For any reaction, `2A rarr B`, rate constant of reaction is `0.231" min"^(–1)`. Time (in sec) when 25% of A will remain unreacted.

Options

A. 150
B. 180
C. 200
D. 140

Explanation

To solve the problem, we need to determine the time when 25% of reactant A remains unreacted in the reaction \(2A \rightarrow B\) with a rate constant \(k = 0.231 \, \text{min}^{-1}\).

### Step 1: Understand the Reaction Order
The reaction given is \(2A \rightarrow B\), which is a first-order reaction with respect to A. The rate constant is given in \( \text{min}^{-1} \).

### Step 2: Determine the Initial and Final Concentrations
Let the initial concentration of A be \(C_0\). If 25% of A remains unreacted, then 75% of A has reacted. Thus, the final concentration \(C_t\) when 25% remains is:
\[ C_t = C_0 - 0.75C_0 = 0.25C_0 \]

### Step 3: Use the First-Order Kinetics Equation
For a first-order reaction, the integrated rate law is given by:
\[ \ln \left( \frac{C_t}{C_0} \right) = -kt \]
Substituting the values we have:
\[ \ln \left( \frac{0.25C_0}{C_0} \right) = -kt \]
This simplifies to:
\[ \ln(0.25) = -kt \]

### Step 4: Calculate \(\ln(0.25)\)
We know that:
\[ \ln(0.25) = \ln\left(\frac{1}{4}\right) = \ln(1) - \ln(4) = 0 - \ln(4) = -\ln(4) \]
Using the property of logarithms, we can express \(\ln(4)\) as:
\[ \ln(4) = 2\ln(2) \approx 2 \times 0.693 = 1.386 \]
Thus:
\[ \ln(0.25) \approx -1.386 \]

### Step 5: Substitute the Values into the Equation
Now substituting back into the equation:
\[ -1.386 = -0.231 \cdot t \]
Solving for \(t\):
\[ t = \frac{1.386}{0.231} \approx 6.0 \, \text{minutes} \]

### Step 6: Convert Time to Seconds
To convert minutes into seconds:
\[ t = 6.0 \times 60 = 360 \, \text{seconds} \]

### Final Answer
The time when 25% of A will remain unreacted is approximately **360 seconds**.

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