Question

Consider the following reaction: CO2(g)+CCl4(g)⇌2COCl2(g). Calculate ΔG for this reaction at 25 ∘C under the following conditions PCO2= 0.140 atm PCCl4= 0.160 atm PCOCl2= 0.735 atm.

Answer 1

∆G = -RT lnKp.

So first we have to calculate Kp.

Calculation of Kp

Given reaction is CO2(g)+CCl4(g)⇌2COCl2(g)

Given that the equilibrium partial pressures are

PCO2= 0.140 atm PCCl4= 0.160 atm PCOCl2= 0.735 atm

Equilibrium constant Kp = ( PCOCl2)²/ (PCO2)(PCCl4)

   = (0.735)² / (0.14) (0.16)

   = 24.12

Kp =  24.12

Calculation of ∆G

∆G = -RT lnKp

R = 8.314 J/K/mol

given that T=25 °C = 25 +273 K = 298 K

Hence,

∆G = -RT lnKp

= - (8.314 J/K/mol)  (298 K) In (24.12)

= -7886.2 J/mol

∆G = -7886.2 J/mol

Therefore,

∆G at 25oC =  -7886.2 J/mol