A substance 'A' decomposes by a first order reaction starting initially with `[A]= 2.00M` and after 200 min, [A] becomes `0.15M`. For this reaction, `t1//2` is

Options

A. `53.49 min`
B. `50.49 min`
C. `48.45 min`
D. `46.45 min`

Explanation

To find the half-life (\(t_{1/2}\)) of a first-order reaction given the initial concentration and the concentration after a certain time, we can follow these steps:

### Step 1: Write down the first-order rate equation
The rate constant \(K\) for a first-order reaction can be calculated using the formula:
\[ K = \frac{2.303}{T} \log \left( \frac{[A_0]}{[A]} \right) \]
where:
- \([A_0]\) = initial concentration
- \([A]\) = concentration after time \(T\)

### Step 2: Substitute the known values
Given:
- \([A_0] = 2.00 \, M\)
- \([A] = 0.15 \, M\)
- \(T = 200 \, \text{min}\)

Substituting these values into the equation:
\[ K = \frac{2.303}{200} \log \left( \frac{2.00}{0.15} \right) \]

### Step 3: Calculate the logarithm
Calculate \(\frac{2.00}{0.15}\):
\[ \frac{2.00}{0.15} = 13.33 \]
Now, calculate \(\log(13.33)\):
\[ \log(13.33) \approx 1.13 \]

### Step 4: Substitute back to find \(K\)
Now substitute \(\log(13.33)\) back into the equation for \(K\):
\[ K = \frac{2.303}{200} \times 1.13 \]
Calculating \(K\):
\[ K \approx \frac{2.303 \times 1.13}{200} \approx \frac{2.60539}{200} \approx 0.01303 \, \text{min}^{-1} \]

### Step 5: 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 \(K\):
\[ t_{1/2} = \frac{0.693}{0.01303} \approx 53.3 \, \text{min} \]

### Final Answer
Thus, the half-life (\(t_{1/2}\)) of the reaction is approximately:
\[ t_{1/2} \approx 53.3 \, \text{minutes} \]