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 9.5 atm of H2, 5.4 atm of F2, and 0.24 atm of HF.

____kJ

Answer 1

H2 + F2 --> 2 HF

we know that

dGo rxn = dGof products - dGof reactants

in this case

dGo rxn = ( 2 x dGo HF) - (dGo H2) - ( dGo F2)

the standard free energy values are

dGo H2 = 0

dGo F2 = 0

dGo HF = -275.4 kJ/mol

so

dGo rxn = ( 2 x -275.4) - ( 0) - (0)

dGo rxn = -550.8 kJ/mol

2)

we know that

dG = dGo + RT lnQ

now

consider the reaction

H2 + F2 --> 2 HF

the reaction quotient is given by

Q = (pHF)^2 / (pH2) (pF2)

Q = (0.24)^2 / ( 9.5) (5.4)

Q = 1.1228 x 10-3

now

dG = dGo + RT ln Q

dG = ( -550.8 x 1000) + ( 8.314 x 298 x ln 1.1228 x 10-3)

dG = -567.627 x 1000 J/mol

dG = -567.627 kJ/mol

so

the value of dG is -567.627 kJ/mol