Chemical kinetics Lecture-4
Kinetic reaction profile
Rate of reaction
or
Rate of reaction in terms of reactants
$=-\frac{\Delta[ClO^-]}{\Delta t}$ $=-\frac{\Delta[Br^-]}{\Delta t}$
$\frac{\Delta[ClO^-]}{\Delta t}$ $=\frac{c_2 - c_1}{t_3 - t_1}=\frac{-ve}{+ve}=-ve$
Rate of reaction in terms of products
General reaction
aA+bB+……$\rarr pP+qQ+….$
A,B,C..$\quad$ $\rArr$ reactants
a b c $\quad$ $\rArr$ stoichiometric coefficients
P,Q,R $\quad$ $\rArr$ products
p q r $\quad$ $\rArr$ stoichiometric coefficients
For reactants - $\quad$ $\nu$ $\quad$is $\quad$-ve
For products - $\quad$ $\nu$ $\quad$is $\quad$+ve
$\nu_A$ $\quad$ = $\quad$ -a
$\nu_B$ $\quad$ = $\quad$ -b
$\nu_p$ $\quad$ = $\quad$ p
$\nu_q$ $\quad$ = $\quad$ q
$aA+bB \rarr pP+qQ$
Degree of advancement of a reaction $\rightarrow$ $\xi$
$n_i=n_i^0 + \nu_i \xi……(1)$
$n_i$ number of moles of no chemical species i
$n_i^0=$ the number of moles of the chemical species ‘i’ when $\xi= \leftarrow$ degree of advancement 0
$\nu_i$ stoichiometric coefficients
$n_i=n^0_i + \nu_i \xi……(1)$
Differentiate (1) with respect to time
$\frac{dn_i}{dt}$= $\frac{dn_i^0}{dt}+\frac{d(\nu_i \xi)}{dt}……(2)$
Because $n_i^0$ is a constant , hence $\frac{dn_i^0}{dt}= 0$
$\frac{d(\nu_i \xi)}{dt}=\nu_i \frac{d\xi}{dt}$
$\frac{dn_i}{dt}=\nu_i \frac{d\xi}{dt}$
$\frac{d\xi}{dt}=\frac{1}{\nu_i}\frac{dni}{dt}$…….(3)
Rate of advancement of the reaction $\rightarrow$ rate of the reaction
$i \rarr A$
$n_A=n_A^0+\nu_A \xi$
$n_A=n^0_A - a \xi$
$\frac{dn_A}{dt}=\frac{dn_A^0}{dt} + \frac{d(-a \xi)}{dt}$
$\frac{dn_A}{dt}=-a\frac{d\xi}{dt}$
$\frac{d\xi}{dt}=-\frac{1}{a}\frac{dn_A}{dt}$
Rate of reaction in terms of the change in the number of moles of ‘A’
$i\rarr B$
$n_B=n^0_B + \nu_B \xi$
$n_B=n^0_B - b \xi$
$\frac{dn_B}{dt}=\frac{dn^0_B}{dt} + \frac{d(-b \xi)}{dt}$
$\frac{dn_B}{dt}=-b\frac{d\xi}{dt}$
$\frac{d\xi}{dt}=-\frac{1}{b} \frac{dn_B}{dt}$
$i\rarr P$
$n_p=n^0_P+ \nu_P \xi$
$n_p=n^0_P+ p \xi$
$\frac{dn_P}{dt}$=$\frac{dn^0_P}{dt}+\frac{d(P\xi)}{dt}$
$\frac{dn_P}{dt}=P\frac{d\xi}{dt}$ $∴ \frac{d\xi}{dt}=\frac{1}{P}\frac{dn_P}{dt}$
$\frac{d\xi}{dt}$ $=-\frac{1}{a}\frac{dn_A}{dt}=-\frac{1}{b}\frac{dn_B}{dt}$*
$=\frac{1}{p}\frac{Qn_p}{dt}=\frac{1}{q}\frac{dn_Q}{dt}$
Stoichiometric coefficients
Reactant $\quad$ : $\quad$ Put a ‘-ve’ sign
$\qquad\qquad\quad$ before the coefficient
Product $\quad$ : $\quad$ Put a ‘+ve’ sign
Example of stoichiometric coefficient
$2N_2O_5\rarr4NO_2 + O_2$
$\frac{d\xi}{dt}$ $=-\frac{1}{2}\frac{dn_{N_2 O_5}}{dt}$ $=\frac{1}{4} \frac{dn_{NO_2}}{dt} =\frac{dn_{O_2}}{dt}$
Reaction
Since most of the reactions are done under constant volume conditions , then
from-(1) $\rarr n_i=n^0_i + \nu_i \xi……(1)$
$\frac{d\xi}{dt}=\frac{1}{\nu_i}\frac{dn_i}{dt}……(3)$
$\frac{1}{V}\quad \frac{d\xi}{dt}$ $=\frac{1}{V}\quad \frac{1}{\nu_i}\frac{dn_i}{dt}$
V(volume) is constant
$\frac{d(\xi V)}{dt}$ $=\frac{1}{\nu_i}\frac{d(n_i/V)}{dt}$
$\frac{d(\xi /V)}{dt}$ $=\frac{1}{\nu_i}\frac{d[i]}{dt}……(4)$
[i] $\rArr$ concentration of species ‘i’
(rate of reaction)
Example
$CH_3CHO(g) \rarr CH_4(g)+CO(g)$
The rate of this reaction can be followed by measuring the pressure in the system of constant volume and temperature
Assume ideal gas behaviour of the gases
Initial $\quad$ $n^0_{CH_3CHO} \quad\quad 0 \quad\quad\quad\quad 0$
$n^0_{CH_3CHO}-\xi \quad \quad \quad \quad\xi \quad \quad \quad\quad \xi $
$n_{CH_3CHO}=n^0_{CH_3CHO}+\nu_i\xi$
$n_{CH_3CHO}=n^0_{CH_3CHO}-\xi$
$n_{CH_4}=n^0_{CH_4}+\nu_i \xi$
$n_{CH_4}=n^0_{CH_4}+\xi$
$n_{CH_4}=\xi \hspace{3 mm}[n^0_{CH_4=0}]$
$n_{CO}=n^0_{CO}(=0) + \nu_i \xi (+1)$
$n_{CO}=0+\xi$
$n_{CO}=\xi$
Thank you