Chemistry Chemical Kinetics

A → B
t=0: a, 0
t=30min: (a-x), x
given x = 0.2 mole / lit
Average rate = +Δ [B]/Δt = 0.2 mole/lit / (30/60)h = 0.4 mole lit? ¹h? ¹
= 4 * 10? ¹ mole lit? ¹h? ¹

Rate = K [A]?
K = Rate / [A]?
= (mole L? ¹s? ¹) / (mole L? ¹)?
= mole¹? L? ¹ s? ¹
Unit of K: mole¹? L? ¹ s? ¹

X → Y

$\begin{array}{l}{E}_{a_f}-E_{a_b}=-20\\ 30-E_{a_b}=-20\end{array}$

$E_{a_b}=50kJ/mole$

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 = $e^{-\frac{80.9*10^3}{8.314*700}}$ ${}^{\frac{{}^{}}{}}$

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

So, x = 14

$log\left(\frac{k_2}{k_1}\right)=\frac{Ea}{2.303*8.314}\left[\frac{1}{300}-\frac{1}{325}\right]$

log (5) = $\frac{Ea}{2.303*8.314}\left[\frac{1}{300}-\frac{1}{325}\right]$

$\therefore Ea=\frac{0.7*2.303*8.314*300*325}{25}=52271.7$

Ea in kJ/mole = $\frac{52271.7}{1000}=52.2kJ/mol$

the nearest integer is 52.

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