Question

Calculate $\Delta H _{ f }^{0}$ for chloride ion from the following data:
$\frac{1}{2} H _{2}( g )+\frac{1}{2} Cl _{2}( g ) \rightarrow HCl ( g ) ; \Delta H _{ f }^{ \circ }=-92.4 \,kJ$
$HCl ( g )+ nH _{2} O \rightarrow H ^{+}( aq )+ Cl ^{-}( aq ) ; \Delta H _{ f }^{ \circ }=-74.8 \,kJ$
$\Delta H _{ f }^{0} $ of $ H ^{+}( aq )=0.0 \, kJ$

• A. $-167.2 \, kJ$
• B. $-165.2 \, kJ$
• C. $-157.2 \, kJ$
• D. $-147.2 \, kJ$

Answer 1

Answer: (A)

Given,
$\frac{1}{2} H _2 g + aq \rightarrow H ^{+} aq + e ^{-} ; \Delta H _{ f }^{\circ}=0 \,kJ \ldots . i $
$\frac{1}{2} H _2 g +\frac{1}{2} Cl _2 g \rightarrow HClg ; \Delta H _{ f }^{\circ}=-92.4 \,kJ \ldots ii$
$HClg + nH _2 Ol \rightarrow H ^{+} aq + Cl ^{-} aq ; \Delta H ^{\circ}=-74.8 \,kJ \ldots iii$
$ii + iii - i$
$\therefore \frac{1}{2} Cl _2 g + nH _2 O + e ^{-} \rightarrow Cl ^{-} aq ; \Delta H _{ f }^{\circ}=-92.4-74.8 kJ =-167.2 \,kJ.... iv$
Heat of formation for $Cl ^{-}$aq $=-167.2 \,kJ$