For a second order reaction rate at a particular time is X . IF the initial concentration is tripled is tripled the rate will become

Options

A. `3x`
B. `9x^(2)`
C. `9x`
D. `27x`

Explanation

To solve the problem, we need to understand how the rate of a second-order reaction changes with respect to the concentration of the reactant. Let's break it down step by step.

### Step-by-Step Solution:

1. **Understanding the Rate Law for a Second-Order Reaction**:
For a second-order reaction, the rate law can be expressed as:
\[ \text{Rate} = k[A]^2 \]
where \( k \) is the rate constant and \([A]\) is the concentration of the reactant A.

2. **Given Information**:
We are given that the rate at a particular time is \( x \). This means:
\[ x = k[A]^2 \]

3. **Tripling the Initial Concentration**:
If the initial concentration of A is tripled, we can express the new concentration as:
\[ [A]_{\text{new}} = 3[A] \]

4. **Substituting the New Concentration into the Rate Law**:
Now, we substitute the new concentration into the rate law:
\[ \text{Rate}_{\text{new}} = k[3A]^2 \]

5. **Calculating the New Rate**:
Expanding the equation:
\[ \text{Rate}_{\text{new}} = k(3[A])^2 = k \cdot 9[A]^2 \]
This can be rewritten as:
\[ \text{Rate}_{\text{new}} = 9k[A]^2 \]

6. **Relating the New Rate to the Original Rate**:
Since we know from step 2 that \( k[A]^2 = x \), we can substitute this into our new rate equation:
\[ \text{Rate}_{\text{new}} = 9x \]

7. **Conclusion**:
Therefore, when the initial concentration is tripled, the new rate becomes:
\[ \text{Rate}_{\text{new}} = 9x \]

### Final Answer:
The rate will become \( 9x \).

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