Question

Using the given data, determineArGº at 298 K for the precipitation reaction below. Ag+(aq) +r(aq) → AgI(5) Substance AGº(kJ/mol) at 298 K r(aq) 51.59 Ag (aq) 77.12 AgI(s) -66.2 Select one: O a. 62.5 kJ/mol-rxn O b.-40.7 kJ/mol-1xn OC. 91.7 kJ/mol-rxn O d. -91.7 kJ/mol-xn O e. 40.7 kJ/mol-rxn

Answer 1

∆rG° = $\sum$ ∆fG° ( products) - $\sum$ ∆fG° (reactants)

∆rG° = ∆fG° (AgI) - {∆fG°(Ag+) + ∆fG° ( I-) }

= -66.2 kJ/mol - ( 77.12+51.59) kJ/mol

= -66.2 kJ / mol - 128.71 kJ/mol

= -194 .91 kJ/mol

But this option is not given. Once please check the signs of Ag+ and I- ∆fG° values.