How many times will the rate of the elementary reaction 3X + Y → X2Y change if the concentration of the substance X is doubled and that of Y is halved?

Options

A. r2= 4.5r1
B. r2= 5r1
C. r2= 4r1
D. r2= 2r1

Answer

To solve the problem, we need to determine how the rate of the reaction changes when the concentration of substance X is doubled and the concentration of substance Y is halved.

The reaction is given as:
\[ 3X + Y \rightarrow X_2Y \]

### Step 1: Write the rate law for the reaction
For an elementary reaction, the rate law can be expressed as:
\[ r = k [X]^m [Y]^n \]
where \( m \) and \( n \) are the stoichiometric coefficients of the reactants in the balanced equation. Here, \( m = 3 \) for \( X \) and \( n = 1 \) for \( Y \).

Thus, the rate law for this reaction is:
\[ r = k [X]^3 [Y]^1 \]

### Step 2: Define the initial concentrations
Let:
- The initial concentration of \( X \) be \( [X] \)
- The initial concentration of \( Y \) be \( [Y] \)

The initial rate \( r_1 \) can be expressed as:
\[ r_1 = k [X]^3 [Y] \]

### Step 3: Determine the new concentrations
According to the problem:
- The concentration of \( X \) is doubled: \( [X] \) becomes \( 2[X] \)
- The concentration of \( Y \) is halved: \( [Y] \) becomes \( \frac{1}{2}[Y] \)

### Step 4: Write the new rate expression
The new rate \( r_2 \) with the changed concentrations can be expressed as:
\[ r_2 = k (2[X])^3 \left(\frac{1}{2}[Y]\right) \]

### Step 5: Simplify the new rate expression
Now, simplify \( r_2 \):
\[ r_2 = k (2^3 [X]^3) \left(\frac{1}{2}[Y]\right) \]
\[ r_2 = k (8 [X]^3) \left(\frac{1}{2}[Y]\right) \]
\[ r_2 = k \cdot 8 [X]^3 \cdot \frac{1}{2} [Y] \]
\[ r_2 = k \cdot 4 [X]^3 [Y] \]

### Step 6: Relate the new rate to the initial rate
Now we can relate \( r_2 \) to \( r_1 \):
\[ r_2 = 4 (k [X]^3 [Y]) \]
\[ r_2 = 4 r_1 \]

### Conclusion
Thus, the rate of the reaction changes by a factor of 4 when the concentration of \( X \) is doubled and the concentration of \( Y \) is halved.

### Final Answer
The rate of the reaction changes by a factor of 4.

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