The reaction between X and Y is first order with respect to X and second order with respect to Y. If the concentration of X is halved and the concentration of Y is doubled, the rate of reaction will be

Options

A. the same as the initial value
B. three times the initial value
C. double the initial value
D. half the initial value.

Explanation

To solve the problem, we need to analyze the effect of changing the concentrations of reactants X and Y on the rate of the reaction.

### Step-by-Step Solution:

1. **Write the Rate Law Expression**:
The rate of a reaction can be expressed using the rate law. For the given reaction, the rate \( r \) can be expressed as:
\[ r = k [X]^1 [Y]^2 \]
where \( k \) is the rate constant, \( [X] \) is the concentration of X, and \( [Y] \) is the concentration of Y.

2. **Define Initial Concentrations**:
Let the initial concentrations of X and Y be \( [X] = x \) and \( [Y] = y \). Thus, the initial rate \( r_0 \) can be written as:
\[ r_0 = k [X]^1 [Y]^2 = k x^1 y^2 = kxy^2 \]

3. **Change the Concentrations**:
According to the problem, the concentration of X is halved and the concentration of Y is doubled. Therefore, the new concentrations will be:
\[ [X] = \frac{x}{2}, \quad [Y] = 2y \]

4. **Calculate the New Rate**:
Substitute the new concentrations into the rate law expression to find the new rate \( r' \):
\[ r' = k \left(\frac{x}{2}\right)^1 (2y)^2 \]
Simplifying this gives:
\[ r' = k \left(\frac{x}{2}\right)(4y^2) = k \frac{4xy^2}{2} = 2kxy^2 \]

5. **Relate New Rate to Initial Rate**:
From the initial rate expression \( r_0 = kxy^2 \), we can see that:
\[ r' = 2r_0 \]
This means the new rate \( r' \) is twice the initial rate \( r_0 \).

### Conclusion:
Thus, if the concentration of X is halved and the concentration of Y is doubled, the rate of reaction will be **double the initial rate**.

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