For an elementary reaction , `X(g)toY(g)+Z(g)`
the half life period is 10 min. In what period of time would the concentration of X be reduced to 10% of original concentration?

Options

A. 20 min
B. 33 min
C. 15 min
D. 25 min

Answer

To solve the problem of how long it takes for the concentration of X to be reduced to 10% of its original concentration in the reaction \( X(g) \to Y(g) + Z(g) \) with a half-life of 10 minutes, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction Order**:
The reaction is an elementary reaction, which means it follows first-order kinetics.

2. **Use the Half-Life Formula**:
The half-life (\( T_{1/2} \)) for a first-order reaction is given by the formula:
\[ T_{1/2} = \frac{0.693}{k} \]
where \( k \) is the rate constant.

3. **Calculate the Rate Constant \( k \)**:
Given that the half-life is 10 minutes:
\[ k = \frac{0.693}{T_{1/2}} = \frac{0.693}{10 \text{ min}} = 0.0693 \text{ min}^{-1} \]

4. **Determine the Time for 10% Concentration**:
We want to find the time \( T \) when the concentration of \( X \) is reduced to 10% of its original concentration. If we denote the initial concentration as \( A \), then:
\[ A - X = 0.1A \implies X = 0.9A \]

5. **Use the First-Order Kinetics Equation**:
The first-order kinetics equation is given by:
\[ T = \frac{2.303}{k} \log\left(\frac{A}{A - X}\right) \]
Substituting the values:
\[ T = \frac{2.303}{0.0693} \log\left(\frac{A}{0.1A}\right) = \frac{2.303}{0.0693} \log(10) \]

6. **Calculate \( \log(10) \)**:
Since \( \log(10) = 1 \):
\[ T = \frac{2.303}{0.0693} \times 1 \approx 33.23 \text{ minutes} \]

7. **Final Answer**:
Therefore, the time required for the concentration of \( X \) to be reduced to 10% of its original concentration is approximately **33.23 minutes**.