The rate of a first-order reaction is `0.04 "mol L"^(-1) s^(-1)` at `10` seconds and `0.03 "mol L"^(-1) s^(-1)` at `20` seconds after initiation of the reaction. The hlaf-life period of the reaction is :

Options

A. 54.1 s
B. 24.1 s
C. 34.1 s
D. 44.1 s

Answer

To solve the problem, we need to follow these steps:

### Step 1: Write the rate law for a first-order reaction
The rate of a first-order reaction is given by the equation:
\[ \text{Rate} = k \cdot [A] \]
where \( k \) is the rate constant and \([A]\) is the concentration of the reactant.

### Step 2: Set up equations based on the given rates
We know the rates at two different times:
- At \( t = 10 \) seconds, the rate is \( 0.04 \, \text{mol L}^{-1} \text{s}^{-1} \).
- At \( t = 20 \) seconds, the rate is \( 0.03 \, \text{mol L}^{-1} \text{s}^{-1} \).

Using the rate law, we can write two equations:
1. \( 0.04 = k \cdot [A]_{10} \) (Equation 1)
2. \( 0.03 = k \cdot [A]_{20} \) (Equation 2)

### Step 3: Divide the two equations to eliminate \( k \)
Dividing Equation 1 by Equation 2 gives:
\[ \frac{0.04}{0.03} = \frac{[A]_{10}}{[A]_{20}} \]
This simplifies to:
\[ \frac{4}{3} = \frac{[A]_{10}}{[A]_{20}} \]

### Step 4: Use the integrated rate law for first-order reactions
For a first-order reaction, the relationship between concentration and time is given by:
\[ \ln \left( \frac{[A]_0}{[A]} \right) = kt \]
We can express this in terms of the concentrations at 10 seconds and 20 seconds:
\[ \ln \left( \frac{[A]_{10}}{[A]_{20}} \right) = k(t_{20} - t_{10}) = k(20 - 10) = 10k \]

### Step 5: Substitute the concentration ratio into the equation
Using the ratio we found:
\[ \ln \left( \frac{4}{3} \right) = 10k \]
Now, calculate \( k \):
\[ k = \frac{1}{10} \ln \left( \frac{4}{3} \right) \]

### Step 6: Calculate \( k \)
Using a calculator:
\[ \ln \left( \frac{4}{3} \right) \approx 0.2877 \]
Thus,
\[ k \approx \frac{0.2877}{10} = 0.02877 \, \text{s}^{-1} \]

### Step 7: Calculate the half-life of the reaction
The half-life \( t_{1/2} \) for a first-order reaction is given by:
\[ t_{1/2} = \frac{0.693}{k} \]
Substituting the value of \( k \):
\[ t_{1/2} = \frac{0.693}{0.02877} \approx 24.1 \, \text{s} \]

### Final Answer
The half-life period of the reaction is approximately \( 24.1 \, \text{seconds} \).

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