Chemistry Chemical Kinetics

For a first order reaction.

   = 3*10^-3 S-1

Fraction of molecule having energy $E_a$ or greater than $E_a=e^{\frac{-E_a}{RT}}$  

$t_{99.9}=\frac{2.303}{k}\log ?\frac{100}{100-99.9}=3*\frac{2.303}{k}$

$t_{50}=\frac{0.693}{k}$

$\frac{t_{99.9}}{t_{50}}=\frac{3*2.303*k}{k*0.693}=\frac{9}{96}\approx 10$

$r_1=k{\left[x\right]}^n$ … (i)

$$ $2r_1=k{\left[4x\right]}^n$  … (ii)

$\left(ii\right)÷\left(i\right)$

$\Rightarrow \frac{2r_1}{r_1}=\frac{k{\left[4x\right]}^n}{k{\left[x\right]}^n}$

$\Rightarrow 2={\left(2\right)}^{2n}$

$\Rightarrow n=\frac{1}{2}$

For a first order reaction.

$k=\frac{0.693}{t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=\frac{0.693}{231}=3*10^{-3}{s}^{-1}$   

t1/2 = [A]^1-n

for second order (n = 2) reaction : t1/2  is proportional to 1/ [A ]

$C_{12}{H}_{22}{O}_{11}+H_{2}O\to \underset{Glucose}{C_6{H}_{12}{O}_6}+\underset{Fructose}{C_6{H}_{12}{O}_6}$

$K=\frac{2.303}{t}log\frac{a}{a-x}$

$\frac{kt}{2.303}=log\frac{a}{a-x}$

$\frac{ln2*9}{\frac{10}{3}*2.303}=log\left(\frac{1}{f}\right)$

$log\left(\frac{1}{f}\right)=81.24*10^{-2}=81*10^{-2}$

K = 3.3 * 10^-4 s^-1

Time for 40% completion ; t

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

3.3 * 10^-4 = $\frac{2.303}{t}lo{g}_{10}\frac{{\left[R\right]}_0}{0.6{\left[R\right]}_0}$  

  $t=\frac{2.303}{3.3}*10^4*0.22\Rightarrow t=25.58mins$

so; the nearest integer is 26.

Fraction of molecules having enough energy to form product  =   $e^{-Ea/RT}$

 Fraction of molecules having enough energy to form product =   $e^{-\frac{80.9*10^3}{8.314*700}}$

$=e^{-13.8}\cong e^{-14}$

So, x = 14