A reaction is first order in A and second order in B.
\((i)\) The differential rate equation will be
\(-\frac {d[R]}{dt }= k[A][B]^2\)
\((ii)\) If the concentration of B is increased three times, then
\(-\frac {d[R]}{dt }= k[A][3B]^2\)
= \(9. k[A][B]^2\)
Therefore, the rate of reaction will increase 9 times.
\((iii)\) When the concentrations of both A and B are doubled,
\(-\frac {d[R]}{dt} = k[A][B]^2\)
= \(k[2A][2B]^2\)
= \(8. k[A][B]^2\)
Therefore, the rate of reaction will increase 8 times.
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.