Step 1: Understanding the Concept:
The rate law of a chemical reaction is an experimentally derived mathematical equation that establishes a direct relationship between the reaction rate and the molar concentrations of the reactants. The exponent to which a reactant's concentration is raised in the rate law is called the order of the reaction with respect to that reactant.
Step 2: Key Formula or Approach:
The initial rate law given for the reaction is:
$$ \text{Rate}_1 = k[\text{A}]^2 $$
This exponent of 2 tells us that the reaction follows second-order kinetics with respect to reactant $\text{A}$. If we change the initial concentration to a new value $[\text{A}]'$, the new rate expression becomes:
$$ \text{Rate}_2 = k([\text{A}]')^2 $$
Step 3: Detailed Explanation:
Let's calculate the exact mathematical impact of doubling the concentration of reactant $\text{A}$:
1. Let the original concentration be $[\text{A}] = a$. The initial rate is:
$$ \text{Rate}_1 = k(a)^2 = ka^2 $$
2. The prompt states that the concentration of $\text{A}$ is doubled. Therefore, the new concentration is:
$$ [\text{A}]' = 2a $$
3. Substitute this doubled value into the rate law equation to find the new rate ($\text{Rate}_2$):
$$ \text{Rate}_2 = k(2a)^2 $$
$$ \text{Rate}_2 = k(4a^2) = 4(ka^2) $$
4. Replace the term $ka^2$ with our original rate value ($\text{Rate}_1$):
$$ \text{Rate}_2 = 4 \times \text{Rate}_1 $$
Thus, doubling the concentration of a second-order reactant causes the overall reaction rate to increase by a factor of four. This matches option (B).
Step 4: Final Answer:
The rate of the reaction will become four times the original rate.