Why in the redox titration of $\mathrm{KMnO}_4$ vs oxalic acid, we heat oxalic acid solution before starting the titration?
As we know with increase in temperature rate of reaction increases, Hence, we heat oxalic acid solution before starting of titration to increase the rate of decolourisation.
Why can't molecularity of any reaction be equal to zero?
Molecularity of the reaction is the number of molecules taking part in an elementary step. For this we require at least a single molecule leading to the value of minimum molecularity of one. Hence, molecularity of any reaction can never be equal to zero.
Why molecularity is applicable only for elementary reactions and order is applicable for elementary as well as complex reactions?
A complex reaction occurs through a number of steps i.e., elementary reactions. Number of molecules involved in each elementary reaction may be different, i.e., the molecularity of each step may be different. Therefore, it is meaningless to talk of molecularity of the overall complex reaction.
On the other hand, order of complex reaction depends upon the molecularity of the slowest step. Hence, it is not meaningless to talk of the order of a complex reaction.
Why can we not determine the order of a reaction by taking into consideration the balanced chemical equation?
Balanced chemical equation often leads to incorrect order or rate law. e.g., the following reaction seems to be a tenth order reaction
$$\mathrm{KClO}_3+6 \mathrm{FeSO}_4+3 \mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{KCl}+3 \mathrm{H}_2 \mathrm{O}+3 \mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3$$
This is actually a second order reaction. Actually the reaction is complex and occurs in several steps. The order of such reaction is determined by the slowest step in the reaction mechanism.
Order is determined experimentally and is confined to the dependence of observed rate of reaction on the concentration of reactants.
Match the graph given in Column I with the order of reaction given in Column II. More than one item in Column I may link to the same item of Column II.
| Column I | Column II | ||
|---|---|---|---|
| A. |  |
||
| B. |  |
1. | First order |
| C. |  |
2. | |
| D. |  |
A. $\rightarrow$ (1) B. $\rightarrow$ (2) C. $\rightarrow$ (2) D. $\rightarrow$ (1)
For zero order reaction rate equation may be written as
$$[R]=-k t+\left[R_0\right]\quad\text{.... (i)}$$
Which denotes a straight line equation similar to $y=m x+c$
On transforming (i)
$$\begin{aligned} \frac{[R]-\left[R_0\right]}{t} & =-k \\ k & =\frac{\left[R_0\right]-[R]}{t} \end{aligned}$$
$$\begin{aligned} k & =\text { Rate } \\ \text { Rate } & =k \cdot[t]^0 \\ \text { Rate } & \propto[t]^0 \end{aligned}$$
For a first order reaction $\frac{d x}{d t} \propto[\text{concentration}]^{1}$
$\therefore$ Graph between rate and concentration may be drawn as
$$\begin{aligned} k & =\frac{2.303}{t} \log \frac{[R]_0}{[R]} \\ \frac{k t}{2.303} & =\log \left[\frac{R^{\circ}}{R}\right] \\ \frac{k t}{2.303} & =\log [R]_0-\log [R]\\ \log [R]&=\underset{\text { Slope }}{\left(\frac{-k}{2.303}\right)} t+\underset{\text { Intercept }}{\log [R]_0} \end{aligned}$$