State a condition under which a bimolecular reaction is kinetically first order reaction.
Presence of one of the reactants in excess, as in such a condition, its concentration remains constant and rate of such reaction depends upon concentration of one reactant only and reaction is known as pseudo first order reaction e.g., acid catalysed hydrolysis of ethyl acetate.
$$\underset{\text { Ethyl acetate }}{\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5}+\mathrm{H}_2 \mathrm{O} \xrightarrow[\text { Acetic acid }]{\mathrm{H}^{+}} \mathrm{CH}_3 \mathrm{COOH}+\underset{\text { Ethyl alcohol }}{\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}}$$
This reaction is bimolecular but is found to be of first order as experimentally it is observed that rate of reaction depends upon the concentration of ethyl acetate not on water as it is present in excess.
Write the rate equation for the reaction $2 \mathrm{~A}+\mathrm{B} \longrightarrow \mathrm{C}$ if the order of the reaction is zero.
For reaction $2 A+B \longrightarrow C$ if the rate of reaction is zero then it can be represented as
$$\text { Rate }=k[A]^0[B]^0=k$$
i.e., rate of reaction is independent of concentration of $A$ and $B$.
How can you determine the rate law of the following reaction?
$$2 \mathrm{NO}(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_2(\mathrm{~g})$$
We can determine the rate of this reaction as a function of initial concentrations either by keeping the concentration of one of the reactants constant and changing the concentration of the other reactant or by changing the concentration of both the reactants. e.g., for the given reaction,
(i) Keeping $\left[\mathrm{O}_2\right]$ constant, if the concentration of NO is doubled, rate is found to become four times. This shows that,
$$\text { Rate } \propto[\mathrm{NO}]^2$$
(ii) Keeping [NO] constant, if the concentration of $\left[\mathrm{O}_2\right]$ is doubled, rate is also found to become double. This shows that,
$$\text { Rate } \propto\left[\mathrm{O}_2\right]^2$$
Hence, overall rate law will be
$$\begin{aligned} \text { Rate } =k\left[\mathrm{NO}^2\left[\mathrm{O}_2\right]\right. \\ \text{Rate law expression}\quad -\frac{1}{2} \frac{\Delta[\mathrm{NO}]}{\Delta t} & =-\frac{\Delta\left[\mathrm{O}_2\right]}{\Delta t} \\ & =\frac{1}{2} \frac{\Delta\left[\mathrm{NO}_2\right]}{\Delta t} \end{aligned} $$
For which type of reactions, order and molecularity have the same value?
If the reaction is elementary reaction then order and molecularity have same value because elementary reaction proceeds in a single step.
In a reaction if the concentration of reactant $A$ is tripled, the rate of reaction becomes twenty seven times. What is the order of the reaction?
Rate of any elementary reaction can be represented as
$$r=k[A]^n$$
After changing concentration to its triple value $A=3 A, r$ becomes $27 r$
$$\begin{aligned} & 27 r=k[3 A]^n \\ & \frac{r}{27 r}=\frac{k[A]^n}{k[3 A]^n} \\ & \frac{1}{27}=\left[\frac{1}{3}\right]^n \Rightarrow\left[\frac{1}{3}\right]^3=\left[\frac{1}{3}\right]^n \end{aligned}$$
Hence, $n=3$
Order of reaction is three.