Question

For the reaction: \[ 2A \rightarrow Products \] the rate law is: \[ \text{Rate}=k[A]^2 \] If the concentration of \(A\) is doubled, the rate of reaction will become:

• A. Doubled
• B. Four times
• C. Half
• D. Unchanged

Hint:

For reactions: \[ \text{Rate}\propto[A]^n \] If concentration becomes \(m\) times: \[ \text{New rate}=m^n \] times the original rate.

Answer 1

The Correct Option is B

Concept: The rate law of a reaction expresses how the rate depends upon concentration of reactants. For a reaction: \[ \text{Rate}=k[A]^n \] where:
• \(k\) = rate constant
• \(n\) = order of reaction with respect to \(A\) If concentration changes, the rate changes according to the power of concentration.

Step 1: Write the given rate law.
Given: \[ \text{Rate}=k[A]^2 \] This means reaction is second order with respect to \(A\).

Step 2: Determine new concentration.
Suppose initial concentration is: \[ [A] \] New concentration after doubling: \[ [A]'=2[A] \]

Step 3: Substitute into rate equation.
New rate: \[ \text{Rate}' = k(2[A])^2 \] \[ \text{Rate}' = k(4[A]^2) \] \[ \text{Rate}' = 4k[A]^2 \] But: \[ k[A]^2=\text{original rate} \] Therefore: \[ \text{New rate}=4\times \text{original rate} \] Hence, the rate becomes four times. Therefore, the correct answer is: \[ \boxed{\text{(B) Four times}} \]