The rate for a first order reaction is `0.6932 xx 10^(-2) mol L^(-1) "min"^(-1)` and the initial concentration of the reactants is `1M`, `T_(1//2)` is equal to

Options

A. 6.932 min
B. 100 min
C. `0.6932 xx 10^(-3) min`
D. `0.6932 xx 10^(-2) min`

Answer

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

### Step 1: Identify the formula for half-life of a first-order reaction
The half-life of a first-order reaction is given by the formula:
\[ T_{1/2} = \frac{0.693}{k} \]
where \(k\) is the rate constant.

### Step 2: Determine the rate constant \(k\)
We know that the rate of a reaction for a first-order process can be expressed as:
\[ R = k \cdot [A] \]
where \(R\) is the rate of the reaction and \([A]\) is the concentration of the reactant.

Given:
- Rate \(R = 0.6932 \times 10^{-2} \, \text{mol L}^{-1} \text{min}^{-1}\)
- Initial concentration \([A] = 1 \, \text{mol L}^{-1}\)

We can rearrange the equation to find \(k\):
\[ k = \frac{R}{[A]} = \frac{0.6932 \times 10^{-2}}{1} = 0.6932 \times 10^{-2} \, \text{min}^{-1} \]

### Step 3: Substitute \(k\) into the half-life formula
Now that we have \(k\), we can substitute it into the half-life formula:
\[ T_{1/2} = \frac{0.693}{0.6932 \times 10^{-2}} \]

### Step 4: Calculate \(T_{1/2}\)
Calculating the above expression:
\[ T_{1/2} = \frac{0.693}{0.6932} \times 10^{2} \]
\[ T_{1/2} \approx 1 \times 10^{2} = 100 \, \text{minutes} \]

### Final Answer
Thus, the half-life \(T_{1/2}\) of the reaction is approximately **100 minutes**.

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