Chemical kineticsLecture-8
Chemical kinetics Lecture-8
Zero order reaction
$A \rightarrow P$
${ rate } = k[A]^0 = k $—(1)
$-\frac{d[A]}{d t} = k$—(2)
$d[A] = - kdt $ —(3)
$\int_{[A]_0}^{[A]_t} d[A]=-\int_0^t k d t \cdots \text { (4) }$
$k$ is a constant and can be kept out side the integral
$[A]_t - [A]_0 = - k(t-0)$
${[A]_t-[A]_0=-k t} \ $
${[A]_t=[A]_0-k t} $—(5)
$y=m x+c$
If plot of $[A]_t$ versus ’ $t$ ’ is a straight line, then this is the signature of a zero order reaction
Half-life for a zero order reaction
$[A]0$, $\hspace{2mm}$ $t\frac{1}{2}$ $\rightarrow$ $\frac{1}{2}[A]_0$
$[A]_t = [A]_0 - kt \dots (5)$
$\frac{1}{2}[A]_t = [A]0 - kt\frac{1}{2} \dots (6)$
$kt_{1 / 2} =[A]_0-\frac{1}{2}[A]_0$
$ k t_{1 / 2} =\frac{1}{2}[A]_0 $
$t_{1 / 2} =\frac{[A]_0}{2 k} $—(7)
Half-life is proportional to the concentration of the reactant
Higher concentration
Higher, half life
As a reaction proceeds, then the half-life also decreases
Half-life being used as a preliminary check
First order reaction
$A \rightarrow P $
${ rate } = k[A]…(8) $
$-\frac{d[A]}{d t}=k[A] \cdots(9)$
$\therefore-\frac{d[A]}{[A]} =k d t$
$\int_{[A]_0}^{[A_t]}\frac{d[A]}{[A]}$ =-k $\int_0^t d t$
$\ln [A]_t-\ln [A]_0=-k t$….(10)
$\ln [A]_t=\ln [A]_0-k t$…. (11)
$\ln \frac{[A]_t}{[A]_0}$ = -k t…. (12)
$\frac{[A]_t}{[A]_0}$ =$e^{-k t}$….(13)
$[A]_t=[A]_0 e^{-k t}\ $ ,
$[A]_t=[A]_0 \exp (-k t)$…(14)
$\ln [A]_t=ln [A]_0-k t \cdots(11)$
If the experimental data fit first order kinetics, then the plot of $\ ln [A]_t$ versus time will be a straight line that is linear
Half-life for first order reaction
$t_\frac{1}{2}$ $\Rightarrow$ $\frac{1}{2}[A]_0$
$\ln \frac{[A]_t}{[A]_0}=-k t$…(12)
At $\quad t_{1 / 2} \quad \ln \frac{[A_{t\frac {1}{2}}]}{[A_0]}$=$-k t_{1 / 2}$ …..(15)
$\ln\frac{\frac{1}{2}[A]_0}{[A]_0} = -kt\frac{1}{2 }$
$\ln \frac{1}{2}$ = $-k t_{1 / 2}$ …. (16)
$t_{1 / 2}=\frac{\ln 2}{k} $…..(17)
$ t_{1 / 2}=\frac{0.693}{k}$…..(18)
The half-life for a first order reaction is independent of the concentration of the reactant
$t_{1 / 2}$ takes the same value no matter what the extent of reaction is
Example of first order reaction
$2N_2 O_5(g)\rightarrow 4NO_2(g)+\mathrm{O}_2(\mathrm{g})$
$ rate = k[\mathrm{N}_2{O}_5]$
$\mathrm { rate } = -\frac{1}{2} \frac {{d}[\mathrm{N}_2 \mathrm{O}_5]}{ d t}$
$= - \frac{1}{2} \frac {{d}[\mathrm{N}_2 \mathrm{O}_5]}{ d t}$
$-\frac{1}{2} \frac{d\left[\mathrm{N}_2 \mathrm{O}_5\right]}{d t}=k\left[\mathrm{N}_2{O}_5\right]$——-(19)
$ {\int_{[N_2 O_5]}^{[N_2 O_5]} \frac {d[N_2 O_5]}{[N_2 O_5]}} $
$=-\int_{t=0}^t 2 k d t$
$\ln \left[\mathrm{N}_2 \mathrm{O}_5\right]_t-\ln \left[\mathrm{N}_2 \mathrm{O}_5\right]_0=-2 R t$
$\ln \left[\mathrm{N}_2 \mathrm{O}_5\right]_t=\ln \left[\mathrm{N}_2 \mathrm{O}_5\right]_0-2 k t\ldots (20)$
To generalise-
$ 2 A \rightarrow P$
$a A \rightarrow P$
$\text { rate }=-\frac{1}{a} \frac{d[A]}{d t}=k[A]$
$\ln [A]_{t}=\ln [A]_0-a k t$ …(21)
Relaxation time $(\tau)$ $\rightarrow tau$
$[A]_0 \rightarrow \frac{1}{e}[A]_0$
Relaxation time
${[A]_{t=\tau}=0.368[A]_0}$
at $t=\tau$ (relaxation time)
Reaction has gone to 63.2%
Completion and still has 36.8% to go
${[A]_t } =[A]_0 e^{-k t} \cdots \text { (14) }$
$t=\tau$ $\quad \frac{1}{e}[A]_0 =[A]_0 e^{-k t}$
$\frac{1}{e} = e^{-k t}$
$e^{-1} = e^{-k t}$
$e^{-1} = e^{-k \tau}$…22
$e^{1}$ = $e^{(k\tau)}$
ln e = k $\tau$
k $\tau$ = 1
$\tau$ = $\frac{1}{k} \quad \ldots(23)$
Relaxation times are only applicable for first order or pseudo first order reaction
${[A]_t=[A]_0 e^{-k t}}$
$t_{1 / 2}=\frac{0.693}{k}$
Thank you