If `(dx)/(dt)=k[H_(3)O^(+)]n`
and rate becomes 100 times when pH changes from 2 to 1 Hence order of reactions is

Options

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

Explanation

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

### Step-by-Step Solution:

1. **Understanding pH and [H₃O⁺] Relationship**:
The pH is defined as:
\[ \text{pH} = -\log[H^+] \]
Therefore, the concentration of hydronium ions \([H_3O^+]\) can be expressed as:
\[ [H_3O^+] = 10^{-\text{pH}} \]

2. **Calculating [H₃O⁺] at pH 2**:
When pH = 2:
\[ [H_3O^+] = 10^{-2} \, \text{M} \]

3. **Calculating [H₃O⁺] at pH 1**:
When pH = 1:
\[ [H_3O^+] = 10^{-1} \, \text{M} \]

4. **Determining the Rate Change**:
According to the problem, the rate of reaction becomes 100 times greater when the pH changes from 2 to 1:
\[ R_2 = 100 \times R_1 \]

5. **Expressing Rates in Terms of [H₃O⁺]**:
The rate of reaction can be expressed as:
\[ R = k[H_3O^+]^n \]
Thus, we have:
- For pH 2 (where \([H_3O^+] = 10^{-2}\)):
\[ R_1 = k(10^{-2})^n \]
- For pH 1 (where \([H_3O^+] = 10^{-1}\)):
\[ R_2 = k(10^{-1})^n \]

6. **Setting Up the Equation**:
From the rate change, we can set up the equation:
\[ k(10^{-1})^n = 100 \times k(10^{-2})^n \]

7. **Simplifying the Equation**:
Dividing both sides by \(k\) (assuming \(k \neq 0\)):
\[ (10^{-1})^n = 100 \times (10^{-2})^n \]
This simplifies to:
\[ 10^{-n} = 100 \times 10^{-2n} \]

8. **Expressing 100 as a Power of 10**:
We know that:
\[ 100 = 10^2 \]
Thus, we can rewrite the equation as:
\[ 10^{-n} = 10^2 \times 10^{-2n} \]
Which simplifies to:
\[ 10^{-n} = 10^{2 - 2n} \]

9. **Equating Exponents**:
Since the bases are the same, we can equate the exponents:
\[ -n = 2 - 2n \]

10. **Solving for n**:
Rearranging gives:
\[ -n + 2n = 2 \]
\[ n = 2 \]

### Conclusion:
The order of the reaction is **2**.