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. `0.6932 times 10^-2 min`
B. `0.6932 times 10^-3 `min
C. 100min
D. 6932min

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 given data
- Rate of the reaction, \( \text{Rate} = 0.6932 \times 10^{-2} \, \text{mol L}^{-1} \text{min}^{-1} \)
- Initial concentration, \( [A_0] = 1 \, \text{M} \)

### Step 2: Use the relationship for the rate of a first-order reaction
For a first-order reaction, the rate can be expressed as:
\[ \text{Rate} = k[A_0] \]
where \( k \) is the rate constant.

### Step 3: Solve for the rate constant \( k \)
Substituting the given values into the equation:
\[ 0.6932 \times 10^{-2} = k \times 1 \]
Thus, we find:
\[ k = 0.6932 \times 10^{-2} \, \text{min}^{-1} \]

### Step 4: Use the formula for half-life of a first-order reaction
The half-life \( T_{1/2} \) for a first-order reaction is given by:
\[ T_{1/2} = \frac{0.693}{k} \]

### Step 5: Substitute \( k \) into the half-life formula
Substituting the value of \( k \):
\[ T_{1/2} = \frac{0.693}{0.6932 \times 10^{-2}} \]

### Step 6: Simplify the expression
Calculating the above expression:
\[ T_{1/2} \approx \frac{0.693}{0.6932} \times 10^{2} \]
Since \( \frac{0.693}{0.6932} \) is approximately equal to 1, we have:
\[ T_{1/2} \approx 100 \, \text{minutes} \]

### Final Answer
Thus, the half-life \( T_{1/2} \) of the reaction is:
\[ T_{1/2} = 100 \, \text{minutes} \]