If doubling the concentration of a reactant `X` in a reaction `X+Y rarr` Products, increases the rate four times and tripling its concentration increases the rate nine times, this indicates that the rate of reaction is proportional to the ………….of the concentration of.............and thus rate is given by........... .

Answer

To solve the problem, we need to analyze how the rate of reaction changes with varying concentrations of reactant X.

### Step-by-Step Solution:

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

2. **Doubling the Concentration of X**:
When the concentration of X is doubled, we have:
\[ \text{Rate}_1 = k (2[X])^n [Y]^m \]
The problem states that this increases the rate four times:
\[ \text{Rate}_1 = 4 \times \text{Rate}_0 \]
So we can write:
\[ 4 \times \text{Rate}_0 = k (2[X])^n [Y]^m \]
Dividing both sides by \(\text{Rate}_0 = k [X]^n [Y]^m\) gives:
\[ 4 = \frac{(2[X])^n}{[X]^n} = 2^n \]
This implies:
\[ 2^n = 4 \implies n = 2 \]

3. **Tripling the Concentration of X**:
Next, when the concentration of X is tripled, we have:
\[ \text{Rate}_2 = k (3[X])^n [Y]^m \]
The problem states that this increases the rate nine times:
\[ \text{Rate}_2 = 9 \times \text{Rate}_0 \]
So we can write:
\[ 9 \times \text{Rate}_0 = k (3[X])^n [Y]^m \]
Dividing both sides by \(\text{Rate}_0\) gives:
\[ 9 = \frac{(3[X])^n}{[X]^n} = 3^n \]
This implies:
\[ 3^n = 9 \implies n = 2 \]

4. **Conclusion about Y**:
Since the rate changes with the concentration of X but does not depend on Y (as seen from the calculations), we can conclude that the order with respect to Y is zero:
\[ m = 0 \]

5. **Final Rate Law**:
Therefore, the rate law can be expressed as:
\[ \text{Rate} = k [X]^2 \]

### Final Answers:
- The rate of reaction is proportional to the **square** of the concentration of **X**.
- Thus, the rate is given by: **Rate = k [X]^2**.