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Question

2 AB (g) + B2 (g) → 2AB2 (g); H° = +492.0 kJ/mol, S° = +236.9 J/Kmol

The molar mass for AB for part (a) is 149.88

Answer the following questions regarding the reaction represented above.

(a) Calculate the quantity of heat released when 83.9 g of AB (g) is converted to AB2 (g).

(b) For the reaction at 25°C, the value of the standard free-energy change, G°, is +421.4 kJ/mol.

(i) Calculate the value of the equilibrium constant, K, for the reaction at 25°C. 

(ii) Indicate whether the value of G° would become more positive, less positive, or remain unchanged as the temperature is increased. Justify your answer. 

Answer 1

2AB(g) + B2(g) ==> 2AB2(g) ∆H = 492 kJ/mol; ∆S = 236.9 kJ/Kmol

(a) 83.9 g AB x 1 mol AB / 149.88 x 1 mol 2 mols AB2 / 2 mols AB x 492 kJ/mol = 275 kJ

(b)

(i) ∆Gº = -RT ln K

ln K = ∆G / -RT = 421.4 kJ/mol / (-0.008314 kJ/Kmol)(298K) = - 170

K = 1.48x10^-74

(ii) ∆Gº = ∆Hº - T∆Sº

As T increases, the value of T∆S also increases because ∆Sº is positive. This will result in the difference between ∆Hº and this product becoming greater, and eventually, will become a negative value, so at high enough temperatures, ∆Gº will become negative,

T∆S = 984 kJ

T = 984 / 0.118 = ~ 8340K. So at around this temperature, ∆Gº would be negative