Question

Rate of the reaction $\text{A}+\text{B}\rightarrow$ product is $3.6\times10^{-2}\text{ mol dm}^{-3}\text{s}^{-1}$ and rate law is $r=k[\text{A}][\text{B}]^{2}$. What is rate constant of the reaction if $[\text{A}]=0.2$ M and $[\text{B}]=0.1$ M?

• A. $18\text{ mol}^{-2}\text{dm}^{6}\text{s}^{-1}$
• B. $10\text{ mol}^{-2}\text{dm}^{6}\text{s}^{-1}$
• C. $24\text{ mol}^{-2}\text{dm}^{6}\text{s}^{-1}$
• D. $4.8\text{ mol}^{-2}\text{dm}^{6}\text{s}^{-1}$

Hint:

General formula for the units of a rate constant $k$ for an $n^{th}$ order reaction:
Units = $(\text{mol L}^{-1})^{1-n} \text{s}^{-1}$
Here, order $n = 3$, so $(\text{mol L}^{-1})^{-2} \text{s}^{-1} = \text{mol}^{-2} \text{L}^2 \text{s}^{-1}$ (where L is $\text{dm}^3$, so $\text{L}^2 = \text{dm}^6$).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The rate constant ($k$) can be calculated by rearranging the rate law equation and substituting the given rate and concentrations.

Step 2: Formula Application:
$k = \frac{r}{[A][B]^2}$

Step 3: Explanation:
$k = \frac{3.6 \times 10^{-2}}{(0.2) \times (0.1)^2}$ $k = \frac{0.036}{0.2 \times 0.01} = \frac{0.036}{0.002}$ $k = 18$.

Step 4: Final Answer:
The rate constant is 18 mol$^{-2}$ dm$^6$ s$^{-1}$.