All energetically effective collisions do not result in a chemical change. Explain with the help of an example.
Only effective collision lead to the formation of products. It means that collisions in which molecules collide with sufficient kinetic energy (called threshold energy = activation energy + energy possessed by reacting species).
And proper orientation lead to a chemical change because it facilitates the breaking of old bonds between (reactant) molecules and formation of the new ones i.e., in products. e.g., formation of methanol from bromomethane depends upon the orientation of the reactant molecules.
$$\mathrm{CH}_3 \mathrm{Br}+\mathrm{OH}^{-} \longrightarrow \mathrm{CH}_3 \mathrm{OH}+\mathrm{Br}^{-}$$
The proper orientation of reactant molecules leads to bond formation whereas improper orientation makes them simply back and no products are formed. To account for effective collisions, another factor $P$ (probability or steric factor) is introduced $K=P z_{A B} e^{-E a / R T}$.
What happens to most probable kinetic energy and the energy of activation with increase in temperature?
Kinetic energy is directly proportional to the absolute temperature and the number of molecules possessing higher energies increases with increase in temperature, i.e., most probable kinetic energy increases with increase in temperature.
Energy of activation is related to temperature by the following Arrhenius equation
$$k=A e^{-E_a / R T}$$
Thus, it also shows an increase with rise in temperature.
Describe how does the enthalpy of reaction remain unchanged when a catalyst is used in the reaction?
A catalyst is a substance which increases the speed of a reaction without itself undergoing any chemical change.
According to "intermediate complex formation theory" reactants first combine with the catalyst to form an intermediate complex which is short-lived and decomposes to form the products and regenerating the catalyst.
The intermediate formed has much lower potential energy than the intermediate complex formed between the reactants in the absence of the catalyst.
Thus, the presence of catalyst lowers the potential energy barrier and the reaction follows a new alternate pathway which require less activation energy.
We know that, lower the activation energy, faster is the reaction because more reactant molecules can cross the energy barrier and change into products.
Enthalpy, $\Delta H$ is a state function. Enthalpy of reaction, i.e., difference in energy between reactants and product is constant, which is clear from potential energy diagram. Potential energy diagram of catalysed reaction is given as
Explain the difference between instantaneous rate of a reaction and average rate of a reaction.
The difference between instantaneous rate of reaction and average rate of a reaction are as below
| Instantaneous rate of reaction | Average rate of reaction | |
|---|---|---|
| (i) | It occurs within a short span of time. | It occurs during a long interval of time. |
| (ii) | It can't be calculated for multistep reaction. | It can be calculated for multistep reaction. |
| (iii) | It can be calculated for elementary reaction. | It can be calculated for elementary reaction. |
With the help of an example explain what is meant by pseudo first order reaction.
A reaction in which one reactant is present in large amount and its concentration does not get altered during the course of the reaction, behaves as first order reaction. Such reaction is called pseudo first order reaction.
e.g., (i) hydrolysis of ethyl acetate
$$\text { Rate of reaction }=k\left[\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5\right]$$
where, $k=k^{\prime}\left[\mathrm{H}_2 \mathrm{O}\right]$
e.g., (ii) inversion of cane sugar
$$\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\mathrm{H}_2 \mathrm{O} \xrightarrow{\mathrm{H}^{+}} \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6$$
Rate of reaction $=k\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right]$
where $\quad k=k^{\prime}\left[\mathrm{H}_2 \mathrm{O}\right]$