Chemistry | VITEEE 2015 | Discussion Page

For the given reaction,

$H_{2}(g)+Cl_{2}{}(g)\rightarrow 2H^{+}(aq)+2Cl^{-}(aq).$

$\triangle G^{0}=-262.4kJ$

The value of free energy of formation

( $\triangle G^{0}_{f}$ ) for the ion Cl^-1 (aq) . therefore will be

$-131.2 kJ mol^{-1}$

$+131.2 kJ mol^{-1}$

$-262.4 kJ mol^{-1}$

$+262.4 kJ mol^{-1}$

Option A

$(\triangle G^{0}) reaction =\triangle G^{0}_{f}$ (products ) - $\triangle G^{0}_{f}$ (reactants)

$\therefore$ $264.4=[2\triangle G^{0}_{f} (H^{+})+2 \triangle G^{0}_{f}(C^{1-})]$

or $264.4=-[\triangle G^{0}_{f} (H_{2})+ \triangle G^{0}_{f}(Cl_{2})]$

$=[0+2\triangle G^{0}_{f} (Cl^{-})]+[0+0]$

or $-262.4=2\triangle G^{0}_{f}(Cl^{-})$

or $-\triangle G^{0}_{f}(Cl^{-})=-131.2 kJmol^{-1}$

Discussion Forum