Question

use the following equilibria to calculate kc for the reaction.
S(s)+O2(g) ><SO2(g)K'c=4.2×10^-52

2S(s)+3O2(g)><2SO2(g)Kc=9.8×10^128
calcuate kc for the reation

2SO2(g)+O2><2SO3(g)

Answer 1

consider the first reaction

S (s) + 02 ----> S02

solids are not considered to equilibrium constant expression

so

Kc1 = [S02] / [02]

now

consider the second reactioon

2S (s) + 302 ---. 2S03

so

the equilibrium constant is given by

Kc2 = [S03]^2 / [02]^3

now

consider the final reaction

2S02 + 02 ---> 2 S03

the equilibrium constant is given by

Kc = [S03]^2 / [S02]^2 [02]

from the three equations

we can see that

Kc = Kc2 / ( Kc1)^2

given

Kc1 = 4.2 x 10-52

Kc2 = 9.8 x 10^128

using those values

we get

Kc = 9.8 x 10^128 / ( 4.2 x 10-52)^2

Kc = 0.555 x 10^232

Kc = 5.55 x 10^231

so

Kc for the required reaction is 5.55 x 10^231