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 solve the problem, we need to determine the half-life (t1/2) of a first-order reaction given the initial concentration of substance A and its concentration after a certain time.

### Step 1: Write the first-order rate equation
For a first-order reaction, the relationship between the concentration of the reactant and time can be expressed using the following equation:

\[ \ln \left( \frac{[A]_0}{[A]} \right) = kt \]

where:
- \([A]_0\) is the initial concentration,
- \([A]\) is the concentration at time \(t\),
- \(k\) is the rate constant,
- \(t\) is the time.

### Step 2: Substitute the known values
From the problem, we know:
- \([A]_0 = 2.00 \, M\)
- \([A] = 0.15 \, M\)
- \(t = 200 \, \text{min}\)

Substituting these values into the equation gives:

\[ \ln \left( \frac{2.00}{0.15} \right) = k \times 200 \]

### Step 3: Calculate the left side
Now, calculate the value of \(\frac{2.00}{0.15}\):

\[ \frac{2.00}{0.15} = 13.33 \]

Now take the natural logarithm:

\[ \ln(13.33) \approx 2.5649 \]

### Step 4: Solve for the rate constant \(k\)
Now we can solve for \(k\):

\[ 2.5649 = k \times 200 \]

\[ k = \frac{2.5649}{200} \approx 0.0128245 \, \text{min}^{-1} \]

### Step 5: Calculate the half-life \(t_{1/2}\)
The half-life for a first-order reaction is given by the formula:

\[ t_{1/2} = \frac{0.693}{k} \]

Substituting the value of \(k\):

\[ t_{1/2} = \frac{0.693}{0.0128245} \approx 54.1 \, \text{min} \]

### Final Answer
The half-life \(t_{1/2}\) of the reaction is approximately **54.1 minutes**.

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