Question

Consider the reactions so2(g) → o2(g) + s(s) kc = 2.5 × 10−53 so3(g) → 1 2 o2(g) + so2(g) kc = 4.0 × 10−13 calculate kc for the reaction 2 s(s) + 3 o2(g) → 2 so3(g)

Answer 1

Answer is: Kc for the reaction is 1·10¹³⁰.
Chemical reaction 1: SO₂(g) → O₂(g) + S(s); Kc₁ = 2,5 × 10⁻⁵³.
Chemical reaction 2: SO₃(g) → 1/2 O₂(g) + SO₂(g); Kc₂ = 4,0 ×10⁻¹³.
Chemical reaction 3:  2S(s) + 3O₂(g) → 2SO₃(g); Kc₃ = ?.
We can get reaction 3 by summing reverse reaction  multiply with and reverse reactio 2 multiply with two.
Kc₃ = (1/Kc₁²) · (1/Kc₂²) = 1/(2,5 × 10⁻⁵³)² · 1/(4,0 ×10⁻¹³)².
Kc₃ = 1,6·10¹⁰⁵ · 6,25·10²⁴ = 1·10¹³⁰.

Answer 2

Answer:  
SO2(g) ---> O2(g) + S(s)
 Kc=2.5 x 10^-53
 we get ::
 2 O2(g) + 2 S(s) ---> 2 SO2(g)
 Kc1 = (1 / (2.5 x 10^-53))^2
 
 SO3(g) ---> 1/2 O2(g) + SO2(g)
 Kc=4.0 x 10^-13
 we get :
 O2(g) + 2 SO2(g) --> 2 SO3
 Kc2 = (1 / 4.0 x 10^-13) ^2  
 2 O2(g) + 2 S(s) ---> 2 SO2(g) , Kc1
 O2(g) + 2 SO2(g) --> 2 SO3 , Kc2
 ----------- ----------------- ---------- -------- +
 2 S(s) + 3 O2(g) ---> 2 SO3(g) Kc = Kc1 . Kc2 
 Kc = (1 / (2.5 x 10^-53))^2 (1 / 4.0 x 10^-13)^2
 Kc = 1.0 10^130