For a reaction initial rate is given as : `R_(0)=k[A_(0)]^(2)[B_(0)]`. By what factor, the initial rate of reaction will increase if initial concentration is taken 1.5 times and B is tripled?

Options

A. 4.5
B. 2.25
C. 6.75
D. None of these

Answer

To solve the problem, we need to analyze how the initial rate of reaction changes when the concentrations of reactants are altered. The initial rate of the reaction is given by the equation:

\[ R_0 = k [A_0]^2 [B_0] \]

Where:
- \( R_0 \) is the initial rate of reaction.
- \( k \) is the rate constant.
- \( [A_0] \) is the initial concentration of reactant A.
- \( [B_0] \) is the initial concentration of reactant B.

### Step 1: Define the new concentrations
According to the problem:
- The initial concentration of A is increased to 1.5 times, so:
\[ [A] = 1.5 [A_0] = \frac{3}{2} [A_0] \]

- The initial concentration of B is tripled, so:
\[ [B] = 3 [B_0] \]

### Step 2: Substitute the new concentrations into the rate equation
Now, we can substitute these new concentrations into the rate equation:

\[ R' = k \left(\frac{3}{2} [A_0]\right)^2 (3 [B_0]) \]

### Step 3: Simplify the equation
Calculating \( \left(\frac{3}{2} [A_0]\right)^2 \):
\[ \left(\frac{3}{2} [A_0]\right)^2 = \frac{9}{4} [A_0]^2 \]

Now substituting this back into the rate equation:
\[ R' = k \left(\frac{9}{4} [A_0]^2\right) (3 [B_0]) \]

### Step 4: Combine the terms
Now we can combine the terms:
\[ R' = k \cdot \frac{9}{4} \cdot 3 \cdot [A_0]^2 \cdot [B_0] \]
\[ R' = k \cdot \frac{27}{4} [A_0]^2 [B_0] \]

### Step 5: Relate the new rate to the original rate
Now we can relate \( R' \) to \( R_0 \):
\[ R' = \frac{27}{4} R_0 \]

### Step 6: Calculate the factor of increase
To find the factor by which the initial rate of reaction increases, we can write:
\[ \text{Factor of increase} = \frac{R'}{R_0} = \frac{27}{4} \]

### Final Answer
Thus, the initial rate of reaction will increase by a factor of \( \frac{27}{4} \) or 6.75.

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