] = [A]_0 e^{-kt} \)