Question

The rate of A + B \(\rightarrow\) P is \(3.6 \times 10^{-2}\) mol/\(dm^3\)/s. When [A] = 0.2 M and [B] = 0.1 M. Calculate the rate constant if reaction is first order in B and second order in A.

Hint:

Always sum the individual orders to find the overall order (\(2+1=3\)). This determines the units of the rate constant.

Answer 1

Step 1: Understanding the Concept:
The rate law expresses the relationship between the reaction rate and the concentrations of the reactants.
Step 2: Key Formula or Approach:
\[ Rate = k [A]^2 [B]^1 \]
Step 3: Detailed Explanation:
Given:
\(Rate = 3.6 \times 10^{-2} \text{ mol L}^{-1} \text{ s}^{-1}\)
\([A] = 0.2 \text{ M}\)
\([B] = 0.1 \text{ M}\)
Substitute into the rate law:
\[ 3.6 \times 10^{-2} = k (0.2)^2 (0.1) \]
\[ 3.6 \times 10^{-2} = k (0.04) (0.1) \]
\[ 3.6 \times 10^{-2} = k (0.004) \]
\[ k = \frac{3.6 \times 10^{-2}}{4 \times 10^{-3}} = \frac{36 \times 10^{-3}}{4 \times 10^{-3}} \]
\[ k = 9 \]
Units for 3rd order: \(M^{-2} s^{-1} = L^2 mol^{-2} s^{-1}\).
Step 4: Final Answer:
The rate constant is \(9 \text{ L}^2 \text{ mol}^{-2} \text{ s}^{-1}\).