Conisder a reaction `X + Y rarr` Products. If the initial concentration of `X` increased to four times of its original value, keeping the concentration of `Y` constant, the rate of reaction increases four-fold. When the concentration of both `X` and `Y` becomes four times their original values the rate of reaction becomes `16` times its original values. The observed rate law is

Options

A. `k[X]^(2)[Y]^(2)`
B. `k[X]^(1)[Y]^(2)`
C. `k[X]^(1)[Y]^(1)`
D. `k[X]^(2)[Y]^(1)`

Explanation

To solve the problem, we need to determine the rate law for the reaction \( X + Y \rightarrow \text{Products} \) based on the information provided.

### Step-by-Step Solution:

1. **Understanding the Rate Law**:
The rate of a reaction can be expressed by the rate law equation:
\[ \text{Rate} = k [X]^A [Y]^B \]
where \( k \) is the rate constant, \( [X] \) and \( [Y] \) are the concentrations of reactants \( X \) and \( Y \), and \( A \) and \( B \) are the orders of the reaction with respect to \( X \) and \( Y \), respectively.

2. **Analyzing the First Condition**:
The first condition states that when the concentration of \( X \) is increased to four times its original value (keeping \( [Y] \) constant), the rate increases four-fold.
- Let the initial concentration of \( X \) be \( [X]_0 \) and that of \( Y \) be \( [Y]_0 \).
- The initial rate can be expressed as:
\[ \text{Rate} = k [X]_0^A [Y]_0^B \]
- When \( [X] \) becomes \( 4[X]_0 \):
\[ \text{New Rate} = k (4[X]_0)^A [Y]_0^B = k 4^A [X]_0^A [Y]_0^B \]
- According to the problem, this new rate is four times the original rate:
\[ k 4^A [X]_0^A [Y]_0^B = 4 \cdot k [X]_0^A [Y]_0^B \]
- Dividing both sides by \( k [X]_0^A [Y]_0^B \):
\[ 4^A = 4 \]
- This implies:
\[ A = 1 \]

3. **Analyzing the Second Condition**:
The second condition states that when both \( [X] \) and \( [Y] \) are increased to four times their original values, the rate becomes 16 times its original value.
- The new concentrations are \( [X] = 4[X]_0 \) and \( [Y] = 4[Y]_0 \):
\[ \text{New Rate} = k (4[X]_0)^A (4[Y]_0)^B = k 4^A [X]_0^A 4^B [Y]_0^B = k 4^{A+B} [X]_0^A [Y]_0^B \]
- According to the problem, this new rate is 16 times the original rate:
\[ k 4^{A+B} [X]_0^A [Y]_0^B = 16 \cdot k [X]_0^A [Y]_0^B \]
- Dividing both sides by \( k [X]_0^A [Y]_0^B \):
\[ 4^{A+B} = 16 \]
- Since \( 16 = 4^2 \):
\[ A + B = 2 \]

4. **Finding the Value of B**:
We already found \( A = 1 \). Substituting this into the equation \( A + B = 2 \):
\[ 1 + B = 2 \implies B = 1 \]

5. **Final Rate Law**:
Substituting the values of \( A \) and \( B \) back into the rate law:
\[ \text{Rate} = k [X]^1 [Y]^1 = k [X] [Y] \]
Therefore, the observed rate law is:
\[ \text{Rate} = k [X][Y] \]

### Conclusion:
The observed rate law for the reaction \( X + Y \rightarrow \text{Products} \) is:
\[ \text{Rate} = k [X][Y] \]