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Solve the following.
For the reaction A + B rightleftharpoons Product, following data was obtained: Experiment number…
For the reaction $\text{A} + \text{B} \rightleftharpoons \text{Product}$, following data was obtained:
Calculate the following:
(i) The overall order of the reaction
(ii) The rate law equation
(iii) The value of rate constant
| Experiment number | Initial concentration of $[\text{A}]$ ($\text{mol L}^{-1}$) | Initial concentration of $[\text{B}]$ ($\text{mol L}^{-1}$) | Initial Rate ($\text{mol L}^{-1}\text{ min}^{-1}$) |
|---|---|---|---|
| 1 | 0·15 | 0·15 | $9\cdot6 \times 10^{-2}$ |
| 2 | 0·30 | 0·15 | $3\cdot84 \times 10^{-1}$ |
| 3 | 0·15 | 0·30 | $1\cdot92 \times 10^{-1}$ |
| 4 | 0·30 | 0·30 | $7\cdot68 \times 10^{-1}$ |
Answer
Answer
Orders: A: 2, B: 1, overall: 3
Rate law: rate = k[A]^2[B]
Rate constant: k = 28.44 L^2 mol^-2 s^-1
Let rate $= k[\mathrm{A}]^x[\mathrm{B}]^y$.
Experiments 1 and 2 ([B] constant): $\dfrac{3.84\times10^{-1}}{9.6\times10^{-2}} = 4 = 2^x$, so $x = 2$.
Experiments 1 and 3 ([A] constant): $\dfrac{1.92\times10^{-1}}{9.6\times10^{-2}} = 2 = 2^y$, so $y = 1$.
(i) Overall order $= 2 + 1 = 3$
(ii) Rate law: rate $= k[\mathrm{A}]^2[\mathrm{B}]$
(iii) $k = \dfrac{\text{rate}}{[\mathrm{A}]^2[\mathrm{B}]} = \dfrac{9.6\times10^{-2}}{(0.15)^2(0.15)} = 28.44\ \mathrm{L^2\,mol^{-2}\,min^{-1}}$
Final answer: 28.44 L^2 mol^-2 min^-1