Class 11th

(a) $\Delta G=\Delta G°+RT\mathcal{l}nQ$

$\Delta G°=\Delta H°-T\Delta S°..............\left(1\right)$                      

At equilibrium $\Delta G=0,Q=K_{eq}$

  $\therefore \Delta G°=-RT\mathcal{l}n{K}_{eq}.................\left(2\right)$                   

$\therefore \Delta H°-T\Delta S°=-RT\mathcal{l}n{K}_{eq}$                      

$\mathcal{l}nKeq=-\left(\frac{\Delta H°-T\Delta S°}{RT}\right)$         

(b) $W_{rev}=-nRT\mathcal{l}n\left(\frac{V_f}{V_i}\right)$  

(c) $\Delta G=-T\Delta S_{Total}$ (at constant P)

$\therefore \frac{\Delta G}{\Delta S_{Total}}=-T$        

(d) $\Delta G°=-RT\mathcal{l}nKeq$

$\therefore Keq=e^{\left(-\Delta G°/RT\right)}$  

Electrophilic addition of bromine to an alkene is anti-addition, in which cis-alkene gives two enantiomers and trans – alkene gives meso form

Here; trans-but-2-ene will give meso products

Electrophilic addition of bromine to an alkene is anti-addition, in which cis-alkene gives two enantiomers and trans – alkene gives meso form

Here; trans-but-2-ene will give meso products

Structure (I) is anti conformer.

Structure (II) is fully eclipsed conformer.

Structure (III) is skew or gauche conformer.

Structure (IV) is partially eclipsed.

$\therefore$ order of stability

(I) > (III) > (IV) > (II)

Order of P.E. is (II) > (IV) > (III) > (I).

Ethylene glycol (HO – CH₂ – CH₂ – OH) and terephthalic acid

2700 kJ energy released from 180 gm (1 mole) of glucose

$\therefore$ 1 kJ energy released from $\frac{180}{2700}$ gm of glucose

$\therefore$ 10000 kJ energy released from   $\left(\frac{180*10000}{2700}\right)$ gm of glucose

Amount of glucose = $\frac{18000}{27}=666.66\approx 667gm$

$Area=\frac{1}{2}\left(\sqrt[]{5}-1\right)\frac{9-\sqrt[]{5}}{4}-{\displaystyle \underset{1}{\overset{\sqrt[]{5}}{\int }}\frac{1}{2}\sqrt[]{5-x^2}dx}$

$=\frac{1}{8}\left(-14+10\sqrt[]{5}\right)-\frac{1}{2}{\displaystyle \underset{co{s}^{-1}\frac{1}{\sqrt[]{5}}}{\overset{0}{\int }}\left(-si{n}^2\theta \right)d\theta }$

$A=\frac{-14}{8}+\frac{5}{4}\sqrt[]{5}-\left(\frac{5}{4}co{s}^{-1}\frac{1}{\sqrt[]{5}}-\frac{1}{2}\right)$

$=\frac{5}{4}\sqrt[]{5}-\frac{5}{4}-\frac{5}{4}co{s}^{-1}\frac{1}{\sqrt[]{5}}$

$\alpha =\frac{5}{4},\beta =\frac{-5}{4},\gamma =\frac{-5}{4}$

$\left|\alpha +\beta +\gamma \right|=\frac{5}{4}$

$R_1\to R_1-R_2,R_2\to R_2-R_3$

$\Delta =\left|\begin{array}{ccc}\hfill a-c+1\hfill & \hfill b-c\hfill & \hfill a-b\hfill \\ \hfill b-d-1\hfill & \hfill c-d\hfill & \hfill b-c\hfill \\ \hfill x-b+d\hfill & \hfill x+d\hfill & \hfill x+c\hfill \end{array}\right|$

$C_1\to C_1-C_2\&C_2\to C_2-C_3$

$\Delta =\left|\begin{array}{ccc}\hfill a+1-b\hfill & \hfill 2b-c-a\hfill & \hfill a-b\hfill \\ \hfill b-1-c\hfill & \hfill 2c-b-d\hfill & \hfill b-c\hfill \\ \hfill -b\hfill & \hfill d-c\hfill & \hfill x+c\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill -\lambda +1\hfill & \hfill 0\hfill & \hfill -\lambda \hfill \\ \hfill -\lambda -1\hfill & \hfill 0\hfill & \hfill -\lambda \hfill \\ \hfill -b\hfill & \hfill \lambda \hfill & \hfill x+c\hfill \end{array}\right|,R_1\to R{}_1-R_2$

$=\left|\begin{array}{ccc}\hfill 2\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill -\lambda -1\hfill & \hfill 0\hfill & \hfill -\lambda \hfill \\ \hfill -b\hfill & \hfill \lambda \hfill & \hfill x+c\hfill \end{array}\right|=2{\lambda }^2=2\Rightarrow {\lambda }^2=1$

0 = -16 m + c . (i)

$\frac{\left|m\left(-10\right)+C\right|}{\sqrt[]{m^2+1}}=2.............\left(ii\right)$

(i) & (ii) Þ 36 m² = 4 (m² + 1)

$m=\frac{1}{2\sqrt[]{2}},C=\frac{8}{\sqrt[]{2}}$

$4\sqrt[]{2}\left(m+c\right)=34$