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The degree of dissociation of SO_3 is a at equilibrium pressure
P_0.
Kp for 2SO3(g)⇌2SO2(g)+O2(g)2{SO}_{3}(g) \rightleftharpoons 2{SO}_{2}(g) + O_{2}(g) is

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2SO_3(g) ⇌\rightleftharpoons 2SO_2 (g) + O_2(g)
t = 0 a 0 0
t =t a(1-α) aα a(α2)a\left( \frac{\alpha}{2} \right)
Total mole at eq. =a(1+α2)= a\left( 1 + \frac{\alpha}{2} \right)
PSO3=(1−α1+(α/2))P0=[2(1−α)2+α]×P0P_{SO_{3}} = \left( \frac{1 - \alpha}{1 + (\alpha/2)} \right)P_{0} = \left\lbrack \frac{2(1 - \alpha)}{2 + \alpha} \right\rbrack \times P_{0}
PSO2=(α1+(α/2))P0=(2α2+α)×P0P_{SO_{2}} = \left( \frac{\alpha}{1 + (\alpha/2)} \right)P_{0} = \left( \frac{2\alpha}{2 + \alpha} \right) \times P_{0}
PO2=(α21+(α2))P0{PO}_{2} = \left( \frac{\frac{\alpha}{2}}{1 + \left( \frac{\alpha}{2} \right)} \right)P_{0}
KP=4α2(P∘)2(2+α)2×(α2+α)×P04(1−α)2[2+α]2×(P0)2=[α3p0(2+α)(1−α)2]K_{P} = \frac{\frac{4\alpha^{2}\left( P^{\circ} \right)^{2}}{(2 + \alpha)^{2}} \times \left( \frac{\alpha}{2 + \alpha} \right) \times P^{0}}{\frac{4(1 - \alpha)^{2}}{\lbrack 2 + \alpha\rbrack^{2}} \times \left( P_{0} \right)^{2}} = \left\lbrack \frac{\alpha^{3}p^{0}}{(2 + \alpha)(1 - \alpha)^{2}} \right\rbrack

Source: Grass-G / CH-24 Chemistry Paper 31 Dec..docx

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