Question

Consider the following reaction: CO2(g)+CCl4(g)⇌2COCl2(g). Calculate ΔG for this reaction at 25 ∘C under the following conditions. Part A PCO2= 0.115 atm PCCl4= 0.180 atm PCOCl2= 0.745 atm please, be fast because it will soon be over by 11:59pm

Answer 1

The given reaction is

CO₂ (g) + CCl₄ (g) <======> 2 COCl₂ (g)

Assume the partial pressures given are the equilibrium pressures and obtain the equilibrium constant K as

K = (P_COCl2)²/(P_CO2)(P_CCl4) = (0.745 atm)²/(0.115 atm).(0.180 atm) = 26.8128

Use the relation between ΔG and K to find ΔG as

ΔG = -R*T*ln K where T = 25°C = (25 + 273) K = 298 K

Plug in values and get

ΔG = -(8.314 J/mol.K)*(298 K)*ln (26.8128) = -(2477.572 J/mol)*(3.2889) = -8148.4865 J/mol = (-8148.4865 J/mol)*(1 kJ/1000 J) = -8.1484865 kJ/mol ≈ -8.15 kJ/mol (ans).