Question

Consider the following threeequilibria at 400oC

2SO2(g) + O2(g) ⇌ 2SO3(g) K_p=3.5×105

SO3(g) +H2O(g)⇌H2SO4(g) K_p =8.0×10^−12

H2O(g)⇌2H2(g) + O2(g) K_p=1.3×10^−33

What is Kc at 400oC for the following equilibrium?

SO2(g) +O2(g)+ H2(g)⇌H2SO4(g)

Answer 1

Answer – In this one there are given three at 400^oC

2SO₂(g) + O₂(g) <----> 2SO₃(g) K_p=3.5×10⁵ ……..1

SO₃(g) +H₂O(g) <-----> H₂SO₄(g) K_p =8.0×10^-12….2

2 H₂O(g) <---> 2H₂(g) + O₂(g) K_p=1.3×10^-33……….3

We need to calculate Kc for the reaction

SO₂(g) +O₂(g)+ H₂(g) <----> H₂SO₄(g)

Now for this reaction we need to divided by 2 with equation number 1 and divide equation 3 by 2 and reverse so the new reaction with Kp are as follow –

SO₂(g) + ½ O₂(g) <----> SO₃(g) K_p=591.61 ……..4

SO₃(g) +H₂O(g) <-----> H₂SO₄(g) K_p =8.0×10^-12….5

H₂(g) + ½ O₂(g) <---> H₂O(g) K_p=2.77×10¹⁶……….6

SO₂(g) +O₂(g)+ H₂(g) <----> H₂SO₄(g) , Kp = 1.31*10⁸

We know

Kp = Kc *(RT)^∆n

1.31*10⁸ = Kc * (0.0821*298 )^-2

Kc = 1.31*10⁸ / (0.0821*298 )^-2

     = 7.85*10¹⁰