Half life of a certain zero order reaction, A `rightarrow` P is 2 hour when the initial concentration of the reactant, 'A' is 4 mol `L^(-1)`. The time required for its concentration to change from 0.40 to 0.20 mol `L^(-1)` is

Options

A. 0.75 min
B. 2 min
C. 12 min
D. 8 min

Answer

To solve the problem, we need to find the time required for the concentration of reactant A to change from 0.40 mol L⁻¹ to 0.20 mol L⁻¹ in a zero-order reaction. Here’s a step-by-step solution:

### Step 1: Understand the Zero-Order Reaction
For a zero-order reaction, the rate of reaction is constant and does not depend on the concentration of the reactant. The integrated rate equation for a zero-order reaction is given by:
\[ [A]_t = [A]_0 - k \cdot t \]
where:
- \([A]_t\) = concentration at time \(t\)
- \([A]_0\) = initial concentration
- \(k\) = rate constant
- \(t\) = time

### Step 2: Calculate the Rate Constant \(k\)
We know the half-life (\(t_{1/2}\)) of the reaction is 2 hours when the initial concentration \([A]_0\) is 4 mol L⁻¹. The relationship for half-life in a zero-order reaction is:
\[ t_{1/2} = \frac{[A]_0}{2k} \]
From this, we can rearrange to find \(k\):
\[ k = \frac{[A]_0}{2 \cdot t_{1/2}} \]

Substituting the values:
\[ k = \frac{4 \, \text{mol L}^{-1}}{2 \cdot 2 \, \text{hours}} = \frac{4}{4} = 1 \, \text{mol L}^{-1} \text{hour}^{-1} \]

### Step 3: Set Up the Equation for the Concentration Change
Now we need to find the time required for the concentration to change from 0.40 mol L⁻¹ to 0.20 mol L⁻¹. Using the integrated rate equation:
\[ [A]_t = [A]_0 - k \cdot t \]

Here, \([A]_0 = 0.40 \, \text{mol L}^{-1}\) and \([A]_t = 0.20 \, \text{mol L}^{-1}\). Plugging in the values:
\[ 0.20 = 0.40 - 1 \cdot t \]

### Step 4: Solve for Time \(t\)
Rearranging the equation to solve for \(t\):
\[ t = 0.40 - 0.20 \]
\[ t = 0.20 \, \text{hours} \]

### Step 5: Convert Time to Minutes
Since the answer is required in minutes, we convert hours to minutes:
\[ t = 0.20 \, \text{hours} \times 60 \, \text{minutes/hour} = 12 \, \text{minutes} \]

### Final Answer
The time required for the concentration of A to change from 0.40 mol L⁻¹ to 0.20 mol L⁻¹ is **12 minutes**.

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