The Keq for the equilibrium below is 0.113 at 973K
SO2 (g) + 1/2 O2 (g) <-> SO3 (g)
At this temperature, what is the value of Keq for the following reaction?
2 SO2 (g) + O2 (g) <-> 2SO3 (g)
I know to multiply by 2 to get the second equation but I don't get the correct answer if I do 2(Keq) = 2(0.113). What am I doing wrong?
In simple words you should square it. Let me show you why.
SO2 (g) + 1/2 O2 (g) <-> SO3 (g). Keq = 0.113
The equation wanted is 2 SO2 (g) + O2 (g) <-> 2SO3 (g) Keq = ?
So let's wrote Keq for the first one: It is
Keq = 0.113 = (SO3)/(SO2)(O2)1/2
For the second one; i.e. the one multiplied by 2 it is
Keq = ? = (SO3)^2/(SO2)^2(O2)^1. You can see that all of the reactants and products are just squared so Keq for 0.113 will be squared or (0.113)^2
BTW, for the REVERSE direction of the first one it is 1/Keq.