Chemical equilibrium — why multiplication, rather than addition?
For the 8th grader, hmmm...  

Let suppose this funny equilibrium equation:

$$\ce{2 S &lt;=&gt; H}$$

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.