Question

For the reaction $\mathrm{A + B \rightarrow \text{product}}$, the rate of reaction is $3.6 \times 10^{-2}\ \mathrm{mol\ dm^{-3}\ sec^{-1}}$. When $[\mathrm{A}] = 0.2\ \mathrm{mol\ dm^{-3}}$ and $[\mathrm{B}] = 0.1\ \mathrm{mol\ dm^{-3}}$, find the rate constant of the reaction if it is second order with respective to both reactants.

• A. $18\ \mathrm{mol^{-3}\ dm^9\ sec^{-1}}$
• B. $90\ \mathrm{mol^{-3}\ dm^9\ sec^{-1}}$
• C. $72\ \mathrm{mol^{-3}\ dm^9\ sec^{-1}}$
• D. $36\ \mathrm{mol^{-3}\ dm^9\ sec^{-1}}$

Hint:

Pay close attention to phrases like "second order with respective to both reactants." This means the exponent for each individual reactant is 2, making the overall reaction order $2 + 2 = 4$. Always convert decimals into powers of 10 to prevent simple calculation mistakes with decimal places!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given the experimental rate of a reaction ($r = 3.6 \times 10^{-2}\ \mathrm{mol\ dm^{-3}\ sec^{-1}}$) along with the initial concentrations of the reactants ($[\mathrm{A}] = 0.2\ \mathrm{mol\ dm^{-3}}$ and $[\mathrm{B}] = 0.1\ \mathrm{mol\ dm^{-3}}$). The reaction is explicitly specified as being second-order with respect to both reactants, and we need to solve for the rate constant ($k$).

Step 2: Key Formula or Approach:
Based on the experimental orders given, the differential rate law expression is formulated as: $$r = k [\mathrm{A}]^2 [\mathrm{B}]^2$$ Rearranging this algebraic setup to solve for the rate constant ($k$) yields: $$k = \frac{r}{[\mathrm{A}]^2 [\mathrm{B}]^2}$$

Step 3: Detailed Explanation:
Substitute the given values directly into our rearranged rate law expression: $$k = \frac{3.6 \times 10^{-2}}{(0.2)^2 \times (0.1)^2}$$ First, compute the squared concentrations in the denominator: $$(0.2)^2 = 0.04 = 4 \times 10^{-2}$$ $$(0.1)^2 = 0.01 = 1 \times 10^{-2}$$ Multiply the denominator values together: $$\text{Denominator} = (4 \times 10^{-2}) \times (1 \times 10^{-2}) = 4 \times 10^{-4}$$ Now, divide the reaction rate by this product to find $k$: $$k = \frac{3.6 \times 10^{-2}}{4 \times 10^{-4}} = \frac{3.6}{4} \times 10^2 = 0.9 \times 100 = 90$$ Let's quickly check the units: $$\text{Units} = \frac{\mathrm{mol\ dm^{-3}\ sec^{-1}}}{(\mathrm{mol\ dm^{-3}})^2 \times (\mathrm{mol\ dm^{-3}})^2} = \frac{\mathrm{mol\ dm^{-3}\ sec^{-1}}}{\mathrm{mol^4\ dm^{-12}}} = \mathrm{mol^{-3}\ dm^9\ sec^{-1}}$$

Step 4: Final Answer:
The rate constant of the reaction is $90\ \mathrm{mol^{-3}\ dm^9\ sec^{-1}}$, which corresponds to option (B).