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%

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22-01-2020
Chemistry

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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%

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lucianoangelini92 lucianoangelini92 · Answer 1 · 2020-01-23T18:52:17+01:00

Answer:

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

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%)