Chemical kineticsLecture-10
Chemical kinetics Lecture-10
Chemical kinetics
Hydrolysis of ethylacetate
$\rightarrow $ acid catalyzed
$\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5+\mathrm{H}_2 \mathrm{O} \xrightarrow{\mathrm{H}^{+}} $
$\mathrm{CH}_3 \mathrm{COOH}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} $
$r=k\left[\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5\right]\left[\mathrm{H}_2 \mathrm{O}\right]$
Example of pseudo order rate equation
$r=k\left[\mathrm{H}_2 \mathrm{O}\right]\left[\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5\right] $
$r=k^{\prime}\left[\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5\right] $
$k^{\prime}=k\left[\mathrm{H}_2 \mathrm{O}\right] $
$k^{\prime} \longrightarrow $pseudo-first order rate constant
$C_6 H_5 N_2 Cl (aq)+H_2 O(l) $
$\rightarrow\left.C_6 H_5 O H\left(aq)\right)+N_2(g)+H C I(aq)\right) $
$r= k\left[C_6 H_5 N_2(l)\right]\left[H_2 O\right] $
$r= k^{\prime}\left[C_6 H_5 N_2(l)\right. $
$k^{\prime}=k\left[H_2 O\right]$
Initial rate method
$a A+b B \rightarrow P $
$r=k[A]^\alpha[B]^\beta$
Initial rate $\rightarrow r_0$
$r_0=k[A]_0^\alpha[B]_0^\beta \longrightarrow(1)$
$\qquad \qquad\downarrow$
$\qquad\qquad\downarrow$
$r_0={k[A]_0^\alpha} [ B]_0^{\beta} \longrightarrow (1) $
$r_0=k^{\prime}[B]_0^\beta \longrightarrow (2) $
$k^{\prime}=k[A]_0^\alpha$(pseudo order rate constant)
$[A]_0$ is held constant
Experiment
$\space$
$r_0=k\left[\mathrm{ClO^-}\right]_0^\alpha\left[\mathrm{Br}^{-}\right]_0^\beta…. (3)$
$r_0={k\left[\mathrm{Br}^-\right]_0^\beta}\left[\mathrm{ClO}^{-}\right]_0^\alpha $
$r_0=k^{\prime}\left[\mathrm{ClO}^{-}\right]_0^\alpha…. (4)$
$r=k\left[\mathrm{ClO}^{-}\right][\mathrm{Br}^-] \cdots \text { (5) } $
$r_0=k^{\prime}\left[\mathrm{ClO}^{-}\right]_0^\alpha \cdots \text { (4) } $
$\frac{r_0}{[\mathrm{ClO}-]_0^\alpha}=k^{\prime} $
$\text { if } \alpha=1 \quad \frac{r_0}{\left[\mathrm{ClO} ^{-}\right]_0}=k^{\prime} \cdots \text { (6) }$
Experiment-1
$r_0=3.19 \times 10^{-6} \mathrm{mol} L^{-1} \mathrm{~s}^{-1} $
${\left[\mathrm{ClO}^{-}\right]_0=3.23 \times 10^{-3} \mathrm{mol}L^{-1}} $
$\frac{r_0}{\left[\mathrm{ClO^-}\right]_0}=\frac{3.19 \times 10^{-6} \mathrm{mol}L^{-1} \mathrm{~s}^{-1}}{3.23 \times 10^{-3} \mathrm{mol}L^{-1}}$
$={9.88 \times 10^{-4} \mathrm{S}^{-1}}=k^{\prime} $
Experiment-2
$r_0=5.98 \times 10^{-6} \mathrm{mol} L^{-1} \mathrm{~s}^{-1} $
${\left[\mathrm{ClO}^{-}\right]_0=6.07 \times 10^{-3} \mathrm{mol}L^{-1}} $
$\frac{r_0}{\left[\mathrm{ClO^-}\right]_0}=\frac{5.98 \times 10^{-6} \mathrm{mol}L^{-1} \mathrm{~s}^{-1}}{6.07 \times 10^{-3} \mathrm{mol}L^{-1}} $
$={9.85 \times 10^{-4} \mathrm{s}^{-1}}=k^{\prime} $
Experiment-3
$r_0=9.14 \times 10^{-6} \mathrm{mol} L^{-1} \mathrm{~s}^{-1} $
${\left[\mathrm{ClO}^{-}\right]_0=9.25 \times 10^{-3} \mathrm{mol}L^{-1}} $
$\frac{r_0}{\left[\mathrm{ClO^-}\right]_0}={9.88 \times 10^{-4} \mathrm{~s}^{-1}}$
$=k^{\prime} $
Temperature dependence of reaction rates
$\quad\downarrow$
Example
Expression arrhenius
$k=A e^{-E_a / R T} \cdots 1 a $
$k=A \exp \left(-E_a / R T\right) \cdots 1 b \longrightarrow$ Arrhenius equation
$k=A_{e^{-}} E_a / R T$
A : pre-exponential factor or frequency factor or Arrhenius factor
Ea : Activation Energy Arrhenius Activation Energy
$k$ : rate constant
$R$ : gas constant
Origin of chemical kinetics
1850 - 1910 $\longrightarrow$ A lot of work being done to understand the temperature dependence
1904 $\longrightarrow$ Ostwald
Temperature dependence of reaction rates is “one of the darkest chapters in chemical mechanics”.
Van’t Hoff
$\left(\frac{\partial \ln K c}{\partial T}\right)_p=\frac{\Delta U^0}{R T^2}….(2)$
$K_c$ : concentration equilibrium constant
$\Delta U^{\circ}$ : standard internal energy change
Reaction
$A+B \xrightleftharpoons [{k_{-1}}]{k_1} P+Q $
$r_{\text {forward }}=k_1[A][B] $
$r_{\text {backward }}=k_{-1}[P][Q]$
At equilibrium
$r_{\text {forward }} =r_{\text {backward }} $
$k_1[A][B] =k_{-1}[P][Q] …..(3)$
$\frac{[P][Q]}{[A][B]} =K_C=\frac{k_1}{k_{-1}}……(4)$
$\frac{d ln K_c}{d T} =\frac{\Delta U^0}{R T^2} \cdots \text { (2) } $
$K_c =\frac{K_1}{K_{-1}} \cdots \text { (4) }$
using (4) in (2) we have
$\frac{d \ln k_1}{d T}-\frac{d \ln k_{-1}}{d T}=\frac{\Delta U^0}{R T^2} \cdots(5) $
$\frac{d \ln k_1}{d T}=\frac{E_1}{R T^2} \cdots(6) $
$\frac{d \ln k_{-1}}{d T}=\frac{E_{-1}}{R T^2} \cdots \text { (7) }$
$\Delta U^0 =E_1-E_{-1} \cdots(8) $
$\frac{d \ln k}{d T} =\frac{E}{R T^2} \cdots(9) $
$\ln k =\text { constant } - \frac {E}{RT}$
$k =A e^{-E / R T}$
Thank you