Chemistry Chemical Kinetics

$\frac{{\left(t_{1/2}\right)}_l}{{\left(t_{1/2}\right)}_{II}}={\left(\frac{P_I}{P_{II}}\right)}^{1-n}\Rightarrow n=1$

t1/2 = 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

$k=A.e^{-\left(\frac{Ea}{RT}\right)}$

$\mathcal{l}nk=\mathcal{l}nA-\frac{Ea}{RT}$

$\Rightarrow \mathcal{l}nk=\mathcal{l}nA+\left(-\frac{Ea}{1000R}\right).\frac{1000}{T}$

Slope = $\frac{-Ea}{1000R}=-18.5$  

=> Ea = 18.5 * 1000 * 8.31 = 153.735 * 10³ J = 154 KJ

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

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

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

CN is a strong field ligand, so pairing occurs.

1. ${\left[Fe{\left(CN\right)}_6\right]}^{4-}$

So; it is diamagnetic

2. ${\left[Fe{\left(CN\right)}_6\right]}^{3-}$

So; it is paramagnetic.

3. ${\left[Ti{\left(CN\right)}_6\right]}^{3-}$

Ti³⁺ = 4s⁰3d¹

 It is paramagnetic

4. ${\left[Ni{\left(CN\right)}_4\right]}^{2-}$

It is diamagnetic

5. ${\left[Co{\left(Cn\right)}_6\right]}^{3-}$

Hence,  ${\left[Fe{\left(CN\right)}_6\right]}^{3-}and{\left[Ti{\left(CN\right)}_6\right]}^{3-}$ ${{}_{}}^{}{{}_{}}^{}$ are paramagnetic.

$M{n}^{2+}\stackrel{}{\to }{t}_{2g^{111}}{e_g}^{11}$

(Number of unpaired electron = 5)

${\mu }_S=\sqrt[]{35}=5.91\cong 6$

Most basic oxide V₂O3

Here V has +3 O.S. Hence V⁺³ $\left[Ar\right]3d^2$

two unpaired e^- in d- subshell

$\mu =\sqrt[]{2\left(2+2\right)}=\sqrt[]{8}=2.84BM\approx 3$

t_1/2 = 200 sec and it is 1st order reaction

$K=\frac{2.303log2}{200}=\frac{2.303}{t}log\frac{Co}{0.2Co}$

$\frac{log2}{200}=\frac{1}{t}log5$

$t=\frac{7}{3}*200=466.67sec=467sec$