The reaction `X+Y rarrZ` 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 formation of Z changes when the concentrations of X and Y are altered. The reaction is given as:

\[ X + Y \rightarrow Z \]

The rate law for this reaction is expressed as:

\[ \text{Rate} = k [X]^m [Y]^n \]

Where:
- \( m = 1 \) (first order with respect to X)
- \( n = 2 \) (second order with respect to Y)

Thus, the rate law becomes:

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

### Step 1: Calculate the initial rate with given concentrations

Given:
- \([X] = 0.40 \, \text{mol dm}^{-3}\)
- \([Y] = 0.30 \, \text{mol dm}^{-3}\)
- Initial rate of formation of Z = \( R \, \text{mol dm}^{-3} \text{sec}^{-1} \)

Substituting the values into the rate equation:

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

Calculating \((0.30)^2\):

\[ (0.30)^2 = 0.09 \]

Thus, the equation becomes:

\[ R = k (0.40)(0.09) \]

### Step 2: Modify the concentrations of X and Y

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: Calculate the new rate with modified concentrations

Substituting the new concentrations into the rate equation:

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

Substituting the new values:

\[ \text{Rate}_{\text{new}} = k (0.20)^1 (0.60)^2 \]

Calculating \((0.60)^2\):

\[ (0.60)^2 = 0.36 \]

Thus, the new rate becomes:

\[ \text{Rate}_{\text{new}} = k (0.20)(0.36) \]

### Step 4: Relate the new rate to the initial rate

Now, we can express the new rate in terms of the initial rate:

From the initial rate calculation, we had:

\[ R = k (0.40)(0.09) \]

Now, substituting back into the new rate:

\[ \text{Rate}_{\text{new}} = k (0.20)(0.36) \]

Now, we can express \( \text{Rate}_{\text{new}} \) in terms of \( R \):

\[ \text{Rate}_{\text{new}} = k (0.20)(0.36) = k (0.20)(0.36) = k (0.20)(0.36) \]

### Step 5: Calculate the ratio of new rate to initial rate

Now we can find the ratio of the new rate to the initial rate:

\[ \frac{\text{Rate}_{\text{new}}}{R} = \frac{k (0.20)(0.36)}{k (0.40)(0.09)} = \frac{(0.20)(0.36)}{(0.40)(0.09)} \]

Calculating the right side:

\[ = \frac{0.20 \times 0.36}{0.40 \times 0.09} = \frac{0.072}{0.036} = 2 \]

Thus, we find:

\[ \text{Rate}_{\text{new}} = 2R \]

### Final Answer

The new initial rate of formation of Z will be \( 2R \, \text{mol dm}^{-3} \text{sec}^{-1} \).

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