ThermochemistryHard

Question

If the enthalpy of formation and enthalpy of solution of HCl (g) are - 92.3 kJ/mol and - 75.14 kJ/mol respectively then find enthalpy of formation of Cl- (aq) :

Options

A.-17.16 kJ/mol
B.-167.44 kJ/mol
C.17.16 kJ/mol
D.None of these

Solution

HCI(g) + aq → H+ (aq.) + CI- (aq.);
ᐃrH = - 75.14 = ᐃfH (H+, aq) + ᐃfH (CI-, g) - ᐃfH (HCI, g)
ᐃfH (H+, aq) = 0
ᐃfH (CI- , aq) = - 75.14 - 92.3 = - 167.44 kJ/mol

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