Consider the decomposition of SO3 to SO2 and oxygen 2 SC); (g)艹2502 (g) + O2 (g) yso,-(1-2e)/(1 + ε) yo2/(1+8) (b) Starting from pure SO3, find the equilibrium composition at 1100 K and 1 bar (10 points) The equilibrium relationship is Substituting in the values of the equilibrium constant and pressure, and the expressions for the mole fractions, this becomes 2.677 1-28)-(1+8 This is a cubic equation, and is therefore not easy to solve. Because the quantity on the right hand size is bigger than one, we expect to be closer to its upper limit of 0.5 than to its lower limit of 0. Thus, we may be best off solving for the ε in (1-2c) in the denominator of the above equation. Doing so gives 48 1-2E)-12.677(1 + ε) 22.677(1+E) Iterating on the above, we find that it converges to -0.3787, from which yso3-0.176 so20.549 y02 = 0.275

We don't know the pressures, but we can use the mole fractions, which are in the same ratios as the pressures. I will use x instead of ϵ because it's easier to type.

Assume an initial unit pressure.

2SO₃ ⇌ 2SO₂ + O₂

I: 1 0 0

C: -2x +2x +x

E: 1 - 2x 2x x

Total = 1 - 2x + 2x + x = 1 + x

χₛₒ₃ = (1-2x)/(1 + x)

χₛₒ₂ = 2x/(1 + x)

χₒ₂ = x/(1 + x)

The equilibrium relationship is

[tex]K = \dfrac{\chi_{\text{SO2}}^{2} \chi_{\text{O2}} }{\chi_{\text{SO3}}^{2}} = \dfrac{\left ({\frac{2x}{1+x}}\right )^{2}{\left (\frac{x }{1+x}\right)}}{\left (\frac{1-2x }{1 + x}\right )^{2}}=\dfrac{4x^{3}}{(1-2x)^{2}(1+x)} = 2.667[/tex]

This is a cubic equation and not easy to solve. In the question, it is solved by successive approximations. I instead used an online calculator to get

x₁ = -1.2345, x₂ = 0.3785, and x₃ = 0.8559

We reject x₁, because x cannot be negative.

We reject x₃, because x cannot be greater than 0.5.

∴ x = 0.3785

χₛₒ₃ = (1-2x)/(1 + x}) = (1 - 2(0.3785))/(1+ 0.3785) = (1 - 0.7571)/1.3785 = 0.2429/1.37854 = 0.176

χₛₒ₂ = 2x/(1 + x) = 0.7571/1.3785 = 0.549

χₒ₂ = x/(1 + x) = 0.3785/1.3785 = 0.275

The equilibrium mole fractions are χₛₒ₃ = 0.176; χₛₒ₂ = 0.549; χₒ₂ = 0.275