Question

A reaction is first order with respective to $\mathrm{A}$ and second order with respective to $\mathrm{B}$. What is the effect on reaction rate if concentration of $\mathrm{B}$ is increased 3 times?

• A. Rate increases 6 times
• B. Rate increases 2 times
• C. Rate increases 9 times
• D. Rate increases 3 times

Hint:

For any reactant of order $n$, increasing its concentration by a factor of $x$ changes the reaction rate by a factor of $x^n$. Since $\mathrm{B}$ is second-order ($n=2$) and its concentration is tripled ($x=3$), the rate must increase by $3^2 = 9$ times!

Answer 1

The Correct Option is C

Step 1: Translate the orders into a rate law.
First order in $A$ and second order in $B$ means $\text{Rate} = k[A][B]^2$.
Step 2: Decide what changes.
Only $[B]$ changes, scaling to $3$ times its value, while $[A]$ stays fixed.
Step 3: Use a ratio to avoid carrying $k$.
Take the new rate over the old rate so $k$ and $[A]$ cancel cleanly: \[ \frac{R_2}{R_1} = \frac{k[A](3[B])^2}{k[A][B]^2}. \]
Step 4: Simplify the $B$ factor.
The $[A]$ and $[B]^2$ terms cancel, leaving $\dfrac{R_2}{R_1} = 3^2 = 9$.
Step 5: Read off the meaning.
Because $B$ is squared in the rate law, tripling $B$ multiplies the rate by $3^2$, not by $3$.
Step 6: State the answer.
The rate becomes $9$ times the original value, option (C).
\[ \boxed{\text{Rate increases } 9 \text{ times (option C)}} \]