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 step by step, we need to determine the time it takes for 50% of a first-order reaction to be completed, given that 60% of the reaction is completed in 60 minutes.

### Step 1: Understand the Reaction Completion
We know that 60% of the reaction is completed in 60 minutes. If we assume the initial concentration (C0) is 100, then after 60 minutes, the remaining concentration (C) is:
\[ C = C_0 - \text{(percentage completed)} = 100 - 60 = 40 \]

### Step 2: Use the First-Order Reaction Formula
For a first-order reaction, the rate constant \( k \) can be calculated using the formula:
\[ k = \frac{2.303}{t} \log \left(\frac{C_0}{C}\right) \]
Where:
- \( t \) = time taken (60 minutes)
- \( C_0 \) = initial concentration (100)
- \( C \) = concentration at time \( t \) (40)

### Step 3: Calculate the Rate Constant \( k \)
Substituting the values into the equation:
\[ k = \frac{2.303}{60} \log \left(\frac{100}{40}\right) \]
\[ k = \frac{2.303}{60} \log(2.5) \]

Using the logarithmic values provided:
\[ \log(2.5) = \log\left(\frac{5}{2}\right) = \log(5) - \log(2) \]
Given \( \log(5) = 0.69 \) and \( \log(2) \approx 0.301 \):
\[ \log(2.5) = 0.69 - 0.301 = 0.389 \]

Now substituting back:
\[ k = \frac{2.303}{60} \times 0.389 \]
\[ k \approx 0.01527 \, \text{min}^{-1} \]

### Step 4: Calculate the Half-Life \( t_{1/2} \)
For a first-order reaction, the half-life is given by:
\[ t_{1/2} = \frac{0.693}{k} \]
Substituting the value of \( k \):
\[ t_{1/2} = \frac{0.693}{0.01527} \]
Calculating this gives:
\[ t_{1/2} \approx 45.3 \, \text{minutes} \]

### Conclusion
Thus, the time it takes for 50% of the reaction to be completed is approximately **45.3 minutes**.

---