Question

The rate law for the reaction \(A+B\rightarrow P\) is found to be rate \(= k[A]^2[B]\).
The rate constant of the reaction at 300 K is \(6.0 \text{M}^{-2}\text{s}^{-1}\). Calculate the rate of the reaction when \([A] = 1 \text{M}\) and \([B] = 0.2 \text{M}\)

• A. \(0.6 \text{M s}^{-1}\).
• B. \(1.2 \text{M s}^{-1}\).
• C. \(1.8 \text{M s}^{-1}\).
• D. \(2.4 \text{M s}^{-1}\).

Hint:

Substitute concentrations into rate = k[A]^2[B].

Answer 1

The Correct Option is B

Step 1: Plug in numbers
$[A]^2 = 1$ and $[B] = 0.2$, so the concentration part is $0.2$. Multiply by $k=6.0$: $6.0 \times 0.2 = 1.2$.

Step 2: Answer
Option (B).

Final Answer:
1.2 M per second. \[ \boxed{\text{(B)}\ 1.2\ \text{M s}^{-1}} \]