The half life of a first order reaction is 6 hours. How long will it take for the concentration of reactant to change from 0.8 M to 0.25 M ?

Options

A. 1.07 hour
B. 5.1 hour
C. 2.7 hour
D. 10.07 hour

Answer

To solve the problem of how long it will take for the concentration of a reactant to change from 0.8 M to 0.25 M in a first-order reaction with a half-life of 6 hours, we can follow these steps:

### Step 1: Determine the rate constant (k) using the half-life formula for first-order reactions.
The half-life (T_half) of a first-order reaction is given by the formula:
\[ T_{1/2} = \frac{0.693}{k} \]
Given that the half-life is 6 hours, we can rearrange the formula to find k:
\[ k = \frac{0.693}{T_{1/2}} = \frac{0.693}{6 \text{ hours}} \approx 0.1155 \text{ hours}^{-1} \]

### Step 2: Use the first-order kinetics equation to find the time required for the concentration change.
The integrated rate law for a first-order reaction is:
\[ \ln\left(\frac{[A]_0}{[A]}\right) = kt \]
Where:
- \([A]_0\) is the initial concentration (0.8 M)
- \([A]\) is the final concentration (0.25 M)
- \(k\) is the rate constant (0.1155 hours\(^{-1}\))
- \(t\) is the time we want to find

Substituting the values into the equation:
\[ \ln\left(\frac{0.8}{0.25}\right) = 0.1155 \cdot t \]

### Step 3: Calculate the ratio of concentrations.
Calculating the ratio:
\[ \frac{0.8}{0.25} = 3.2 \]
Now, we take the natural logarithm:
\[ \ln(3.2) \approx 1.163 \]

### Step 4: Solve for time (t).
Now we can substitute back into the equation:
\[ 1.163 = 0.1155 \cdot t \]
Rearranging gives:
\[ t = \frac{1.163}{0.1155} \approx 10.07 \text{ hours} \]

### Conclusion:
The time required for the concentration of the reactant to change from 0.8 M to 0.25 M is approximately **10.07 hours**.

---