For the reaction `A + B to` products, it is found that order of A is 1 and order of B is 1/2. When concentrations of both A & B are increased four times the rate will increase by a factor

Options

A. 6
B. 8
C. 4
D. 16

Explanation

To solve the problem, we need to determine how the rate of the reaction changes when the concentrations of reactants A and B are increased four times. We know the order of the reaction with respect to A and B, which are given as follows:

- Order of A = 1
- Order of B = 1/2

### Step-by-Step Solution:

1. **Write the Rate Law Expression**:
The rate law for the reaction can be expressed as:
\[ \text{Rate} (R) = k [A]^m [B]^n \]
where \( k \) is the rate constant, \( m \) is the order with respect to A, and \( n \) is the order with respect to B. Given the orders:
\[ R = k [A]^1 [B]^{1/2} = k [A] [B]^{1/2} \]

2. **Determine the Initial Rate**:
Let the initial concentrations of A and B be \( [A] \) and \( [B] \) respectively. The initial rate can be expressed as:
\[ R_0 = k [A] [B]^{1/2} \]

3. **Increase Concentrations**:
When the concentrations of both A and B are increased four times, the new concentrations will be:
\[ [A]' = 4[A] \]
\[ [B]' = 4[B] \]

4. **Calculate the New Rate**:
Substitute the new concentrations into the rate law to find the new rate \( R' \):
\[ R' = k [A]' [B]'^{1/2} = k (4[A]) (4[B])^{1/2} \]
Simplifying this, we have:
\[ R' = k (4[A]) (2\sqrt{4[B]}) = k (4[A]) (2[B]^{1/2}) = k \cdot 8 [A] [B]^{1/2} \]

5. **Compare the New Rate with the Initial Rate**:
Now we can compare the new rate \( R' \) with the initial rate \( R_0 \):
\[ R' = 8 \cdot k [A] [B]^{1/2} = 8 R_0 \]

6. **Conclusion**:
Therefore, the rate increases by a factor of 8 when the concentrations of A and B are increased four times.

### Final Answer:
The rate will increase by a factor of **8**.