For a reaction `(d x)/(d t) = K[H^(+)]^(n)`. If `pH` of reaction medium changes from two to one rate becomes `100` times of value at `pH = 2`, The order of reaction is

Options

A. `1`
B. `2`
C. `0`
D. `3`

Explanation

To solve the problem, we need to determine the order of the reaction based on the change in pH and the corresponding change in the reaction rate.

### Step-by-Step Solution:

1. **Understand the relationship between pH and [H⁺]:**
- The pH of a solution is related to the concentration of hydrogen ions [H⁺] by the formula:
\[ \text{pH} = -\log[H⁺] \]
- Therefore, if pH = 2, then:
\[ [H⁺] = 10^{-2} \, \text{M} \]
- If pH = 1, then:
\[ [H⁺] = 10^{-1} \, \text{M} \]

2. **Write the rate expressions for both pH values:**
- The rate of the reaction is given by:
\[ \frac{dx}{dt} = K[H⁺]^n \]
- For pH = 2:
\[ R_1 = K[H⁺]^{n} = K(10^{-2})^{n} \]
- For pH = 1:
\[ R_2 = K[H⁺]^{n} = K(10^{-1})^{n} \]

3. **Set up the ratio of the rates:**
- According to the problem, the rate at pH = 1 is 100 times the rate at pH = 2:
\[ R_2 = 100 R_1 \]
- Substituting the expressions for \( R_1 \) and \( R_2 \):
\[ K(10^{-1})^{n} = 100 \times K(10^{-2})^{n} \]

4. **Cancel \( K \) from both sides:**
- Since \( K \) is a constant and non-zero, we can cancel it:
\[ (10^{-1})^{n} = 100 \times (10^{-2})^{n} \]

5. **Simplify the equation:**
- Rewrite \( 100 \) as \( 10^{2} \):
\[ (10^{-1})^{n} = 10^{2} \times (10^{-2})^{n} \]
- This simplifies to:
\[ 10^{-n} = 10^{2 - 2n} \]

6. **Set the exponents equal to each other:**
- Since the bases are the same, we can set the exponents equal:
\[ -n = 2 - 2n \]

7. **Solve for \( n \):**
- Rearranging gives:
\[ -n + 2n = 2 \]
\[ n = 2 \]

### Conclusion:
The order of the reaction \( n \) is **2**.

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