A flask contains a mixture of compounds A and B . Both compounds decompose by first - order kinetics . The half - lives for A and B are 300 s and 180 s , respectively . If the concentration of A and B are equal initially , the time required for the concentration of A to be four times that of B (in s ) is : ( Use In 2 : 0.693)

Options

A. 180
B. 900
C. 300
D. 120

Answer

To solve the problem, we need to analyze the first-order kinetics of the decomposition of compounds A and B. Given their half-lives, we can calculate their concentrations over time and determine when the concentration of A is four times that of B.

### Step-by-Step Solution:

1. **Understanding First-Order Kinetics**:
- The half-life (t₁/₂) of a first-order reaction is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
- Where \( k \) is the rate constant.

2. **Calculate Rate Constants**:
- For compound A:
\[ t_{1/2} = 300 \, \text{s} \implies k_A = \frac{0.693}{300} \approx 0.00231 \, \text{s}^{-1} \]
- For compound B:
\[ t_{1/2} = 180 \, \text{s} \implies k_B = \frac{0.693}{180} \approx 0.00385 \, \text{s}^{-1} \]

3. **Initial Concentrations**:
- Let the initial concentration of both A and B be \( C \).
- At time \( t = 0 \):
\[ [A]_0 = C, \quad [B]_0 = C \]

4. **Concentration as a Function of Time**:
- The concentration of A at time \( t \):
\[ [A] = C e^{-k_A t} \]
- The concentration of B at time \( t \):
\[ [B] = C e^{-k_B t} \]

5. **Condition for A to be Four Times B**:
- We need to find the time \( t \) when:
\[ [A] = 4[B] \]
- Substituting the expressions for concentrations:
\[ C e^{-k_A t} = 4 \left( C e^{-k_B t} \right) \]
- Dividing both sides by \( C \):
\[ e^{-k_A t} = 4 e^{-k_B t} \]
- Taking the natural logarithm of both sides:
\[ -k_A t = \ln(4) - k_B t \]
- Rearranging gives:
\[ (k_B - k_A) t = \ln(4) \]
- Thus, we can solve for \( t \):
\[ t = \frac{\ln(4)}{k_B - k_A} \]

6. **Substituting Values**:
- We know \( \ln(4) = 2 \ln(2) = 2 \times 0.693 = 1.386 \).
- Now substituting \( k_A \) and \( k_B \):
\[ t = \frac{1.386}{0.00385 - 0.00231} = \frac{1.386}{0.00154} \approx 900 \, \text{s} \]

### Final Answer:
The time required for the concentration of A to be four times that of B is **900 seconds**.