The half life of a first order reaction is 10 minutes. If initial amount is 0.08 mole/litre and concentration at some instant ‘t’ is 0.01 mol/litre, then the value of ‘t’ is :

Options

A. 10 min.
B. 30 min.
C. 20 min.
D. 40 min.

Explanation

To solve the problem step by step, we will use the first-order reaction kinetics equations.

### Step 1: Identify the given data
- Half-life (t₁/₂) = 10 minutes
- Initial concentration (A₀) = 0.08 mol/L
- Concentration at time t (A) = 0.01 mol/L

### Step 2: Calculate the rate constant (k)
For a first-order reaction, the half-life is related to the rate constant (k) by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
Rearranging this gives:
\[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{10 \text{ min}} = 0.0693 \text{ min}^{-1} \]

### Step 3: Use the first-order rate equation
The first-order rate equation is given by:
\[ k t = 2.303 \log \frac{A_0}{A} \]
Substituting the values we have:
\[ 0.0693 \cdot t = 2.303 \log \frac{0.08}{0.01} \]

### Step 4: Calculate the logarithmic term
First, calculate the ratio:
\[ \frac{0.08}{0.01} = 8 \]
Now, find the logarithm:
\[ \log 8 = \log (2^3) = 3 \log 2 \]
Using \(\log 2 \approx 0.301\):
\[ \log 8 \approx 3 \cdot 0.301 = 0.903 \]

### Step 5: Substitute and solve for t
Now substitute back into the equation:
\[ 0.0693 \cdot t = 2.303 \cdot 0.903 \]
Calculating the right side:
\[ 2.303 \cdot 0.903 \approx 2.08 \]
Now solve for t:
\[ t = \frac{2.08}{0.0693} \approx 30.0 \text{ minutes} \]

### Final Answer
The value of \(t\) is approximately **30 minutes**.

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