To determine the factor by which the initial rate of the reaction increases when the initial concentrations of [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are changed, let's start with the given initial rate formula:
[tex]\[ R_0 = k[A]_0^2[B]_0 \][/tex]
Here [tex]\( R_0 \)[/tex] is the initial rate, [tex]\( k \)[/tex] is the reaction rate constant, [tex]\([A]_0\)[/tex] is the initial concentration of [tex]\( A \)[/tex], and [tex]\([B]_0\)[/tex] is the initial concentration of [tex]\( B \)[/tex].
### Step 1: Define Initial Concentrations and Rate
Let's assume the initial concentrations of [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are given by:
[tex]\[ [A]_0 = 1 \text{ mol/L} \][/tex]
[tex]\[ [B]_0 = 1 \text{ mol/L} \][/tex]
Hence, the initial rate [tex]\( R_0 \)[/tex] will be:
[tex]\[ R_0 = k (1^2) (1) = k \][/tex]
### Step 2: New Concentrations
Now, we are given that the concentration of [tex]\( A \)[/tex] is increased by 1.5 times and the concentration of [tex]\( B \)[/tex] is tripled:
[tex]\[ [A]_{new} = 1.5 \times [A]_0 = 1.5 \text{ mol/L} \][/tex]
[tex]\[ [B]_{new} = 3 \times [B]_0 = 3 \text{ mol/L} \][/tex]
### Step 3: Calculate New Rate
Using the new concentrations, the new rate [tex]\( R_{new} \)[/tex] is:
[tex]\[ R_{new} = k [A]_{new}^2 [B]_{new} \][/tex]
Substitute the new concentrations:
[tex]\[ R_{new} = k (1.5^2) (3) \][/tex]
[tex]\[ R_{new} = k (2.25) (3) \][/tex]
[tex]\[ R_{new} = 6.75k \][/tex]
### Step 4: Factor of Increase
Finally, the factor by which the rate has increased is the ratio of the new rate to the initial rate:
[tex]\[ \text{Rate Increase Factor} = \frac{R_{new}}{R_0} \][/tex]
[tex]\[ \text{Rate Increase Factor} = \frac{6.75k}{k} \][/tex]
Since [tex]\( k \)[/tex] cancels out, we get:
[tex]\[ \text{Rate Increase Factor} = 6.75 \][/tex]
Therefore, the rate of reaction increases by a factor of [tex]\( 6.75 \)[/tex]. The correct answer is:
[tex]\[ \boxed{6.75} \][/tex]
