Question

Consider the reaction 2NO2(g)?N2O4(g)

Calculate ?G at 298 K if the partial pressures of NO2 and N2O4 are 0.38atm and 1.64atm , respectively.

Answer 1

2NO2(g) ? N2O4(g)

Kp = P(products)/P(reactants) = (P(N2O4))/((P(NO2))^2)
Kp = (1.64 atm)/((0.38 atm)^2) = 11.36 atm^-1

Next we need to convert Kp to Kc:

Kp = Kc(RT)^?ngas
So
Kc = Kp/((RT)^?ngas)

?ngas = Mole of gas (product) - Mole of Gas (Reactant) = 1 - 2 = -1
R = 0.08206 L atm K^-1 mol^-1
T = 298 K

Kc = Kp/((RT)^?ngas) = (11.36 atm^-1)/(((0.08206 L atm K^-1 mol^-1)*(298 K))^-1)
Kc = 277.8 L mol-1

Now we use this to calculate ?G.

?G = -RT(lnKeq)

R = 0.008314 kJ mol^-1 K^-1
T = 298 K
Keq = Kc = 277.8

?G = -RT(lnKeq)
?G = -(0.008314 kJ mol^-1 K^-1)*(298 K)*(ln(277.8))
?G = 13.94 kJ mol^-1