MCQ
Which of the following is optically active
- AEthylene glycol
- BOxalic acid
- CGlycerol
- ✓Tartaric acid
So it is optically active.
$\mathop {\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
{\,\,\,\,OH} \\
| \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,H - {C^*} - COOH\,\,\,} \\
|
\end{array}} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,H - {C^*} - COOH} \\
| \\
{\,\,\,OH}
\end{array}}\limits_{{\text{Tartaricacid}}} $
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$r = \frac{{ - d\left[ A \right]}}{{dt}} = K\,{\left[ A \right]^{\frac{1}{2}}}\,{\left[ B \right]^{\frac{1}{2}}}\,{\left[ C \right]^{\frac{1}{4}}}$
The order of reaction is
$(1) Li < Be < B < C (IE_1)$
$(2) Li < Na < K < Rb < Cs$ (Reducing power in gaseous state)
$(3) Li^+ < Na^+ < K^+ < Rb^+ < Cs^+$ (Ionic mobility in aqueous solution)
$(4) S > Se > Te > O [EA]$
$\left(1 F =96,500\, C\, mol ^{-1}\right)$
Reason : The carbon atom carrying negative charge has an octet of electrons.
At $ 298 K : \Delta_{ f } H ^{\circ}\left( SnO _2( s )\right)=-581.0 kJ mol ^{-1}, \Delta_{ fH } H ^{\circ}\left( CO _2( g )\right)=-394.0 kJ mol ^{-1}$
$S ^{\circ}\left( SnO _2( s )\right)=56.0 J K ^{-1} mol ^{-1}, S ^{\circ}( Sn ( s ))=52.0 J K ^{-1} mol ^{-1}$
$S ^{\circ}( C ( s ))=6.0 J K ^{-1} mol ^{-1}, S ^{\circ}\left( CO _2( g )\right)=210.0 J K ^{-1} mol ^{-1}$
Assume that the enthalpies and the entropies are temperature independent.