In a first order reaction, the concentration of the reactants is reduced to `25%` in one hour. The half-life periof of the reactions is

Options

A. `2 hr`
B. `4 hr`
C. `1//2 hr`
D. `1//4 hr`

Explanation

To find the half-life period of a first-order reaction where the concentration of the reactants is reduced to 25% in one hour, we can follow these steps:

### Step 1: Understand the Given Information
- The concentration of the reactant is reduced to 25% of its initial value in 1 hour.
- This means that if we assume the initial concentration \( A_0 = 100 \), then after 1 hour, the concentration \( A_t = 25 \).

### Step 2: Use the First-Order Reaction Rate Equation
For a first-order reaction, the rate constant \( k \) can be calculated using the formula:
\[ k = \frac{2.303}{t} \log\left(\frac{A_0}{A_t}\right) \]
where:
- \( t = 1 \) hour
- \( A_0 = 100 \)
- \( A_t = 25 \)

### Step 3: Substitute the Values
Substituting the values into the equation:
\[ k = \frac{2.303}{1} \log\left(\frac{100}{25}\right) \]

### Step 4: Simplify the Logarithm
Calculate the ratio:
\[ \frac{100}{25} = 4 \]
Now, we can substitute this into the logarithm:
\[ k = 2.303 \log(4) \]

### Step 5: Further Simplify the Logarithm
We can express \( \log(4) \) as:
\[ \log(4) = \log(2^2) = 2 \log(2) \]
Thus, we have:
\[ k = 2.303 \times 2 \log(2) \]

### Step 6: Substitute the Value of \( \log(2) \)
Using the known value \( \log(2) \approx 0.3010 \):
\[ k = 2.303 \times 2 \times 0.3010 \]
Calculating this gives:
\[ k \approx 2.303 \times 0.602 = 1.3864 \, \text{h}^{-1} \]

### Step 7: 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 \( k \):
\[ t_{1/2} = \frac{0.693}{1.3864} \]
Calculating this gives:
\[ t_{1/2} \approx 0.5 \, \text{hours} \]

### Final Answer
The half-life period of the reaction is **0.5 hours**.

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