Chemistry Chemical Kinetics

t_1/2 = 0.301 min

t = 2 min

$K=\frac{2.303}{t}\mathcal{l}og\left(\frac{Co}{Ct}\right)$  

$\frac{0.693}{0.301}=\frac{2.303}{2}\mathcal{l}og\left(\frac{Co}{Ct}\right)$  

$\frac{2.303*0.301}{0.301}=\frac{2.303}{2}\mathcal{l}og\left(\frac{Co}{Ct}\right)$  

$\therefore 2=\mathcal{l}og\left(\frac{Co}{Ct}\right)$  

$\frac{Co}{Ct}=10^2=100$  

              Ans. 100

$T_{90\mathrm{\%}}=\frac{2.303}{k}log\frac{100}{10}$

$T_{50\mathrm{\%}}=\frac{2.303}{k}log\frac{100}{50}$

$\frac{T_{90\mathrm{\%}}}{T_{50\mathrm{\%}}}=\frac{log10}{log2}=\frac{1}{0.301}=3.32$

For first order reaction, using

$K=\frac{2.303}{t}lo{g}_{10}\frac{{\left[R\right]}_0}{\left[R\right]}$              

$K=\frac{2.303}{60}\left(0.477-0.301\right)mi{n}^{-1}$              

$K=6.7*10^{-3}mi{n}^{-1}$              

Ans. is 7 (the nearest integer)

Kt = 2.303 log $\frac{\left[A_0\right]}{\left[A_t\right]}$  

Here, $\left[A_0\right]=100,\left[A_t\right]=90andt=1min$  

$\therefore K*1=2.303lo{g}_{10}\frac{100}{90}$

$\therefore K=2.303\left[1-2*0.477\right]=0.105938=105.938*10^{-3}mi{n}^{-1}$                             

Rounded = 106

$PC{l}_5\left(g\right)\to PC{l}_3\left(g\right)+C{l}_2\left(g\right)$ (first order reaction)

$K=\frac{2.303}{120}log\left(\frac{50}{10}\right)$      

$\therefore K=\frac{2.303*log5}{120min}=\frac{2.303*0.6989}{120}$ = 0.0134 min^-1

Rate constant at 300K = 1.34 * 10^-2min^-1 [the nearest integer = 1.0]

$2N{O}_2\left(g\right)?N_2{O}_4\left(g\right)$

$\Delta S=-176J{K}^{-1}=-0.176kJ{K}^{-1}$  

$\Delta H=-57.8KJmo{l}^{-1}$               

 T = 298 K

Using ,   $\Delta G=\Delta H-T\Delta S$

$=-57.8+\left(298*0.176\right)KJmo{l}^{-1}$              

$=-5.3KJmo{l}^{-1}$               

Hence, magnitude of  $\Delta G$ in kJmol^-1 is 5 (nearest integers)

log k = 20.35 - $\frac{\left(2.47*10^3\right)}{T}$  

Comparing with,

$logk=logA-\frac{Ea}{2.303RT}$

$\frac{Ea}{2.303R}=2.47*10^3$

= 47.29 kJ/mole

Ans. = 47

$\Delta H={\left(E_{act}\right)}_f-{\left(E_{act}\right)}_b$

= x – (x + y) = -y

              = -45 KJ/mol

              Ans. = 45

$K_{700}=6.36*10^{-3}{s}^{-1}$

$K_{600}=x*10^{-6}{s}^{-1}$

E_a = 209 KJ/mol

$log\left(\frac{K_2}{K_1}\right)=\frac{-Ea}{2.303R}\left[\frac{1}{T_2}-\frac{1}{T_1}\right]$

$log{K}_{600}=-4.79$           

$K_{600}=10^{-4.79}=1.62*10^{-5}$

= 16.2 * 10^-6

x = 16 (the nearest integer)

t1/2 = 340 secP0 = 55.5 kPa

t1/2 = 170 sec P0 = 27.8 kPa

$\therefore t_{1/2}\propto {\left(P_0\right)}^{1-n}$

$\therefore 1-n=1n=0$

$\therefore$ zero order reaction