A first order reaction is `75%` completed in 40 minutes. Calculate its `t_(1//2)`.
(Given `log2=0.3010,log4=0.6021`)

Answer

To solve the problem of finding the half-life (\(t_{1/2}\)) of a first-order reaction that is 75% completed in 40 minutes, we can follow these steps:

### Step 1: Understand the completion percentage
Since the reaction is 75% completed, it means that 25% of the reactant remains. If we assume the initial concentration (\(A_0\)) is 100 units, then:
- Remaining concentration (\(A_t\)) = \(A_0 - 75\% \text{ of } A_0 = 100 - 75 = 25\)

### Step 2: Use the first-order rate equation
For a first-order reaction, the rate constant \(K\) can be calculated using the formula:
\[ K = \frac{2.303}{t} \log \left( \frac{A_0}{A_t} \right) \]
Where:
- \(t\) = time taken for the reaction (40 minutes)
- \(A_0\) = initial concentration (100)
- \(A_t\) = concentration at time \(t\) (25)

### Step 3: Substitute the values into the equation
Substituting the values into the equation:
\[ K = \frac{2.303}{40} \log \left( \frac{100}{25} \right) \]
Calculating the logarithm:
\[ \frac{100}{25} = 4 \quad \text{and} \quad \log(4) = 0.6021 \]
Now substituting this back into the equation for \(K\):
\[ K = \frac{2.303}{40} \times 0.6021 \]

### Step 4: Calculate \(K\)
Calculating \(K\):
\[ K = \frac{2.303 \times 0.6021}{40} = \frac{1.3844}{40} = 0.03461 \text{ min}^{-1} \]

### Step 5: Calculate the half-life (\(t_{1/2}\))
The half-life for a first-order reaction is given by the formula:
\[ t_{1/2} = \frac{0.693}{K} \]
Substituting the value of \(K\):
\[ t_{1/2} = \frac{0.693}{0.03461} \]

### Step 6: Perform the calculation
Calculating \(t_{1/2}\):
\[ t_{1/2} \approx 20.0 \text{ minutes} \]

### Final Answer
The half-life (\(t_{1/2}\)) of the reaction is approximately **20 minutes**.

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