The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:
| Experiment | A/ mol L-1 | B/ mol L-1 | Initial rate/mol L-1 min-1 |
| I | 0.1 | 0.1 | 2.0x10-2 |
| II | - | 0.2 | 4.0x10-2 |
| III | 0.4 | 0.4 | - |
| IV | - | 0.2 | 2.0x10-2 |
The given reaction is of the first order with respect to A and of zero order with respect to B.
Therefore, the rate of the reaction is given by,
Rate = k [A]1[B]0
⇒ Rate = k[A]
From experiment I, we obtain
2.0 x 10-2 mol L-1min-1= k (0.1 mol L-1)
⇒ k= 0.2 min-1
From experiment II, we obtain
4.0 x 10-2 mol L-1min-1= 0.2 min-1[A]
⇒ [A] = 0.2 mol L-1
From experiment III, we obtain
Rate = 0.2 min-1 x 0.4 mol L-1
= 0.08 mol L-1 min-1
From experiment IV, we obtain
2.0 x 10-2 mol L-1 min-1= 0.2 min-1 [A]
⇒ [A] = 0.1 mol L-1
Rate law for a reaction between $A$ and $B$ is given by $\mathrm{R}=\mathrm{k}[\mathrm{A}]^{\mathrm{n}}[\mathrm{B}]^{\mathrm{m}}$. If concentration of A is doubled and concentration of B is halved from their initial value, the ratio of new rate of reaction to the initial rate of reaction $\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)$ is
For $\mathrm{A}_{2}+\mathrm{B}_{2} \rightleftharpoons 2 \mathrm{AB}$ $\mathrm{E}_{\mathrm{a}}$ for forward and backward reaction are 180 and $200 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. If catalyst lowers $\mathrm{E}_{\mathrm{a}}$ for both reaction by $100 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Which of the following statement is correct?
The Order of reaction refers to the relationship between the rate of a chemical reaction and the concentration of the species taking part in it. In order to obtain the reaction order, the rate equation of the reaction will given in the question.