Question

The rate law equation for a reaction between A, B and C is $r = k [A] [B] [C]^2$, what will be the rate of reaction if concentration of both A and B are doubled?

• A. $2r$
• B. $4r$
• C. $6r$
• D. $8r$

Hint:

The order of the reaction with respect to A is 1, and with respect to B is 1. Doubling A multiplies the rate by $2^1 = 2$. Doubling B multiplies the rate by $2^1 = 2$. Together, the net change is a factor of $2 \times 2 = 4$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The initial rate equation of a reaction involving components A, B, and C is provided. We need to determine how the final rate shifts relative to the initial rate $r$ when the concentration terms for both A and B are simultaneously doubled while C remains unchanged.

Step 2: Key Formula or Approach:
Write the initial rate expression as: $$r_{\text{initial}} = r = k [A] [B] [C]^2$$ Set up a secondary rate expression ($r_{\text{new}}$) substituting the modified concentrations $[A]' = 2[A]$ and $[B]' = 2[B]$.

Step 3: Detailed Explanation:
Substitute the new concentration values into the rate law formula: $$r_{\text{new}} = k [2A] [2B] [C]^2$$ Pull out the constant numeric scaling coefficients: $$r_{\text{new}} = k \cdot 2[A] \cdot 2[B] \cdot [C]^2$$ $$r_{\text{new}} = 4 \cdot \left(k [A] [B] [C]^2\right)$$ Since the expression inside the parenthesis is equal to our initial rate $r$: $$r_{\text{new}} = 4r$$

Step 4: Final Answer:
The rate of the reaction increases to $4r$, matching option (B).