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.
a) H2(g) + F2(g) -----------------> 2 HF(g)
Go = Go prdocuts - Go reactnats
= 2 x GoHF - (GoH2 + GoF2)
= 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
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