[Big-Daddy]

I have learnt the theory - as a fact - that when you want to combine equilibria you "add" them or subtract them, or manipulate as you will, and then the overall equilibrium constant for the expression you end up with is equivalent to the constants for each reaction you had before, to the power of their coefficient in my manipulation of them. For example:

MoS₄2– + H2O(l)    MoOS₃2– + H₂S(aq) K1 = 1.3×10^–5
MoOS₃2– + H2O(l)    MoO₂S₂2– + H₂S(aq) K2 = 1.0×10^–5
MoO₂S₂2– + H2O(l)     MoO₃S^2– + H2S(aq) K3 = 1.6×10^–5
MoO₃S^2– + H2O(l)      MoO₄2– + H₂S(aq) K4 = 6.5×10^–6

If at equilibrium a solution contains 1×10^–7 M MoO₄2– and 1×10^–6 M H₂S(aq), what would be the concentration of MoS₄2–?

My solution:

You must simply add the reactions together:

R(1)+R(2)+R(3)+R(4) produces R(overall): MoS₄2– + 4 H2O (l)    MoO₄2- + 4 H₂S

K(overall)= ([MoO₄2-]*[H₂S]⁴)/([MoS₄2–]*[H2O (l)]⁴)

H2O is excluded because it is so large that the change is negligible:

K(overall)= ([MoO₄2-]*[H₂S]⁴)/[MoS₄2–]
[MoS₄2–]=([MoO₄2-]*[H₂S]⁴)/K(overall)

So now comes in my point. We have added the reactions plainly, as R(1)+R(2)+R(3)+R(4), so K(overall)=K₁¹*K₂¹*K₃¹*K₄¹. The powers are because the coefficients on the reactions as we add/manipulate them are 1. So now:

[MoS₄2–]=([MoO₄2-]*[H₂S]⁴)/(K₁¹*K₂¹*K₃¹*K₄¹)

Plug in the numbers and get an answer of: 7.396*10^-12 moldm-3, roughly 7.4*10^-12.

What's my question? Why? Why is it then when we take R(1)+R(2)+R(3)+R(4) we get K₁¹*K₂¹*K₃¹*K₄¹; and if we take R(1)+2*R(2)-1/3*R(3)+5*R(4) we get K₁¹*K₂²*K₃^-(1/3)*K₄⁵? The maths behind this bit is unclear to me.

[Stovn0611]

Consider two reactions

A + Z  C + D
E + F  G + H

The equilibrium expression for the first reaction will be ([C][D])/([A][Z])

What happens if we multiply the coefficients by 2?

The reaction becomes: 2A + 2Z  2C + 2D

Now the equilibrium expression will be ([C]²[D]²)/([A]²[Z]²)

This results in the generalization

For nA + nZ  nC + nD,
the equilibrium expression is (([C][D])/[A][Z]))^n

What happens if we add the original two reactions together?

A + Z + E + F  C + D + G + H

The equilibrium expression for this reaction will be ([C][D][G][H])/([A][Z][E][F]) = ([C][D])/([A][Z])*([G][H])/([E][F]) which is just the product of the two equilibrium expressions of the original two reactions.

What happens if we flip a reaction?

C + D  A + Z

The equilibrium expression for this reaction will be ([A][Z])/([C][D]), which is the equilibrium expression of the original reaction raised to the power of -1.

This is how all the rules for manipulating equilibrium expressions are derived and in general you can derive more just by writing out the altered reaction and looking at the new equilibrium expression.

EDIT: Had to use Z instead of B because is bold lol