Question

Consider the reaction: I2(g) + Cl2(g) ↔ 2 ICl (g)

Calculate ΔGrxnfor the reaction at 25oC under each of the following conditions:

a. Standard conditions

b. At equilibrium

c. PICl= 2.55 atm; PI2= 0.325 atm; PCl2= 0.221 atm

Answer 1

a) at 25 C, we will calculate standard enthalpy of reaction and standard entropy of reaction

ΔG⁰rxn = ΔH⁰rxn - T ΔS⁰rxn

ΔH⁰rxn = sum of ΔH⁰formation of products - sum of ΔH⁰formation of reactants

ΔH⁰rxn = [2X ΔH⁰ICl] - [ ΔH⁰I2 + ΔH⁰Cl2]

ΔH⁰rxn    = 2 X 17.78 - [0 +0] = 35.56 KJ / mole

ΔS⁰rxn = sum of ΔS⁰formation of products - sum of ΔS⁰formation of reactants

ΔS⁰rxn = [2X ΔS⁰ICl] - [ ΔS⁰I2 + ΔS⁰Cl2] = [ 2 X 247.44] - [260.58 + 222.97]

= 11.33 J / mol K = 0.01133 KJ / mol K

ΔG⁰rxn = ΔH⁰rxn - T ΔS⁰rxn = 35.56 - 298 X 0.01133 = 32.18 KJ / mol

b) at equilibrium , ΔGrxn = 0

c) Kp = p²ICl / pI2 X pCl2 = (2.55)² / (0.325)(0.221) =

Kp = 90.53

ΔG⁰rxn = -RT ln Kp

R= 8.314 J / mole K

T = 298 K

Kp = 90.53

ΔG⁰rxn = -8.314 X 298 X ln (90.53) = - 11163.15 J / mole = -11.163 KJ / mol