For any acid catalysed reaction, `Aoverset(H^(+))rarrB`
half- life period is independent of concentration of A at given pH. At definite concentration of A half- time is 10min at pH=2 and half- time is 100 min at pH=3. If the rate law expression of reaction is `r=k[A]^(x)[H^(+)]^(y)` then calulate the value of (x+y).

Answer

To solve the problem, we need to analyze the given information about the acid-catalyzed reaction \( A \overset{H^+}{\rightarrow} B \) and the relationship between half-life periods at different pH levels.

### Step-by-Step Solution:

1. **Understanding Half-Life Independence**:
- The problem states that the half-life period is independent of the concentration of \( A \) at a given pH. This characteristic is typical of first-order reactions, where the half-life is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
- Since the half-life is independent of the concentration of \( A \), we can conclude that the reaction with respect to \( A \) is first order. Therefore, we can assign:
\[ x = 1 \]

**Hint**: Recall that for first-order reactions, the half-life does not depend on the concentration of the reactant.

2. **Analyzing Half-Life at Different pH Levels**:
- We are given two half-life periods:
- At \( \text{pH} = 2 \), \( t_{1/2} = 10 \) minutes.
- At \( \text{pH} = 3 \), \( t_{1/2} = 100 \) minutes.
- The pH is related to the concentration of \( H^+ \) ions:
\[ \text{pH} = -\log[H^+] \]
- For \( \text{pH} = 2 \):
\[ [H^+] = 10^{-2} \, \text{M} \]
- For \( \text{pH} = 3 \):
\[ [H^+] = 10^{-3} \, \text{M} \]

3. **Relating Half-Life to \( H^+ \) Concentration**:
- The half-life changes from 10 minutes to 100 minutes when the concentration of \( H^+ \) decreases by a factor of 10 (from \( 10^{-2} \) M to \( 10^{-3} \) M).
- This indicates that the half-life is inversely proportional to the concentration of \( H^+ \):
\[ t_{1/2} \propto \frac{1}{[H^+]} \]

4. **Determining the Order with Respect to \( H^+ \)**:
- The relationship for the half-life in terms of the rate constant \( k \) and the concentration of \( H^+ \) can be expressed as:
\[ t_{1/2} = \frac{k'}{[H^+]^y} \]
- From the observed change in half-life, we can deduce:
- When \( [H^+] \) decreases by a factor of 10, \( t_{1/2} \) increases by a factor of 10.
- This suggests that \( y = 1 \) because the half-life is directly proportional to \( [H^+]^{-1} \).

5. **Final Calculation**:
- We have determined:
\[ x = 1 \quad \text{and} \quad y = 1 \]
- Therefore, the sum \( x + y \) is:
\[ x + y = 1 + 2 = 3 \]

### Conclusion:
The value of \( x + y \) is \( 3 \).