For a first order reaction A `to` B the reaction rate at reactant concentration of 0.01M is found to be `2.0 times10^–5 mol L^–1s^–1`. The half life period of the reaction is:

Options

A. 30sec
B. 220sec
C. 300sec
D. 347sec

Answer

To find the half-life period of a first-order reaction given the rate of reaction and concentration, we can follow these steps:

### Step 1: Identify the given data
- Rate of reaction (R) = \(2.0 \times 10^{-5} \, \text{mol L}^{-1} \text{s}^{-1}\)
- Concentration of reactant A ([A]) = \(0.01 \, \text{M}\)

### Step 2: Write the rate law for a first-order reaction
For a first-order reaction, the rate law can be expressed as:
\[ R = k \cdot [A] \]
where \(k\) is the rate constant.

### Step 3: Rearrange the equation to find the rate constant \(k\)
We can rearrange the rate law to solve for \(k\):
\[ k = \frac{R}{[A]} \]

### Step 4: Substitute the values into the equation
Substituting the given values:
\[ k = \frac{2.0 \times 10^{-5} \, \text{mol L}^{-1} \text{s}^{-1}}{0.01 \, \text{mol L}^{-1}} = \frac{2.0 \times 10^{-5}}{1.0 \times 10^{-2}} = 2.0 \times 10^{-3} \, \text{s}^{-1} \]

### Step 5: Use the half-life formula for a first-order reaction
The half-life (\(t_{1/2}\)) for a first-order reaction is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]

### Step 6: Substitute \(k\) into the half-life formula
Now substituting the value of \(k\):
\[ t_{1/2} = \frac{0.693}{2.0 \times 10^{-3}} = \frac{0.693}{0.002} = 346.5 \, \text{s} \]

### Step 7: Round off the answer
Rounding off \(346.5 \, \text{s}\) gives us approximately \(347 \, \text{s}\).

### Final Answer
The half-life period of the reaction is approximately **347 seconds**.

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