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%)
