‹ Back to the paper
Solve the following.
For the reaction A + B → C + D, the initial rate for different reactions and initial concentration…
For the reaction $\text{A} + \text{B} \rightarrow \text{C} + \text{D}$, the initial rate for different reactions and initial concentration of reactants are given below:
(i) What is the overall order of reaction?
(ii) Write the rate law equation.
| S. No. | Initial Conc. [A] ($\text{mole L}^{-1}$) | Initial Conc. [B] ($\text{mole L}^{-1}$) | Initial rate ($\text{mole L}^{-1}\text{ sec}^{-1}$) |
|---|---|---|---|
| 1 | 1·0 | 1·0 | $2 \times 10^{-3}$ |
| 2 | 2·0 | 1·0 | $4 \times 10^{-3}$ |
| 3 | 4·0 | 1·0 | $8 \times 10^{-3}$ |
| 4 | 1·0 | 2·0 | $2 \times 10^{-3}$ |
| 5 | 1·0 | 4·0 | $2 \times 10^{-3}$ |
Answer
Answer
Orders: A: 1.0, B: 0.0, overall: 1.0
Rate law: rate = k[A]
Rate constant: k = 2e-3 s^-1
Comparing experiments 1, 2 and 3: doubling [A] doubles the rate, so the order with respect to A is 1. Comparing 1, 4 and 5: changing [B] does not change the rate, so the order with respect to B is 0.
(i) Overall order = 1 + 0 = 1
(ii) Rate law: rate $= k[\mathrm{A}]$, with $k = \dfrac{2\times10^{-3}\text{ mol L}^{-1}\text{ s}^{-1}}{1\cdot0\text{ mol L}^{-1}} = 2\times10^{-3}\text{ s}^{-1}$
Final answer: 1