Chemistry Chemical Kinetics

$K=A.e^{-Ea/RT}=\left(6.5*10^2\right)e^{-26000K/T}$

$\frac{Ea}{8.314}=26000$

Ea = 216.164kJ/mol $\cong$ $$ 216

Process is based upon simultaneous disintegration hence,

$\frac{0.693}{100}*t=2.303lo{g}_{10}\frac{A_0}{A_t}$ ………….(i)

and       $\frac{0.693}{50}*t=2.303lo{g}_{10}\frac{B_0}{B_t}$              ………….(ii)

from equation (i) and (ii)

$0.693t\left[\frac{1}{50}-\frac{1}{100}\right]=\left[log\frac{B_0}{B_t}-log\frac{A_0}{A_t}\right]*2.303$  

Here; A₀ = B₀ and $A_t=4*B_t$  

Therefore  $0.693t\left[\frac{1}{100}\right]=2.303\left[log\left(\frac{B_0}{B_t}*\frac{A_t}{A_0}\right)\right]$  

$\therefore t=\frac{2.303*0.3010*2*100}{693}=200s$  

Rate constant, k = 5.5 * 10^-14 s^-1.

$t=\frac{2.303}{k}log\frac{{\left[R\right]}_0}{\left[R\right]}$  

 = $\frac{2.303}{k}log3-(i)$  

$t_{50\mathrm{\%}}=\frac{2.303}{k}log2-(ii)$  

 From (i) & (ii)

$\frac{t_{67\mathrm{\%}}}{t_{50\mathrm{\%}}}=\frac{log3}{log2}$  

= 1.58 t_50%

So;         t_67% is 15.8 * 10^-1 times half life.

X = 16 (the nearest integer)

$K=A.e^{-Ea/RT}=\left(6.5*10^2\right)e^{-26000K/T}$  

$\frac{Ea}{8.314}=26000$

E_a = 216.164kJ/mol $\cong$ 216

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

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

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

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

$\begin{array}{l}t=\frac{t_{1/2}}{log2}.log\frac{V_{\infty }}{V_{\infty }-V_t}\\ 20=\frac{10}{log2}.log\frac{V_{\infty }}{V_{\infty }-25}\end{array}$

$\therefore V_{\infty }=33.33ml$

$\lambda =\frac{v}{\sqrt[]{2}\pi d^{2}n{N}_a}=\frac{RT}{\sqrt[]{2}\pi d^{2}p{N}_a}$

= $\frac{\frac{25}{3}*300}{\sqrt[]{2}*\pi *10^{-20}*10^5*6*10^{23}}$

$x^3=\frac{v}{n{N}_a}$

$={\left(\frac{\frac{25}{3}*300}{10^5*6*10^{23}}\right)}^{1/3}$