For the raction `A+Brarr` products, it is found that order of A is 2 and the order of B is 3. In the rate expression when the concentration of both A and B are doubled the rate will increases by a factor

Options

A. 10
B. 16
C. 32
D. 28

Answer

To solve the problem, we need to analyze how the rate of the reaction changes when the concentrations of the reactants are doubled.

### Step-by-Step Solution:

1. **Write the Rate Law Expression:**
The rate of the reaction can be expressed using the rate law:
\[ \text{Rate} = k [A]^x [B]^y \]
where \( k \) is the rate constant, \( [A] \) is the concentration of reactant A, \( [B] \) is the concentration of reactant B, \( x \) is the order of A, and \( y \) is the order of B.

2. **Substitute the Given Orders:**
From the question, we know that the order of A is 2 and the order of B is 3. Therefore, we can substitute these values into the rate law:
\[ \text{Rate} = k [A]^2 [B]^3 \]

3. **Determine the Initial Rate:**
Let’s denote the initial concentrations of A and B as \( [A] \) and \( [B] \). The initial rate (Rate 1) can be expressed as:
\[ \text{Rate}_1 = k [A]^2 [B]^3 \]

4. **Calculate the New Concentrations:**
When the concentrations of both A and B are doubled, the new concentrations will be:
\[ [A]_{\text{new}} = 2[A] \]
\[ [B]_{\text{new}} = 2[B] \]

5. **Determine the New Rate:**
Now, we can calculate the new rate (Rate 2) using the new concentrations:
\[ \text{Rate}_2 = k (2[A])^2 (2[B])^3 \]

6. **Simplify the New Rate Expression:**
Expanding the expression:
\[ \text{Rate}_2 = k (4[A]^2)(8[B]^3) = k \cdot 32 [A]^2 [B]^3 \]

7. **Relate the New Rate to the Initial Rate:**
Now, we can relate the new rate to the initial rate:
\[ \text{Rate}_2 = 32 \cdot \text{Rate}_1 \]

8. **Conclusion:**
Therefore, when the concentrations of both A and B are doubled, the rate of the reaction increases by a factor of 32.

### Final Answer:
The rate will increase by a factor of **32**.

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