A first-order reaction was 70 percent complete in 20 minutes. What is the rate constant of the reaction?

Options

A. 0.07 min-1
B. 0.06 min-1
C. 0.08 min-1
D. 0.09 min-1

Answer

To find the rate constant of a first-order reaction that is 70% complete in 20 minutes, we can follow these steps:

### Step 1: Understand the Reaction Completion
Given that the reaction is 70% complete, this means that 30% of the reactant remains. If we denote the initial amount of the reactant as \( A_0 \), then:
- Amount remaining, \( A = 30\% \) of \( A_0 \)
- Therefore, \( A = 0.30 A_0 \)

### Step 2: Set Up the First-Order Rate Constant Equation
For a first-order reaction, the rate constant \( k \) can be calculated using the formula:
\[ k = \frac{2.303}{t} \log \left( \frac{A_0}{A} \right) \]
where:
- \( t \) is the time taken for the reaction to reach this completion (20 minutes in this case).
- \( A_0 \) is the initial concentration (which we can take as 100 for simplicity).
- \( A \) is the concentration remaining after the reaction (30).

### Step 3: Substitute the Values into the Equation
Substituting the values into the equation:
- \( A_0 = 100 \)
- \( A = 30 \)
- \( t = 20 \) minutes

The equation becomes:
\[ k = \frac{2.303}{20} \log \left( \frac{100}{30} \right) \]

### Step 4: Calculate the Logarithm
First, calculate \( \frac{100}{30} \):
\[ \frac{100}{30} = \frac{10}{3} \]

Now, calculate the logarithm:
\[ \log \left( \frac{10}{3} \right) \]

Using a calculator, we find:
\[ \log \left( \frac{10}{3} \right) \approx 0.5229 \]

### Step 5: Substitute Back to Find \( k \)
Now substitute the logarithm back into the equation for \( k \):
\[ k = \frac{2.303}{20} \times 0.5229 \]

Calculating this gives:
\[ k \approx \frac{2.303 \times 0.5229}{20} \approx \frac{1.202}{20} \approx 0.0601 \text{ min}^{-1} \]

### Final Result
Thus, the rate constant \( k \) of the reaction is approximately:
\[ k \approx 0.06 \text{ min}^{-1} \]