Calculate the equilibrium concentration of H2O in the second mixture at this temperature.?

For a heterogeneous equilibrium, solid components are not included in the equilibrium expression.

Consider the first mixture:
__________________ 2H₂S(g) _ + _ SO₂(g) _ ⇌ _ 3S(s) _ + _ 2H₂O(g) ____ Kc
Initial (M): __________0.480 ____ 0.080 _________________ 0
Change (M) _________ -2y _______ -y __________________ +2y
Equilibrium (M): __ 0.480 - 2y ___ 0.480 - y ______________ 2y

At equilibrium: [H₂O] = 2y M = 1.0 × 10⁻³ M = 0.0010 M
Hence, 2y = 0.0010 and y = 0.0005

Kc = [H₂O]² / ([H₂S]² [SO₂])
= (2y)² / {(0.480 - 2y)² (0.480 - y)
= (0.0010)² / {(0.480 - 0.0010)² (0.480 - 0.0005)
= (0.0010)² / (0.479² × 0.4795)
= 9.09 × 10⁻⁶

Consider the first mixture:
__________________ 2H₂S(g) _ + _ SO₂(g) _ ⇌ _ 3S(s) _ + _ 2H₂O(g) ____ Kc = 9.09 × 10⁻⁶
Initial (M): __________0.240 ____ 0.325 _________________ 0
Change (M) _________ -2z _______ -z __________________ +2z
Equilibrium (M): __ 0.240 - 2z ___ 0.325 - z ______________ 2z

As Kc is very small, the equilibrium lies to left and only a very small amounts of H₂S and SO₂ react.
We can assume that 0.240 ≫ 2z and 0.325 ≫ z
[H₂S] at equilibrium = (0.240 - 2z) M ≈ 0.240 M
[SO₂] at equilibrium = (0.325 - z) M ≈ 0.325 M

Kc = [H₂O]² / ([H₂S]² [SO₂])
9.09 × 10⁻⁶ = (2z)² / (0.240² × 0.325)
(2z)² = (9.09 × 10⁻⁶) × 0.240² × 0.325
2z = √{(9.09 × 10⁻⁶) × 0.240² × 0.325}
z = 4.13 × 10⁻⁴

[H₂O] at equilibrium in the second mixture = 4.13 × 10⁻⁴ M