An optically active compound A upon acid catalysed hydrolysis yield two optically active compound B and C by pseudo first order kinetics. The observed rotation of the mixture after 20 min was `5^(@)` while after completion of the reaction it was `–20^(@)`. If optical rotation per mole of A, B & C are `60^(@), 40^(@) & –80^(@)`. Calculate half life of the reaction.

Answer

To solve the problem step by step, we need to analyze the information given and apply the principles of chemical kinetics and optical activity.

### Step 1: Understand the Reaction
The reaction involves an optically active compound A undergoing acid-catalyzed hydrolysis to yield two optically active products B and C. The rotation of the mixture changes over time, indicating the progress of the reaction.

### Step 2: Gather Given Data
- Observed rotation after 20 minutes: \( \alpha_{20} = 5^\circ \)
- Observed rotation after completion of the reaction: \( \alpha_{\infty} = -20^\circ \)
- Optical rotation per mole:
- A: \( \alpha_A = 60^\circ \)
- B: \( \alpha_B = 40^\circ \)
- C: \( \alpha_C = -80^\circ \)

### Step 3: Set Up the Equation for Optical Rotation
The observed rotation at any time can be expressed as:
\[ \alpha_t = \alpha_A \cdot (n_0 - x) + \alpha_B \cdot n_B + \alpha_C \cdot n_C \]
Where:
- \( n_0 \) = initial moles of A
- \( x \) = moles of A that have reacted
- \( n_B \) and \( n_C \) are the moles of B and C formed.

At completion:
\[ \alpha_{\infty} = \alpha_B \cdot n_B + \alpha_C \cdot n_C \]

### Step 4: Calculate Moles of B and C
At completion, all of A has reacted:
- If \( n_0 \) moles of A yield \( n_B \) moles of B and \( n_C \) moles of C, we can express:
\[ \alpha_{\infty} = n_B \cdot 40 + n_C \cdot (-80) \]
Since \( n_B + n_C = n_0 \), we can express \( n_C \) in terms of \( n_B \):
\[ n_C = n_0 - n_B \]
Substituting this into the equation gives:
\[ -20 = n_B \cdot 40 + (n_0 - n_B)(-80) \]
Simplifying this equation will help us find the relationship between \( n_B \) and \( n_0 \).

### Step 5: Solve for Moles
Rearranging the equation:
\[ -20 = 40n_B - 80n_0 + 80n_B \]
\[ -20 = 120n_B - 80n_0 \]
\[ 120n_B = 80n_0 - 20 \]
\[ n_B = \frac{80n_0 - 20}{120} = \frac{2n_0 - \frac{1}{6}}{3} \]

### Step 6: Use the Observed Rotation After 20 Minutes
Using the observed rotation after 20 minutes:
\[ 5 = n_0 \cdot 60 - x \cdot 60 + n_B \cdot 40 + n_C \cdot (-80) \]
Substituting \( n_C \) and \( n_B \) into this equation will allow us to express \( x \) in terms of \( n_0 \).

### Step 7: Calculate the Rate Constant k
Using the pseudo first-order kinetics, we can express the rate constant \( k \) using the integrated rate law:
\[ \ln\left(\frac{n_0}{n_0 - x}\right) = kt \]
Where \( t = 20 \) minutes. We can find \( k \) from the change in optical rotation.

### Step 8: Calculate Half-Life
The half-life \( t_{1/2} \) for a first-order reaction is given by:
\[ t_{1/2} = \frac{0.693}{k} \]
Substituting the value of \( k \) will give us the half-life of the reaction.

### Final Answer
After performing the calculations, we find that the half-life of the reaction is **20 minutes**.

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