The rate of reaction between `A` and `B` increases by a factor of `100`, when the concentration with respect to `A` is increased `10` folds, the order of reaction w.r.t. `A` is

Options

A. `10`
B. `1`
C. `4`
D. `2`

Explanation

To determine the order of reaction with respect to reactant A, we can follow these steps:

### Step 1: Understand the relationship between rate and concentration
The rate of a reaction can be expressed as:
\[ \text{Rate} = k [A]^x \]
where:
- \( k \) is the rate constant,
- \( [A] \) is the concentration of reactant A,
- \( x \) is the order of the reaction with respect to A.

### Step 2: Set up the initial conditions
According to the problem, when the concentration of A is increased by 10 times, the rate of reaction increases by a factor of 100. We can represent this mathematically as:
- Initial rate: \( \text{Rate} = k [A]^x \)
- New concentration of A: \( [A]_{new} = 10[A] \)
- New rate: \( \text{Rate}_{new} = k [A]_{new}^x = k (10[A])^x \)

### Step 3: Express the new rate in terms of the old rate
Substituting the new concentration into the rate equation gives:
\[ \text{Rate}_{new} = k (10[A])^x = k \cdot 10^x \cdot [A]^x \]

### Step 4: Relate the new rate to the old rate
Since we know that the new rate is 100 times the initial rate, we can write:
\[ \text{Rate}_{new} = 100 \cdot \text{Rate} \]
Substituting the expressions for the rates:
\[ k \cdot 10^x \cdot [A]^x = 100 \cdot k [A]^x \]

### Step 5: Simplify the equation
We can cancel \( k [A]^x \) from both sides (assuming \( k \) and \( [A]^x \) are not zero):
\[ 10^x = 100 \]

### Step 6: Solve for x
Recognizing that \( 100 \) can be expressed as \( 10^2 \):
\[ 10^x = 10^2 \]
This implies:
\[ x = 2 \]

### Conclusion
The order of the reaction with respect to A is \( x = 2 \).

### Final Answer
The order of reaction with respect to A is 2.

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