Consider the following equilibrium gas phase reaction: PCl 5 (g)-->PCl 3 (g)+Cl
ID: 525059 • Letter: C
Question
Consider the following equilibrium gas phase reaction: PCl5(g)-->PCl3(g)+Cl2(g)
a) Calculate rHo and rSo for this reaction when it is carried out at 298K.
b) Estimate the thermodynamic equilibrium constant for this reaction at 600K
c) Assuming all gases are ideal, calculate the partial pressures of the reactants and products at equilibrium if 10 moles of PCl5(g) is initially placed in a vessel at 600K and the reaction is carried at a constant pressure of 30atm
d) For the reaction above, in which direction (if any) will the equilibrium shift when:
(i) The temperature is decreased
(ii) The volume of the reaction vessel is reduced
(iii) PCl5 is removed from the reaction vessel
(IV) N2(g) is added to the reaction at constant volume
Explanation / Answer
For the given reaction,
a) dHo = dHo(products) - dHo(reactants)
= -287.02 + 347.67 = 60.65 kJ/mol
dSo = dSo(products) - dSo(reactants)
= (311.67 + 222.97) - (364.47)
= 170.17 J/K.mol
dGo = dHo - TdSo
= 60.65 - 298 x 0.17017
= 9.94 kJ/mol
b) dGo = -RTlnKeq
9.94 x 10^3 = -8.314 x 600 lnKeq
Keq = 0.14
c) ICE chart
PCl5 <====> PCl3 + Cl2
I 10 - -
C -x +x +x
E 10-x x x
So,
Keq = [PCl3][Cl2]/[PCl5]
0.14 = x^2/(10-x)
x^2 +0.14x - 1.4 = 0
x = 1.12 mol
So,
Volume of container = nRT/P
= 10 x 0.08205 x 600/30
= 16.41 L
At equilibrium,
Partial pressure of PCl3 = partial pressure of Cl2
= 1.12 x 0.08205 x 600/16.41
= 3.36 atm
Partial pressure of PCl5 = (10 - 1.12) x 0.08205 x 600/16.41
= 26.64 atm
d) Direction of equilibrium shift when,
(i) Temperature is decreased : shift to left handside, endothermic needs energy to proceed reaction.
(ii) Volume of reaction vessel reduced : Pressure goes up, shift to left handside of lower moles
(iii) PCl5 removed from vessel : Shift to left handside
(iv) N2 added to the vessel : No change