Question

Sulfuryl chloride is in equilibrium with sulfur dioxide and chlorine gas: so2cl2(g) so2(g) + cl2(g) a system with a volume of 1.00 l is in equilibrium at a certain temperature with p(so2cl2) = 1.00 bar and p(so2) = p(cl2) = 0.10 bar. by how much will the number of moles of so2cl2 at equilibrium change if the volume is reduced to 0.50 l? (a) increase 1-10% (b) increase 11-50% (c) decrease 1-10% (d) decrease 11-50%

Answer 1

When the volume of a system at equilibrium is decreased, the equilibrium will shift in the direction that reduces the total number of gas particles. In this case, decreasing the volume will shift the equilibrium to the right, towards the side with fewer gas particles. As a result, the number of moles of SO2Cl2 at equilibrium will decrease. (Option d)

Step-by-step explanation:

When the volume of a system at equilibrium is decreased, according to Le Chatelier's principle, the equilibrium will shift in the direction that reduces the total number of gas particles. In this case, the forward reaction produces two gas molecules (SO2 and Cl2), while the reverse reaction produces one gas molecule (SO2Cl2). Therefore, decreasing the volume will shift the equilibrium to the right, towards the side with fewer gas particles. As a result, the number of moles of SO2Cl2 at equilibrium will decrease when the volume is reduced to 0.50 L.

So, the answer is (d) decrease 11-50%.

Answer 2

Sulfuryl chloride decreases by -1/21 (-4.76%) (option c)

Step-by-step explanation:

Denoting

sc= so2cl2(g)

s=so2(g)

c=cl2(g)

Assuming that the compression is an isothermal process , then reaction equilibrium constant in terms of pressure does not change

Kp= psc/ps*pc =

where p= partial pressures

Assuming ideal behaviour , then from Dalton's law,

Xsc₁=psc₁/P₁= psc₁/P₁ = 1 bar/(1 bar + 0.1 bar + 0.1 bar) = 5/6

Xs=ps₁/P₁ = 0.1/1.2=1/12

Xc=pc₁/P₁ = 0.1/1.2=1/12

since Xs=Xc → the reaction started as pure Sulfuryl chloride . Then representing ξ as the extent of reaction and n as the moles

nsc=nsc₀*(1-ξsc) , ns=nsc₀*ξsc , nc=nsc₀*ξsc → n=nsc +ns +nc = nsc₀*(1+ξsc)

therefore

Xs₁=ns₁/n₁=ξsc₁/(1+ξsc₁) → Xs₁*ξsc₁+Xs₁=ξsc₁ → ξsc₁=Xs₁/(1-Xs₁) = (1/12)/(11/12)= 1/11

then from the ideal gas law

ps₁*V₁=ns₁*R*T

after the reduction

ps₂V₂=ns₂*R*T

dividing both equations

(ps₂/ps₁)*(V₂/V₁)=(ns₂/ns₁)=nsc₀*ξsc₂/(nsc₀*ξsc₁) = ξsc₂/ξsc₁

ps₂ = ps₁ * (V₁/V₂) * (ξsc₂/ξsc₁)

since

psc₁*V₁=nsc₁*R*T , psc₂V₂=nsc₂*R*T → psc₂ = psc₁ * (V₁/V₂) * (1-ξsc₂)/(1-ξsc₁)

also knowing that

Kp= psc₁/ps₁² = psc₂/ps₂²

psc₂/ps₂² = psc₁/ps₁² * (V₁/V₂) * (1-ξsc₂)/(1-ξsc₁) / [(V₁/V₂) * (ξsc₂/ξsc₁) ]² =

1 = (V₂/V₁)(1-ξsc₂)*ξsc₁/ [(1-ξsc₁)*ξsc₂]

replacing ξsc₁= 1/11

1 = (V₂/V₁)(1-ξsc₂)/ξsc₂ *(1/10)

10 = (V₂/V₁)* (1/ξsc₂-1) → ξsc₂ = 1/(10*(V₁/V₂)+1)

therefore the extent of reaction varies with the volume reduction according to

ξsc₂ = 1/(10*(V₁/V₂)+1)

since V₁/V₂=2

ξsc₂ = 1/(10*2+1) = 1/21

therefore the decrease in moles of Sulfuryl chloride is

Δnsc/nsc₁ = (ξsc₂-ξsc₁)/(1-ξsc₁) = (1/21-1/11)/(10/11)= (11/21-1)/10 = -1/21 (-4.76%)