If 60 % of a first order reaction was completed in 60 minutes, 50 % of the same reaction would be completed in approximately
[log = 4 = 0.60, log 5 = 0.69].

Options

A. 40 minutes
B. 50 minutes
C. 45 minutes
D. 60 minutes

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Reaction and Given Data
We know that 60% of a first-order reaction is completed in 60 minutes. This means that 40% of the reactant remains.

### Step 2: Define Initial and Final Concentrations
Let the initial concentration of the reactant be \( A_0 = x \). After 60 minutes, since 60% is completed, the remaining concentration \( A \) is:
\[ A = A_0 - 0.6 A_0 = 0.4 A_0 = 0.4x \]

### Step 3: Use the First-Order Reaction Rate Equation
For a first-order reaction, the rate constant \( k \) can be calculated using the formula:
\[ k = \frac{2.303}{t} \log \left( \frac{A_0}{A} \right) \]
Substituting the values we have:
\[ k = \frac{2.303}{60} \log \left( \frac{x}{0.4x} \right) = \frac{2.303}{60} \log \left( \frac{1}{0.4} \right) \]

### Step 4: Simplify the Logarithm
Using the property of logarithms:
\[ \log \left( \frac{1}{0.4} \right) = -\log(0.4) = \log(10) - \log(4) = 1 - 0.60 = 0.40 \]
Now substituting this back into the equation for \( k \):
\[ k = \frac{2.303}{60} \times 0.40 \]

### Step 5: Calculate the Rate Constant \( k \)
Calculating \( k \):
\[ k = \frac{2.303 \times 0.40}{60} = \frac{0.9212}{60} \approx 0.01535 \text{ min}^{-1} \]

### Step 6: Calculate the Half-Life \( T_{1/2} \)
The half-life for a first-order reaction is given by:
\[ T_{1/2} = \frac{0.693}{k} \]
Substituting the value of \( k \):
\[ T_{1/2} = \frac{0.693}{0.01535} \approx 45.2 \text{ minutes} \]

### Conclusion
Thus, approximately 50% of the reaction would be completed in about **45 minutes**.

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