What is half life of first order reaction if time required to decrease concentration of reactants from 0.8M to 0.2M is 12 hrs.

Answer

To find the half-life of a first-order reaction given that the concentration decreases from 0.8 M to 0.2 M in 12 hours, we can follow these steps:

### Step 1: Identify the formula for the rate constant (k) for a first-order reaction.
The formula for the rate constant (k) for a first-order reaction is given by:
\[ k = \frac{2.303}{t} \log \left( \frac{A_0}{A_t} \right) \]
where:
- \( A_0 \) is the initial concentration,
- \( A_t \) is the concentration after time \( t \),
- \( t \) is the time taken.

### Step 2: Substitute the known values into the formula.
Given:
- \( A_0 = 0.8 \, \text{M} \)
- \( A_t = 0.2 \, \text{M} \)
- \( t = 12 \, \text{hours} \)

Substituting these values into the formula:
\[ k = \frac{2.303}{12} \log \left( \frac{0.8}{0.2} \right) \]

### Step 3: Calculate the logarithm.
Calculate \( \frac{0.8}{0.2} \):
\[ \frac{0.8}{0.2} = 4 \]

Now, calculate \( \log(4) \):
Using the property of logarithms:
\[ \log(4) = \log(2^2) = 2 \log(2) \]

We know that \( \log(2) \approx 0.3010 \), so:
\[ \log(4) = 2 \times 0.3010 = 0.6020 \]

### Step 4: Substitute back to find k.
Now substitute \( \log(4) \) back into the equation for k:
\[ k = \frac{2.303}{12} \times 0.6020 \]

Calculating:
\[ k \approx \frac{2.303 \times 0.6020}{12} \]
\[ k \approx \frac{1.384306}{12} \]
\[ k \approx 0.11536 \, \text{hr}^{-1} \]

### Step 5: Use the rate constant to find the half-life (T_half).
The half-life for a first-order reaction is given by:
\[ T_{1/2} = \frac{0.693}{k} \]

Substituting the value of k:
\[ T_{1/2} = \frac{0.693}{0.11536} \]

Calculating:
\[ T_{1/2} \approx 6.01 \, \text{hours} \]

### Final Answer:
The half-life of the first-order reaction is approximately **6 hours**.

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