The half-change time for a first order reaction is `20 min` hence the rate constant for the reaction………….. .

Answer

To find the rate constant \( k \) for a first-order reaction given the half-life \( t_{1/2} \), we can follow these steps:

### Step 1: Write the formula for half-life of a first-order reaction
The half-life \( t_{1/2} \) for a first-order reaction is given by the formula:
\[ t_{1/2} = \frac{\ln 2}{k} \]
where \( k \) is the rate constant.

### Step 2: Rearrange the formula to solve for \( k \)
We can rearrange the formula to solve for \( k \):
\[ k = \frac{\ln 2}{t_{1/2}} \]

### Step 3: Substitute the given half-life into the equation
We know that the half-life \( t_{1/2} \) is \( 20 \) minutes. We can substitute this value into the equation:
\[ k = \frac{\ln 2}{20 \text{ min}} \]

### Step 4: Calculate \( \ln 2 \)
The natural logarithm of \( 2 \) is approximately:
\[ \ln 2 \approx 0.693 \]

### Step 5: Substitute \( \ln 2 \) into the equation
Now we substitute \( \ln 2 \) into the equation for \( k \):
\[ k = \frac{0.693}{20 \text{ min}} \]

### Step 6: Perform the calculation
Now we perform the calculation:
\[ k \approx \frac{0.693}{20} \approx 0.03465 \text{ min}^{-1} \]

### Final Answer
Thus, the rate constant \( k \) for the reaction is:
\[ k \approx 0.03465 \text{ min}^{-1} \]

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