Question

Based upon equations (1) and (2) below, how could one express the equilibrium constant 10 for equation (3)?

(1) S(s) + O2(g) 6 SO2(g) equilibrium constant K1

(2) 2 SO3 6 2 SO2(g) + O2(g) equilibrium constant K2

(3) S(s) + 3/2 O2(g) 6 SO3(g) equilibrium constant K3

Answer 1

The reactions

S + O2 = SO2 K1

2SO3 = 2SO2 + O2 K2

S + 3/2O2 = SO3

clearly, we need to invert reaction (2) in order to have SO3 in the right... also divide by 2, since we have a coefficient of 1

so

S + O2 = SO2 K1

2SO2 + O2= 2SO3 1/K2 (inverting makes an inverse in K as well)

S + O2 = SO2 K1

SO2 + 1/2O2= SO3 (1/K2)^(1/2) (division by 2, is equivalent to powering to the 1/2)

now..

note that if we add reaction 1 + 2 :

S + O2 + SO2 + 1/2O2 = SO2 + SO3

cancelling common terms

S + 3/2O2 = SO3

which is what we wanted

so

K3 can be expressed as

K3 = K1*(1/K2)^(1/2)