The conversion of molecules `X` to `Y` follows second order kinetics. If the concenration of `X` is increased to three times, how will it affect the rate of formation of `Y` ?

Explanation

To solve the problem, we need to analyze how the rate of formation of product `Y` changes when the concentration of reactant `X` is increased to three times its original value, given that the reaction follows second-order kinetics.

### Step-by-Step Solution:

1. **Understand the Rate Law for Second Order Kinetics:**
The rate of a reaction that follows second-order kinetics can be expressed as:
\[ \text{Rate} = k [X]^2 \]
where \( k \) is the rate constant and \([X]\) is the concentration of reactant \( X \).

2. **Define Initial Concentration:**
Let the initial concentration of \( X \) be \( [X] = x \). Therefore, the initial rate of the reaction can be written as:
\[ R_1 = k [X]^2 = k x^2 \]

3. **Increase the Concentration of \( X \):**
According to the problem, the concentration of \( X \) is increased to three times its original value:
\[ [X] = 3x \]

4. **Calculate the New Rate:**
Now, substituting the new concentration into the rate law, we get the new rate \( R_2 \):
\[ R_2 = k [3X]^2 = k (3x)^2 \]
Simplifying this gives:
\[ R_2 = k \cdot 9x^2 = 9k x^2 \]

5. **Compare the New Rate with the Initial Rate:**
To find out how the new rate \( R_2 \) compares to the initial rate \( R_1 \), we can set up the following ratio:
\[ \frac{R_2}{R_1} = \frac{9k x^2}{k x^2} = 9 \]
Thus, we find that:
\[ R_2 = 9 R_1 \]

6. **Conclusion:**
The rate of formation of \( Y \) will increase by a factor of 9 when the concentration of \( X \) is increased to three times its original value.

### Final Answer:
The rate of formation of \( Y \) will increase 9 times.