If doubling the concentration of a reactant `'A'` increases the rate `4` times and tripling the concentration of `'A'` increases the rate `9` times, the rate is proportional to

Options

A. Concentration of 'A'
B. Square of concentration of 'A'
C. Under root of the concentration of 'A'
D. Cube of concentration of 'A'

Answer

To solve the problem, we need to determine the order of the reaction with respect to the reactant 'A'. We are given two conditions regarding how the rate of the reaction changes with the concentration of 'A'.

### Step-by-Step Solution:

1. **Understanding the Rate Law**:
The rate of a reaction can be expressed as:
\[ \text{Rate} = k [A]^n \]
where \( k \) is the rate constant, \([A]\) is the concentration of reactant A, and \( n \) is the order of the reaction with respect to A.

2. **First Condition**:
When the concentration of A is doubled (i.e., \([A] = 2[A_0]\)), the rate increases by 4 times:
\[ 4R = k (2[A_0])^n \]
Simplifying this gives:
\[ 4R = k \cdot 2^n \cdot [A_0]^n \]
Since \( R = k [A_0]^n \), we can substitute:
\[ 4R = 2^n R \]
Dividing both sides by \( R \) (assuming \( R \neq 0 \)):
\[ 4 = 2^n \]
This implies:
\[ 4 = 2^2 \Rightarrow n = 2 \]

3. **Second Condition**:
When the concentration of A is tripled (i.e., \([A] = 3[A_0]\)), the rate increases by 9 times:
\[ 9R = k (3[A_0])^n \]
Simplifying this gives:
\[ 9R = k \cdot 3^n \cdot [A_0]^n \]
Again substituting \( R = k [A_0]^n \):
\[ 9R = 3^n R \]
Dividing both sides by \( R \):
\[ 9 = 3^n \]
This implies:
\[ 9 = 3^2 \Rightarrow n = 2 \]

4. **Conclusion**:
From both conditions, we find that \( n = 2 \). Therefore, the rate is proportional to the square of the concentration of A:
\[ \text{Rate} \propto [A]^2 \]

### Final Answer:
The rate is proportional to the square of the concentration of A, or mathematically:
\[ \text{Rate} \propto [A]^2 \]