Question

Consider the following reaction: CO₂(g)+CCl₄(g)⇌2 COCl₂(g). Calculate ΔG for this reaction at 25^oC under the following conditions.

P_CO2 = 0.115 atm

P_CCl4 = 0.185 atm

P_COCl2 =0.755 atm

Answer 1

ΔGrxn = ΔG°rxn + RTlnQ

ΔGrxn = ΔG°rxn + RTln ( [COCl2]^2 / [CO2] [CCl4])

CO₂(g)+CCl₄(g)⇌2 COCl₂(g)

ΔG°rxn = ΔG°of products - ΔG° of reactants

From the standard ΔG° values;

CO2 (g)= -394.4 kJ/mol
CCl4 (g)= -60.63 kJ/mol
COCl2 (g)= -206.8 kJ/mol

ΔG° = [(2) (-206.80] - [-394.4 & -60.63]
ΔG° = -413.6 - (-455.03)

ΔG° = + 42.43 kJ

Now;

ΔGrxn = ΔG°rxn + RTln ( [COCl2]^2 / [CO2] [CCl4])

ΔGrxn = + 42.43 kJ + (0.008314) (298) ln ( [0.755]² / [0.115] [0.185])

ΔGrxn = + 42.43 kJ + (0.008314) (298) ln (26.79)

ΔGrxn = + 42.43 kJ + (0.008314) (298) (3.288)

ΔGrxn = + 42.43 kJ + 8.1466

ΔGrxn = +50.5766