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
$$\begin{aligned} k & =\frac{2.303}{t} \log \frac{[R]_0}{[R]} \\ \frac{k t}{2.303} & =\log \left[\frac{R^{\circ}}{R}\right] \\ \frac{k t}{2.303} & =\log [R]_0-\log [R]\\ \log [R]&=\underset{\text { Slope }}{\left(\frac{-k}{2.303}\right)} t+\underset{\text { Intercept }}{\log [R]_0} \end{aligned}$$
Match the statements given in Column I and Column II.
Column I | Column II | ||
---|---|---|---|
A. | Catalyst alters the rate of reaction | 1. | Cannot be fraction or zero |
B. | Molecularity | 2. | Proper orientation is not there |
C. | Second half-life of first order reaction | 3. | By lowering the activation energy |
D. | $e^{-E_a/RT}$ | 4. | Is same as the first |
E. | Energetically favourable reactions are sometimes slow | 5. | Total probability is one |
F. | Area under the Maxwell, Boltzmann curve is constant | 6. | Refers to the fraction of molecules with energy equal to or greater than activation energy |
A. $\rightarrow(3)$ B. $\rightarrow$ (1) C. $\rightarrow$ (4) D. $\rightarrow$ (6) E. $\rightarrow$ (2) F. $\rightarrow$ (5)
1. Catalyst alters the rate of reaction by lowering activation energy.
2. Molecularity can't be fraction or zero. If molecularity is zero, then reaction is not possible.
3. Second half-life of first order reaction is same as first because half-life time is temperature independent.
4. $e^{-E_Q / R T}$ refers to the fraction of molecules with kinetic energy equal to or greater than activation energy.
5. Energetically favourable reactions are sometimes slow due to improper orientation of molecule cause some ineffective collision of molecules.
6. Area under the Maxwell, Boltzmann curve is constant because total probability of molecule taking part in a chemical reaction is equal to one.
Match the items of Column I and Column II.
Column I | Column II | ||
---|---|---|---|
A. | Diamond | 1. | Short interval of time |
B. | Instantaneous rate | 2. | Ordinarily rate of conversion is imperceptible |
C. | Average rate | 3. | Long duration of time |
A. $\rightarrow(2)$ B. $\rightarrow$ (1) C. $\rightarrow$ (3)
1. Diamond can't be converted into graphite under ordinary condition.
2. Instantaneous rate of reaction completes at very short span of time.
3. Average rate of reaction occurs to a long duration of time.