Consider the given reaction: X + Y → Z; Rate = k[X]a [Y]b. The initial concentration of X and Y are given below:
| [X]/ mol L−1 | [Y]/ mol L−1 | Rate of formation of Z/ mol L−1s−1 |
| 0.2 | 0.2 | 4.36 × 10−2 |
| 0.4 | 0.2 | 8.72 × 10−2 |
| 0.2 | 0.4 | 17.44 × 10−2 |
The overall order of reaction for the formation of product Z is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Three
The order with respect to each reactant is found from initial rates. Keep the concentration of one reactant fixed, change the other, and see how the rate changes:
\( \displaystyle \text{Rate}=k[\mathrm{X}]^a[\mathrm{Y}]^b \)
If a concentration is doubled, the rate changes by the factor 2order: unchanged for zero order, doubled for first order and four times for second order.
Order with respect to X (experiments 1 and 2: [Y] fixed at 0.2, [X] doubled):
\( \displaystyle \frac{\text{Rate}_2}{\text{Rate}_1}=\frac{8.72\times10^{-2}}{4.36\times10^{-2}}=\left(\frac{0.4}{0.2}\right)^a \)
\( \displaystyle 2=2^a,\quad a=1 \)
Order with respect to Y (experiments 1 and 3: [X] fixed at 0.2, [Y] doubled):
\( \displaystyle \frac{\text{Rate}_3}{\text{Rate}_1}=\frac{17.44\times10^{-2}}{4.36\times10^{-2}}=\left(\frac{0.4}{0.2}\right)^b \)
\( \displaystyle 4=2^b,\quad b=2 \)
So the rate law is Rate = k[X][Y]2, and the order of the reaction is a + b = 1 + 2 = 3.
Shortcut: Doubling [X] doubles the rate (first order); doubling [Y] makes it four times (second order). 1 + 2 = 3.