MCQ
Reaction of phenol with chloroform/sodium hydroxide to give $o$-hydroxy benzaldehyde involves the formation of
- ✓Dichloro carbene
- BTrichloro carbene
- CChlorine atoms
- DChlorine molecules
$O{H^ - } + CHC{l_3}$⇌ $HOH + :CCl_2^ - \to C{l^ - } + :CC{l_2}$
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$Pt(s)|\mathop {{H_2}(g)}\limits_{1\ atm} |HA\,\,\,\,\,\mathop {1\,M}\limits_{({K_a} = {{10}^{ - 7}})} \,\,||\,HB\,\,\,\,\,\mathop {1\,M}\limits_{({K_1} = {{10}^{ - 5}})}$
$\,|\mathop{{H_2}(g)}\limits_{1\ atm} |Pi(s)$
${N_2}(g)\, + \,{O_2}(g)\,\underset{{{k_2}}}{\overset{{{k_1}}}{\longleftrightarrow}}\,2NO(g)$
$C_0=Ce^{-2.1 \times 10^{-3}t}$ for the forward reaction and $C'_0=C'e^{-4.2 \times 10^{-4}t}$ for the backward reaction, hence $K_c$ for the above equilibrium is :-
${{O}_{(g)}}+{{e}^{-}}=O_{(g)}^{-}\Delta {{H}^{o}}=-142\ kJ\,mo{{l}^{-1}}$
$O_{(g)}^{-}+{{e}^{-}}=O_{(g)}^{2-}\Delta {{H}^{o}}=844\ kJ\,mo{{l}^{-1}}$