Question

Consider the gaseous reaction: \[ \text{A}_2 + \text{B}_2 \;\longrightarrow\; 2\,\text{AB} \] The following initial-rate data were obtained for the above reaction (rate of formation of AB): \includegraphics[]{144.png} The value of the rate constant for the above reaction is:

• A. \(1.25\times 10^{-2}\)
• B. \(1.25\times 10^{-3}\)
• C. \(2.5\times 10^{-2}\)
• D. \(2.5\times 10^{-3}\)

Hint:

Always distinguish between the rate of \emph{product formation} and the rate of the \emph{overall reaction}. - For \(\text{A}_2 + \text{B}_2 \to 2\,\text{AB}\), \(\text{Rate(AB)} = 2 \times \text{Rate}\).

Answer 1

The Correct Option is D

Step 1: Determining Reaction Orders from the Data

Assume a rate law of the form:

\[ \text{Rate(AB)} = k' \,[\text{A}_2]^m \,[\text{B}_2]^n, \]

where \(\text{Rate(AB)}\) is the initial rate of formation of \(\text{AB}\).

Comparing experiments 1 and 2:

\[ [A_2] \text{ doubles from }0.1 \text{ to }0.2,\; [B_2]\text{ constant at }0.1 \] \[ \frac{\text{Rate}_2}{\text{Rate}_1} = \frac{2.0\times 10^{-3}}{5.0\times 10^{-4}} = 4 \quad\Longrightarrow\quad 2^m = 4 \quad\Longrightarrow\quad m=2. \]

Comparing experiments 2 and 3:

\[ [B_2]\text{ doubles from }0.1 \text{ to }0.2,\; [A_2]\text{ constant at }0.2 \] \[ \frac{\text{Rate}_3}{\text{Rate}_2} = \frac{1.0\times 10^{-3}}{2.0\times 10^{-3}} = 0.5 \quad\Longrightarrow\quad 2^n = 0.5 \quad\Longrightarrow\quad n=-1. \]

Hence,

\[ \text{Rate(AB)} = k' \,[\text{A}_2]^2 \,[\text{B}_2]^{-1}. \]

Step 2: Calculating the "Apparent" Rate Constant \(k'\)

Using experiment 1 (\([A_2]=0.1\,\text{M}, [B_2]=0.1\,\text{M}\)):

\[ \text{Rate(AB)} = 5.0\times 10^{-4} = k' \,(0.1)^2 \,(0.1)^{-1} = k'\,\bigl(0.01\bigr)\bigl(10\bigr) = k' \,\times\, 0.1, \] \[ k' = \frac{5.0\times 10^{-4}}{0.1} = 5.0\times 10^{-3}. \]

Step 3: Relating \(\text{Rate(AB)}\) to the Reaction Rate and the Final \(k\)

For the overall reaction \(\text{A}_2 + \text{B}_2 \to 2\,\text{AB}\), the rate of consumption of reactants (the "reaction rate") is half the rate of formation of \(\text{AB}\), i.e.,

\[ \text{Rate} = -\frac{d[\text{A}_2]}{dt} = -\frac{d[\text{B}_2]}{dt} = \frac{1}{2}\,\frac{d[\text{AB}]}{dt} = \frac{\text{Rate(AB)}}{2}. \]

Thus, if

\[ \text{Rate(AB)} = k'[\text{A}_2]^2[\text{B}_2]^{-1}, \]

then the actual rate law for reactant consumption is

\[ \text{Rate} = k \,[\text{A}_2]^2\,[\text{B}_2]^{-1} = \frac{\text{Rate(AB)}}{2} = \frac{k'}{2}[\text{A}_2]^2[\text{B}_2]^{-1}. \]

Hence,

\[ k = \tfrac{k'}{2} = \tfrac{5.0\times 10^{-3}}{2} = 2.5\times 10^{-3}. \]

Therefore, the rate constant for the reaction (based on reactant consumption) is \(2.5\times 10^{-3}\).