From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.
(i) Given rate = k [NO]2
Therefore, order of the reaction = 2
Dimension of \(K =\frac {Rate }{ [NO]^2}\)
\(= \frac {mol L^{-1} s^{-1}}{(mol L^{-1})^2}\)
\(= \frac {mol L^{-1} s^{-1}}{mol^2 L^{-2}}\)
\(= L \ mol^{-1} s^{-1}\)
(ii) Given rate = k [H2O2] [I−]
Therefore, order of the reaction = 2
Dimension of \(K = \frac {Rate}{[H_2O_2][I^-]}\)
= \(\frac {mol L^{-1} s^{-1}}{(mol L^{-1})(mol L^{-1})}\)
= \(L\ mol^{-1} s^{-1}\)
(iii) Given rate = \(k [CH_3CHO]^{\frac 32}\)
Therefore, order of reaction =\(\frac 32\)
Dimension of \(K = \frac {Rate}{ [CH_3CHO]^{\frac 32}}\)
= \(\frac {mol L^{-1} s^{-1}}{(mol L^{-1})^{\frac 32}}\)
= \(\frac {mol L^{-1} s^{-1}}{mol^{\frac 32} L^{-\frac 32}}\)
= \(L^{\frac 12} mol^{-\frac 12} s^{-1}\)
(iv) Given rate = \(k [C_2H_5Cl]\)
Therefore,order of the reaction = 1
Dimension of \(k = \frac {Rate}{ [C_2H_5Cl]}\)
= \(\frac {mol L^{-1} s^{-1}}{(mol L^{-1})}\)
= \(s^{-1}\)
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.