Two consecutive irreversible fierst order reactions can be represented by
`Aoverset(k_(1))(rarr)Boverset(k_(2))(rarr)C`
The rate equation for A is readily interated to obtain
`[A]_(t)=[A]_(0).e^(-k_(1^(t)))` , and `[B]_(t)=(k_(1)[A]_(0))/(k_(2)-k_(1))[e^(-k_(1)^(t))-e^(-k_(2)^(t))]`
At what time will B be present in maximum concentration ?

Options

A. `(K_(1))/(K_(2)-K_(1))`
B. `(1)/(k_(1)-k_(2))` In `k_(1))/(k_(2))`
C. `(1)/(k_(2)-k_(1))` In `k_(1))/(k_(2))`
D. none of these

Answer

To find the time at which the concentration of B is at its maximum in the consecutive irreversible first-order reactions represented by:

\[ A \overset{k_1}{\rightarrow} B \overset{k_2}{\rightarrow} C \]

we can follow these steps:

### Step 1: Write the expression for the concentration of B

The concentration of B at time t is given by:

\[ [B]_t = \frac{k_1 [A]_0}{k_2 - k_1} \left( e^{-k_1 t} - e^{-k_2 t} \right) \]

### Step 2: Differentiate the concentration of B with respect to time

To find the time at which B is at maximum concentration, we need to differentiate \([B]_t\) with respect to time \(t\) and set the derivative equal to zero:

\[ \frac{d[B]}{dt} = 0 \]

Differentiating \([B]_t\):

\[ \frac{d[B]}{dt} = \frac{k_1 [A]_0}{k_2 - k_1} \left( -k_1 e^{-k_1 t} + k_2 e^{-k_2 t} \right) \]

### Step 3: Set the derivative equal to zero

Setting the derivative equal to zero gives us:

\[ -k_1 e^{-k_1 t} + k_2 e^{-k_2 t} = 0 \]

Rearranging this equation, we have:

\[ k_1 e^{-k_1 t} = k_2 e^{-k_2 t} \]

### Step 4: Divide both sides by \(e^{-k_2 t}\)

Dividing both sides by \(e^{-k_2 t}\):

\[ k_1 e^{(k_2 - k_1)t} = k_2 \]

### Step 5: Solve for t

Taking the natural logarithm of both sides:

\[ \ln(k_1) + (k_2 - k_1)t = \ln(k_2) \]

Rearranging gives:

\[ (k_2 - k_1)t = \ln(k_2) - \ln(k_1) \]

Thus,

\[ t = \frac{\ln(k_2) - \ln(k_1)}{k_2 - k_1} \]

This can also be expressed as:

\[ t = \frac{1}{k_2 - k_1} \ln\left(\frac{k_2}{k_1}\right) \]

### Final Answer

The time at which B is present in maximum concentration is:

\[ t_{max} = \frac{1}{k_2 - k_1} \ln\left(\frac{k_2}{k_1}\right) \]

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