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 reactants.
Step 2: Key Formula or Approach:
Rate Law: \( Rate = k [A]^x [B]^y \)
Given: \( x = 2 \) (second order in A) and \( y = 1 \) (first order in B).
So, \( Rate = k [A]^2 [B] \).
Step 3: Detailed Explanation:
Substitute the given values:
\( Rate = 3.6 \times 10^{-2} \)
\( [A] = 0.2 \)
\( [B] = 0.1 \)
\[ 3.6 \times 10^{-2} = k (0.2)^2 (0.1) \]
\[ 3.6 \times 10^{-2} = k (0.04) (0.1) \]
\[ 3.6 \times 10^{-2} = k (0.004) \]
\[ k = \frac{3.6 \times 10^{-2}}{4 \times 10^{-3}} \]
\[ k = \frac{0.036}{0.004} = 9 \]
Units: For a 3rd order reaction (\(2+1\)), the unit is \(mol^{-2} dm^6 s^{-1}\).
Step 4: Final Answer:
The rate constant \(k = 9.0 \, mol^{-2} dm^6 s^{-1}\).