Rate of reaction can be expressed by following rate expression .Rate = `K[A]^(2)[B]`, if concentration of A is increased by 3 times and concentration of B is increased by 2 times , how many times will rate of the reaction be increased ?

Options

A. 9 times
B. 27 times
C. 18 times
D. 8 times

Answer

To solve the problem, we need to determine how the rate of reaction changes when the concentrations of reactants A and B are altered. The rate expression given is:

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

### Step 1: Write the initial rate expression
The initial rate of the reaction can be expressed as:

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

### Step 2: Substitute the new concentrations
According to the problem, the concentration of A is increased by 3 times and the concentration of B is increased by 2 times. Therefore, the new concentrations will be:

- New concentration of A = \( 3[A] \)
- New concentration of B = \( 2[B] \)

### Step 3: Write the new rate expression
Substituting the new concentrations into the rate expression, we get:

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

### Step 4: Simplify the new rate expression
Now, we simplify the expression:

\[ \text{Rate}_{\text{new}} = k(3^2[A]^2)(2[B]) \]
\[ \text{Rate}_{\text{new}} = k(9[A]^2)(2[B]) \]
\[ \text{Rate}_{\text{new}} = 18k[A]^2[B] \]

### Step 5: Compare the new rate with the initial rate
Now, we compare the new rate with the initial rate:

- Initial rate: \( \text{Rate}_{\text{initial}} = k[A]^2[B] \)
- New rate: \( \text{Rate}_{\text{new}} = 18k[A]^2[B] \)

### Step 6: Calculate the factor of increase
To find out how many times the rate has increased, we take the ratio of the new rate to the initial rate:

\[ \text{Factor of increase} = \frac{\text{Rate}_{\text{new}}}{\text{Rate}_{\text{initial}}} = \frac{18k[A]^2[B]}{k[A]^2[B]} = 18 \]

### Conclusion
Thus, the rate of the reaction will increase by **18 times**.

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