If `((dx)/(dt))=k[H^(+)]^(n)` and rate becomes 100 times when pH change from 2 to 1. Hence, order is :

Options

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

Answer

To solve the problem, we need to determine the order of the reaction based on the given rate law and the change in pH. Let's break down the steps:

### Step-by-Step Solution:

1. **Understand the Rate Law**:
The rate of the reaction is given by the equation:
\[ \frac{dx}{dt} = k[H^+]^n \]
where \( k \) is the rate constant, \( [H^+] \) is the concentration of hydrogen ions, and \( n \) is the order of the reaction.

2. **Relate pH to Hydrogen Ion Concentration**:
The pH is defined as:
\[ \text{pH} = -\log[H^+] \]
Therefore, the concentration of hydrogen ions can be expressed as:
\[ [H^+] = 10^{-\text{pH}} \]

3. **Calculate Hydrogen Ion Concentrations at Different pH Values**:
- For pH = 2:
\[ [H^+]_2 = 10^{-2} \, \text{M} \]
- For pH = 1:
\[ [H^+]_1 = 10^{-1} \, \text{M} \]

4. **Determine the Change in Rate**:
According to the problem, the rate becomes 100 times greater when the pH changes from 2 to 1. Therefore:
\[ \text{Rate at pH 1} = 100 \times \text{Rate at pH 2} \]

5. **Set Up the Rate Equations**:
Using the rate law for both pH values:
- At pH = 2:
\[ \text{Rate at pH 2} = k(10^{-2})^n \]
- At pH = 1:
\[ \text{Rate at pH 1} = k(10^{-1})^n \]

6. **Equate the Rates**:
From the information given:
\[ k(10^{-1})^n = 100 \times k(10^{-2})^n \]
We can cancel \( k \) from both sides:
\[ (10^{-1})^n = 100 \times (10^{-2})^n \]

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

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

9. **Solve for n**:
Rearranging gives:
\[ -n + 2n = 2 \implies n = 2 \]

10. **Conclusion**:
The order of the reaction is \( n = 2 \), indicating that it is a second-order reaction.

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