In a reaction \(A\) and \(B\) reacts to form product. The initial rate of reaction \((r_0)\) was determined using different initial concentration of \(A\) and \(B\).
| \(A/\, \ mol L^{-1}\) | 0.10 | 0.10 | 0.20 |
| \(B/\, \ mol L^{-1}\) | 0.30 | 0.10 | 0.30 |
| \(r_0/\, \ mol L^{-1}\,s^{-1}\) | \(6.81 \times 10^{-4}\) | \(2.27 \times 10^{-4}\) | \(13.62 \times 10^{-4}\) |
What is the initial rate of reaction \((r_0)\) when initial concentration of \(A\) and \(B\) is \(0.50\, \ mol/L\) and \(0.50\, \ mol/L\) respectively.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 56.75 × 10−4 mol L−1 s−1
Let the rate law be rate = k[A]x[B]y. The order with respect to each reactant comes from experiments in which only that concentration changes.
Order in A (experiments 1 and 3, [B] = 0.30 fixed): [A] doubles and the rate doubles.
\( \displaystyle 2^x = \frac{13.62 \times 10^{-4}}{6.81 \times 10^{-4}} = 2 \;\Rightarrow\; x = 1 \)
Order in B (experiments 2 and 1, [A] = 0.10 fixed): [B] triples and the rate triples.
\( \displaystyle 3^y = \frac{6.81 \times 10^{-4}}{2.27 \times 10^{-4}} = 3 \;\Rightarrow\; y = 1 \)
So rate = k[A][B]. From experiment 2:
\( \displaystyle k = \frac{2.27 \times 10^{-4}}{0.10 \times 0.10} = 2.27 \times 10^{-2}\ \mathrm{L\,mol^{-1}\,s^{-1}} \)
\( \displaystyle r_0 = 2.27 \times 10^{-2} \times 0.50 \times 0.50 = 5.675 \times 10^{-3} = 56.75 \times 10^{-4}\ \mathrm{mol\,L^{-1}\,s^{-1}} \)
So the initial rate is 56.75 × 10−4 mol L−1 s−1.