The reaction `X+YrareZ` is first order with respect to X and second order with respect to Y, initial rate of formation of `Z=R mol " dm"^3 "sec"^-1` when [X] and [Y] are 0.40 mol `dm^(-3)` and 0.30 mol `dm^(-3)` respectively . If [X] is halved and [Y] is doubled , the value of the initial rate would become

Options

A. 4 R
B. `R/4`
C. R
D. 2R

Answer

To solve the problem, we need to determine how the initial rate of the reaction changes when the concentrations of reactants X and Y are altered. The reaction is given as:

\[ X + Y \rightarrow Z \]

The rate law for this reaction, based on the orders provided, is:

\[ \text{Rate} = k [X]^1 [Y]^2 \]

Where:
- \( k \) is the rate constant.
- The reaction is first order with respect to X and second order with respect to Y.

### Step 1: Write the initial rate expression
Given the initial concentrations:
- \([X] = 0.40 \, \text{mol dm}^{-3}\)
- \([Y] = 0.30 \, \text{mol dm}^{-3}\)

The initial rate of formation of Z is given as \( R \, \text{mol dm}^{-3} \text{s}^{-1} \).

Using the rate law, we can express the initial rate as:

\[ R = k [X]^1 [Y]^2 = k (0.40) (0.30)^2 \]

### Step 2: Calculate the new concentrations
Now, we need to find the new rate when:
- \([X]\) is halved: \([X] = \frac{0.40}{2} = 0.20 \, \text{mol dm}^{-3}\)
- \([Y]\) is doubled: \([Y] = 2 \times 0.30 = 0.60 \, \text{mol dm}^{-3}\)

### Step 3: Write the new rate expression
The new initial rate \( R' \) can be expressed as:

\[ R' = k [X']^1 [Y']^2 = k (0.20)^1 (0.60)^2 \]

### Step 4: Substitute the new concentrations into the rate expression
Substituting the values into the rate expression:

\[ R' = k (0.20) (0.60)^2 = k (0.20) (0.36) = k (0.072) \]

### Step 5: Relate the new rate to the original rate
Now we can relate \( R' \) to \( R \):

From the original rate:

\[ R = k (0.40) (0.30)^2 = k (0.40) (0.09) = k (0.036) \]

Now, we can express \( R' \) in terms of \( R \):

\[ R' = \frac{0.072}{0.036} R = 2R \]

### Conclusion
Thus, the new initial rate of formation of Z when the concentrations of X and Y are changed is:

\[ R' = 2R \]

### Final Answer
The value of the initial rate would become \( 2R \).

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