Question

# Consider the following threeequilibria at 400oC 2SO2(g) + O2(g) ⇌ 2SO3(g) Kp=3.5×105 SO3(g) +H2O(g)⇌H2SO4(g) Kp =8.0×10^−12...

Consider the following threeequilibria at 400oC

2SO2(g) + O2(g) ⇌ 2SO3(g) Kp=3.5×105

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

H2O(g)⇌2H2(g) + O2(g) Kp=1.3×10^−33

What is Kc at 400oC for the following equilibrium?

SO2(g) +O2(g)+ H2(g)⇌H2SO4(g)

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

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

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

2 H2O(g) <---> 2H2(g) + O2(g) Kp=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) Kp=591.61 ……..4

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

H2(g) + ½ O2(g) <---> H2O(g) Kp=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

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