The reaction A B, obeys the rate law for pseudo first order kinetics in the presence of a large excess of hydroxide ion. If 90% of A undergoes reaction in half a minute when the reactant concentration is 0.002 M, What is the pseudo first order rate constant in `"min"^(–1)`

Options

A. `4.61`
B. `2.61`
C. `3.61`
D. `5.61`

Answer

To solve the problem, we need to determine the pseudo first-order rate constant (k') for the reaction A → B, given that 90% of A reacts in half a minute (0.5 minutes) when the concentration of A is 0.002 M.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The reaction A → B is said to follow pseudo first-order kinetics in the presence of a large excess of hydroxide ions. This means that the concentration of hydroxide ions remains constant and does not affect the rate significantly.

2. **Identifying the Given Information**:
- Percentage of A that reacts, x = 90%
- Time taken for this reaction, t = 0.5 minutes
- Initial concentration of A, [A]₀ = 0.002 M

3. **Using the First-Order Kinetics Formula**:
For a first-order reaction, the relationship between the percentage decomposition and the rate constant can be expressed as:
\[ t_x = \frac{2.303}{k} \log \left(\frac{100}{100 - x}\right) \]
where \( t_x \) is the time taken for x% decomposition.

4. **Substituting the Values**:
Here, we need to calculate for 90% decomposition:
- \( x = 90 \)
- \( 100 - x = 10 \)
- Substitute into the equation:
\[ 0.5 = \frac{2.303}{k} \log \left(\frac{100}{10}\right) \]

5. **Calculating the Logarithm**:
\[ \log \left(\frac{100}{10}\right) = \log(10) = 1 \]
Therefore, the equation simplifies to:
\[ 0.5 = \frac{2.303}{k} \cdot 1 \]

6. **Rearranging to Find k**:
Rearranging the equation gives:
\[ k = \frac{2.303}{0.5} \]

7. **Calculating k**:
\[ k = 2.303 \times 2 = 4.606 \text{ min}^{-1} \]

### Final Answer:
The pseudo first-order rate constant \( k' \) is **4.606 min\(^{-1}\)**.