Two substances `A(T_((1)/(2))=10" min") & B (T_((1)/(2))="20 min")`follow I order kinetics in such a way that `[A]_(i)=8[B]_(j)`. Time when `[B] = 2[A]` in min is :

Options

A. 20
B. 40
C. 60
D. 80

Answer

To solve the problem step-by-step, we will use the information provided about the first-order kinetics of substances A and B, along with their half-lives.

### Step 1: Identify the given data
- Half-life of substance A, \( T_{1/2} = 10 \) min
- Half-life of substance B, \( T_{1/2} = 20 \) min
- Initial concentrations relationship: \( [A]_0 = 8[B]_0 \)

### Step 2: Calculate the rate constants \( k_1 \) and \( k_2 \)
The rate constant \( k \) for a first-order reaction is given by the formula:
\[ k = \frac{\ln 2}{T_{1/2}} \]
- For substance A:
\[ k_1 = \frac{\ln 2}{10} \]
- For substance B:
\[ k_2 = \frac{\ln 2}{20} \]

### Step 3: Write the concentration equations for A and B
Using the first-order kinetics equation:
\[ [A] = [A]_0 e^{-k_1 t} \]
\[ [B] = [B]_0 e^{-k_2 t} \]
Substituting the initial concentrations:
\[ [A] = 8[B]_0 e^{-k_1 t} \]
\[ [B] = [B]_0 e^{-k_2 t} \]

### Step 4: Set up the equation for the condition \( [B] = 2[A] \)
From the condition given in the problem:
\[ [B] = 2[A] \]
Substituting the expressions for [A] and [B]:
\[ [B]_0 e^{-k_2 t} = 2 \cdot (8[B]_0 e^{-k_1 t}) \]
Canceling \( [B]_0 \) from both sides:
\[ e^{-k_2 t} = 16 e^{-k_1 t} \]

### Step 5: Rearranging the equation
Taking logarithms on both sides:
\[ \ln(e^{-k_2 t}) = \ln(16) + \ln(e^{-k_1 t}) \]
This simplifies to:
\[ -k_2 t = \ln(16) - k_1 t \]
Rearranging gives:
\[ k_1 t - k_2 t = \ln(16) \]
Factoring out \( t \):
\[ t(k_1 - k_2) = \ln(16) \]
Thus:
\[ t = \frac{\ln(16)}{k_1 - k_2} \]

### Step 6: Substitute the values of \( k_1 \) and \( k_2 \)
Substituting \( k_1 \) and \( k_2 \):
\[ t = \frac{\ln(16)}{\frac{\ln 2}{10} - \frac{\ln 2}{20}} \]
Calculating the denominator:
\[ k_1 - k_2 = \frac{\ln 2}{10} - \frac{\ln 2}{20} = \ln 2 \left(\frac{1}{10} - \frac{1}{20}\right) = \ln 2 \left(\frac{2 - 1}{20}\right) = \frac{\ln 2}{20} \]
Now substituting back:
\[ t = \frac{\ln(16)}{\frac{\ln 2}{20}} = 20 \cdot \ln(16) / \ln(2) \]
Since \( \ln(16) = \ln(2^4) = 4 \ln(2) \):
\[ t = 20 \cdot \frac{4 \ln(2)}{\ln(2)} = 80 \text{ min} \]

### Final Answer
The time when \( [B] = 2[A] \) is **80 minutes**.