# Chemical equilibrium — why multiplication, rather than addition? [duplicate]

Why do we multiply (and not take the sum) the concentration of products and reactants when more than one entity is present in either side and rise the power to the coefficient term?

Suppose either

$$\ce{aA + bB <=> cC +dD}$$

described by

$$K_{\mathrm{eq}} = \frac{[\ce{C}]^c \cdot [\ce{D}]^d} {[\ce{A}]^a \cdot [\ce{B}]^b}$$

where we multiply the concentrations and rise the power to the coefficient. Or the dissociation of

$$\ce{A_aB_b <=> aA+ + bB-}$$

described by $$K_{\mathrm{sp}} = [\ce{A}]^a \cdot [\ce{B}]^b$$

again based on a product, and not the sum.

This seeks a clarification of math based on chemistry. How to explain it to an 8th grader?

• Please provide an explicit and specific example; otherwise, it is guess work / asking an oracle if you refer to e.g., the equilibrium constant, or solubility product. Apr 4, 2020 at 14:38
• Please recheck the edited question. Apr 4, 2020 at 14:56
• To render chemical and mathematical equations better, ChemSE is supported by mathjax and mchem. This works well for the questions, answers, and comments. The commands are shown on ChemSE's meta site (chemistry.meta.stackexchange.com/questions/86/…). Your are very welcome to venture out the syntax and use it for future posts. Apr 4, 2020 at 16:47
• There may be a proper answer, but the answer may not be applicable well for the target audience. Apr 5, 2020 at 6:36

Let suppose this funny equilibrium equation:

$$\ce{2 S <=> H}$$

Where $$\ce{S}$$ are ice skaters on a giant skating ground and $$\ce{H}$$ are "collision heaps" of 2 skaters laying on the ice.

Now, what is the frequency of collisions leading to a heap, if there is "skater concentration" $$\ce{ [S] }$$ ? It should be $$k_1 \ce{[S]}$$ as frequency should be proportional to $$\ce{[S]}$$, correct ?

But wait ! When there is more skaters, each skater would collide with other skaters more frequently, proportionally to $$\ce{[S]}$$ as well !

So collision frequency would be like $$k_1 \ce{[S]}^2$$

Now the heap "recovery":

It should be proportional to $$\ce{[H]}$$, as more heaps, more skaters would be raising at the same time. So the heap recovery rate would be like $$k_2 \ce{[H]}$$.

Now imagine the equilibrium on the skating ground, where the rate of collisions is equal to the recovery rate and number of collision heaps is constant.

Then $$k_1 \ce{[S]}^2= k_2 \ce{ [H]}$$

If we express the equilibrium constant $$K=k_1/k_2$$, then $$K = \frac{\ce{[H]}}{\ce{[S]}^2}$$

Do you see ? No sum, but multiplication. Twice as many skaters leads to 4 times more heaps.

You can create similar system, where just for fun collide just boys and girls together.

• Nice analogy! But all all the A's should be replaced with S's, to be consistent with your first equation. And, just to make the origin of the exponent fully explicit, I'd recommend writing $K = \frac{[H]}{[S][S]} = \frac{[H]}{[S]^2}$ Apr 4, 2020 at 16:06
• Thanks, weird I have not notice. :-) Fixed. Apr 4, 2020 at 16:18