Question

The rate for reaction A + B $\rightarrow$ product, is $1.8\times10^{-2}\text{ mol dm}^{-3}\text{s}^{-1}$. Calculate the rate constant if the reaction is second order in A and first order in B. $([A]=0.2\text{ M}; [B]=0.1\text{ M})$

• A. $9.0\text{ mol}^{-2}\text{dm}^{6}\text{s}^{-1}$
• B. $18.0\text{ mol}^{-2}\text{dm}^{6}\text{s}^{-1}$
• C. $4.5\text{ mol}^{-2}\text{dm}^{6}\text{s}^{-1}$
• D. $16.0\text{ mol}^{-2}\text{dm}^{6}\text{s}^{-1}$

Hint:

Logic Tip: To make the division easier without using scientific notation, convert decimals to fractions: $0.2 = 2/10$ and $0.1 = 1/10$. Then $Rate = k \times (4/100) \times (1/10) = k \times (4/1000)$. So, $k = 0.018 \times (1000/4) = 18/4 = 4.5$.

Answer 1

The Correct Option is C

Concept:
The rate law expresses the rate of a reaction as a product of the rate constant ($k$) and the concentrations of the reactants raised to the power of their respective orders. $$\text{Rate} = k [A]^x [B]^y$$ where $x$ and $y$ are the orders with respect to A and B.
Step 1: Write the specific rate law for the given reaction.
The reaction is given as second order in A ($x = 2$) and first order in B ($y = 1$). $$\text{Rate} = k [A]^2 [B]^1$$
Step 2: Substitute the given values into the rate law.
Given data: $\text{Rate} = 1.8 \times 10^{-2}\text{ mol dm}^{-3}\text{s}^{-1}$ $[A] = 0.2\text{ M} = 0.2\text{ mol dm}^{-3}$ $[B] = 0.1\text{ M} = 0.1\text{ mol dm}^{-3}$ Substitute these into the equation: $$1.8 \times 10^{-2} = k \cdot (0.2)^2 \cdot (0.1)^1$$
Step 3: Solve for the rate constant k.
Calculate the concentration terms: $$(0.2)^2 = 0.04 = 4 \times 10^{-2}$$ $$4 \times 10^{-2} \times 0.1 = 0.004 = 4 \times 10^{-3}$$ Now, isolate $k$: $$k = \frac{1.8 \times 10^{-2{4 \times 10^{-3$$ $$k = \frac{1.8}{4} \times 10^{(-2 - (-3))}$$ $$k = 0.45 \times 10^1 = 4.5$$ The overall order is $2 + 1 = 3$. The units for a third-order rate constant are $\text{mol}^{-2}\text{dm}^6\text{s}^{-1}$. $$k = 4.5\text{ mol}^{-2}\text{dm}^6\text{s}^{-1}$$