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: Note the rate law.
The rate is $r = k[A][B][C]^2$. We are told to double both $[A]$ and $[B]$ while $[C]$ stays the same.
Step 2: Track each concentration's effect separately.
Rate depends on $[A]$ to the first power, on $[B]$ to the first power, and on $[C]$ to the second power. Only $A$ and $B$ change.
Step 3: Apply the change to A.
Doubling $[A]$ multiplies the rate by $2^1 = 2$.
Step 4: Apply the change to B.
Doubling $[B]$ multiplies the rate by another $2^1 = 2$.
Step 5: Combine the factors.
Since $[C]$ is unchanged it contributes a factor of $1$. The overall multiplier is $2 \times 2 \times 1 = 4$.
Step 6: State the new rate.
The reaction now proceeds at $4r$, which is option (2).
\[ \boxed{r_{\text{new}} = 4r} \]