If doubling the concentration of a reactant ‘A’ increases the rate 8 times and tripling the concentration of ‘A’ increases the rate 27 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 how the rate of reaction is related to the concentration of reactant A. We will use the information given about how changing the concentration affects the rate.

### Step-by-Step Solution:

1. **Understanding the Rate Law**:
The rate of a reaction can be expressed using the rate law:
\[ r = k [A]^n \]
where \( r \) is the rate, \( k \) is the rate constant, \([A]\) is the concentration of reactant A, and \( n \) is the order of the reaction.

2. **Analyzing the First Condition**:
When the concentration of A is doubled (from \([A]\) to \(2[A]\)), the rate increases 8 times:
\[ 8r = k (2[A])^n \]
Simplifying this gives:
\[ 8r = k \cdot 2^n \cdot [A]^n \]
Dividing both sides by \( r \):
\[ 8 = 2^n \]
Taking logarithm base 2 on both sides:
\[ n = 3 \]

3. **Analyzing the Second Condition**:
When the concentration of A is tripled (from \([A]\) to \(3[A]\)), the rate increases 27 times:
\[ 27r = k (3[A])^n \]
Simplifying this gives:
\[ 27r = k \cdot 3^n \cdot [A]^n \]
Dividing both sides by \( r \):
\[ 27 = 3^n \]
Taking logarithm base 3 on both sides:
\[ n = 3 \]

4. **Conclusion**:
Both conditions consistently indicate that the order of the reaction \( n \) is equal to 3. Therefore, the rate is proportional to the cube of the concentration of A:
\[ r \propto [A]^3 \]

### Final Answer:
The rate is proportional to the cube of the concentration of A, or mathematically:
\[ r \propto [A]^3 \]