Question

Consider the reaction given below.

H2(g) + F2(g) 2 HF(g)

(a) Using thermodynamic data from the course website, calculate G° at 298 K. kJ

(b) Calculate G at 298 K if the reaction mixture consists of 6.6 atm of H2, 3.8 atm of F2, and 0.43 atm of HF.

Answer 1

a) H2(g) + F2(g) -----------------> 2 HF(g)

G^o = Go prdocuts - Go reactnats

          = 2 x G^o_HF - (G^o_H2 + G^o_F2)

          = 2 x 576.56 - (0 +0)

         = 1153 .12 kJ/mol

b)

Q = [HF]^2 / [H2] [F2]

Q = (0.43)^2 / (6.6) (3.8)

Q = 7.37 x 10^-3

G   = Go + RT ln Q

         = 1153.12 + 8.314 x 10^-3 x 298 x ln (7.37 x 10^-3)

         = 1140.96 kJ / mol