Consider the following equilibrium gas phase reaction: PCl 5 (g)-->PCl 3 (g)+Cl

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



Please help with a-d
Thank you!

2. Consider the folowing equilibrium gss pl see readier: pcu, tci Poe cuz ay s Calculate AH and 4,5. far this raa:tion when carried out at 280K,

Explanation / Answer

For the given reaction

a) At 298 K

dHo = dHo(products) - dHo(reactants)

        = (-287.02) - (-342.67) = 55.65 kJ/mol

dSo = dSo(products) - dSo(reactants)

        = (311.67 + 222.97) - (364.47) = 170.17 J/K.mol

b) dGo = dHo - TdSo

            = 55.65 - 298 x 0.17017

            = 4.94 kJ/mol

K is equilibrium constant

dGo = -RTlnK

T = 600 K

4.94 x 1000 = -8.314 x 600 lnK

K = 1

c) For the given reaction,

initial volume = nRT/P

                      = 10 x 0.08205 x 600/30 = 16.4 L

initial [PCl5] = 10 mol/16.41 L = 0.61 M

ICE chart

                  PCl5 <====>   PCl3 + Cl2

I                 0.61                     -           -

C                 -x                      +x         +x

E              10 - x                     x           x

So,

K = [PCl3][Cl2]/[PCl5]

1 = x^2/(0.61 - x)

x^2 + x - 0.61 = 0

x = 0.43 M

Now,

moles PCl3 = 0.43 x 16.41 = 7.06 mol

moles Cl2 = 0.43 x 16.41 = 7.06 mol

moles PCl5 = 0.18 x 16.41 = 2.95 mol

equilibrium partial pressure of,

p[PCl3] = nRT/V = 7.06 x 0.08205 x 600/16.41 = 21.18 atm

p[Cl2] = nRT/V = 7.06 x 16.41 x 0.08205 x 600/16.41 = 21.18 atm

p[PCl5] = nRT/V = 2.95 x 16.41 x 0.08205 x 600/16.41 = 8.85 atm

d) according to the Lechatellier's principle,

(i) decrease in temperature for an endothermic (dH +ve) reaction would shift equilibrium to the reactant side.

(ii) Decrease in volume would shift equilibrium to side which more moles, that is towards product end

(iii) when reactant PCl5 is removed, more product would react to form PCl5 and regain equilibrium, so shift is toward reactant.

(iv) when N2 (inert gas) added, no change would occur

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.