$$A + 2B \rightarrow C$$, the rate equation for the reaction is given as Rate = k[A][B]. If the concentration of A is kept the same but that of B is doubled what will happen to the rate itself?

The rate equation given is Rate = k [A] [B]. This means the rate depends on the concentrations of both A and B.

Initially, let the concentration of A be [A] and the concentration of B be [B]. So the initial rate is:

$$ \text{Rate}_1 = k \times [A] \times [B] $$

Now, the concentration of A is kept the same, so it remains [A]. The concentration of B is doubled, so it becomes 2[B].

The new rate is:

$$ \text{Rate}_2 = k \times [A] \times (2[B]) $$

Simplify this expression:

$$ \text{Rate}_2 = k \times [A] \times 2[B] $$

$$ \text{Rate}_2 = 2 \times (k \times [A] \times [B]) $$

Notice that $$ k \times [A] \times [B] $$ is exactly the initial rate $$\text{Rate}_1$$. So we can write:

$$ \text{Rate}_2 = 2 \times \text{Rate}_1 $$

This shows that the new rate is twice the initial rate. Therefore, the rate doubles.

Hence, the correct answer is Option A.