Ncert Solutions Chemistry Class 11th

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

(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 = $\frac{18000}{27}=666.66\approx 667gm$

Hydrogen peroxide reduces iodine to iodide ion is basic medium as;

$H_2{O}_2+2O{H}^{-}+l_2\to O_2+2l^{-}+2H_{2}O$

$N{a}_{2}O+H_{2}O\to 2NaOH$

w = 20 g

$V_{H_{2}O}=500mL$        

Mole of Na₂O=   $\frac{20}{62}$

$?$ 1 mole of Na₂O gives 2 mole of NaOH

$\therefore \left(\frac{20}{62}\right)$ mole of Na₂O gives $\left(2*\frac{20}{62}\right)$ moles of NaOH

$=\frac{20}{31}mole$    

Molarity of NaOH solution

$=\frac{\frac{20}{31}*100}{500}M=\frac{20}{31}*2M=\frac{40}{31}M$

= 1.29 M

=  $12.9*10^{-1}M\approx 13*10^{-1}M$

Ans. = 13

PV = nRT             PV = constant (at constant T)

Pressure increases & volume decreases, PV remains constant at constant T.