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 step by step, we will derive the rate expression for the reaction and analyze the effect of doubling the concentrations of reactants A and B.

### Step 1: Write the Rate Expression
For the reaction \( A + B \rightarrow \text{products} \), we know the order of A is 2 and the order of B is 3. The rate expression can be written as:

\[ \text{Rate} = k [A]^2 [B]^3 \]

### Step 2: Determine the Initial Rate
Let’s denote the initial concentrations of A and B as \( [A] \) and \( [B] \). The initial rate of the reaction can be expressed as:

\[ \text{Rate}_1 = k [A]^2 [B]^3 \]

### Step 3: Calculate the New Concentrations
If both concentrations are doubled, the new concentrations will be:

\[ [A]_{\text{new}} = 2[A] \]
\[ [B]_{\text{new}} = 2[B] \]

### Step 4: Write the New Rate Expression
Substituting the new concentrations into the rate expression gives us:

\[ \text{Rate}_2 = k (2[A])^2 (2[B])^3 \]

### Step 5: Simplify the New Rate Expression
Now, simplify the expression:

\[ \text{Rate}_2 = k (4[A]^2) (8[B]^3) \]
\[ \text{Rate}_2 = k \cdot 32 [A]^2 [B]^3 \]

### Step 6: Compare the New Rate with the Initial Rate
Now we can compare the new rate \( \text{Rate}_2 \) with the initial rate \( \text{Rate}_1 \):

\[ \text{Rate}_2 = 32 \cdot \text{Rate}_1 \]

### Conclusion
Thus, 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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