If rate `=k[H^(+)]^(n)` and it becomes 100 times when the pH changes from 2 to 1, then the order of reaction is

Options

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

Explanation

To solve the problem, we need to determine the order of the reaction based on the given information about the change in the rate when the pH changes from 2 to 1.

### Step-by-Step Solution:

1. **Understand the relationship between pH and [H⁺]:**
\[ \text{pH} = -\log[H^+] \]
Therefore, the concentration of hydrogen ions \([H^+]\) can be expressed as:
\[ [H^+] = 10^{-\text{pH}} \]

2. **Calculate [H⁺] at pH 2:**
\[ [H^+] \text{ at pH 2} = 10^{-2} \, \text{M} \]

3. **Calculate [H⁺] at pH 1:**
\[ [H^+] \text{ at pH 1} = 10^{-1} \, \text{M} \]

4. **Write the rate expressions for both pH values:**
- For pH 2:
\[ \text{Rate}_1 = k[H^+]^n = k(10^{-2})^n \]
- For pH 1:
\[ \text{Rate}_2 = k[H^+]^n = k(10^{-1})^n \]

5. **Set up the relationship based on the change in rate:**
According to the problem, the rate becomes 100 times when the pH changes from 2 to 1:
\[ \text{Rate}_2 = 100 \times \text{Rate}_1 \]

6. **Substitute the rate expressions:**
\[ k(10^{-1})^n = 100 \times k(10^{-2})^n \]

7. **Cancel \(k\) from both sides:**
\[ (10^{-1})^n = 100 \times (10^{-2})^n \]

8. **Rewrite 100 in terms of powers of 10:**
\[ 100 = 10^2 \]
So the equation becomes:
\[ (10^{-1})^n = 10^2 \times (10^{-2})^n \]

9. **Combine the terms on the right side:**
\[ (10^{-1})^n = 10^{2 - 2n} \]

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

11. **Solve for \(n\):**
\[ -n + 2n = 2 \implies n = 2 \]

12. **Determine the order of the reaction:**
The order of the reaction is equal to \(n\), which we found to be 2.

### Final Answer:
The order of the reaction is **2**.