at 60℃ , N2O4 is 50% dissociated calculate the standard free energy change at this temperature and 1 ATM pressure
Answers
Answered by
76
Dissociation of N2O4, proceeds as,
N2O4 ------> 2NO2
Mole fractions, dissociated is 50% is 0.5,
= (1-0.5) / (1+0.5) = 0.5/1.5 = 0.33 Therefore, p = 0.33 x 1 atm. = 0.33.
Similarly for NO2 = 2 x 0.5/ (1+0.5) = 1/1.5 = p = 0.66
Equilibrium const Kp = p(NO2)2/ p(N2O4) = (0.66)2/(0.33) = 1.33
We know that, ΔG = -RT ln Kp
Substituting the values, = -8.314 x 333 x 2.303 x (0.1239)
= -768.3 kJ/mol
N2O4 ------> 2NO2
Mole fractions, dissociated is 50% is 0.5,
= (1-0.5) / (1+0.5) = 0.5/1.5 = 0.33 Therefore, p = 0.33 x 1 atm. = 0.33.
Similarly for NO2 = 2 x 0.5/ (1+0.5) = 1/1.5 = p = 0.66
Equilibrium const Kp = p(NO2)2/ p(N2O4) = (0.66)2/(0.33) = 1.33
We know that, ΔG = -RT ln Kp
Substituting the values, = -8.314 x 333 x 2.303 x (0.1239)
= -768.3 kJ/mol
Similar questions