# What does it mean for a reaction to favor the reactants/products?

(This example is purely hypothetical.)

You have the reaction $$\ce{H2(g) + O2(g) <=> H2O2(g)}$$

at $T = 500\ \mathrm{K}$. The reaction reaches equilibrium at the following concentrations:

$$\ce{[H2]} = \ce{[O2]} = 5 \times 10^{-3}\ \mathrm{mol\ dm^{-3}}$$ $$\ce{[H2O2]} = 4\times10^{-5}\ \mathrm{mol\ dm^{-3}}$$

This gives $$K_c = \left(\frac{\ce{[H2O2]}}{\ce{[H2][O2]}}\right) = 1.6$$

However, the total concentration of the reactants is $250$ times higher than the concentration of the product.

Still, as $K_c > 1$, per definition, the products are favored.

This seems counterintuitive to me, and my Chemistry teacher couldn't really explain it to me, so I was hoping someone here could explain why, even when the reactants are so much more plentiful, the products are considered to be favored.

## 2 Answers

You may not have had kinetics (reaction rates) yet, but I find it really useful in understanding equilibria.

Kinetics of a reaction

Molecules in solution (gas or liquid) move randomly. Because they need to bump into each other to react, the probability of reaction increases as concentrations of reactants increase. The rate of the reaction is then proportional to the concentrations of reactants:

$$\ce{A + B \rightarrow C}$$ $$\mathrm{rate} = k_\mathrm{forward}\ce{[A][B]}.$$

Likewise, the reverse reaction has a rate dependent on concentration of product:

$$\mathrm{rate} = k_\mathrm{reverse}\ce{[C]}.$$

From Kinetics to Equilibrium

At equilibrium, the forward rate is equal to the reverse rate, and concentrations of both reactants and products remain unchanged. $$k_\mathrm{forward}\ce{[A][B]}=k_\mathrm{reverse}\ce{[C]}$$ We can rearrange to get: $$\frac{k_\mathrm{forward}} {k_\mathrm{reverse}} = \ce{\frac{[C]}{[A][B]}}.$$ We can substitute the right side of the equation to be the equilibrium constant, and we get $$\frac{k_\mathrm{forward}} {k_\mathrm{reverse}}=K_c$$ Thus, when we say that $K_c>1$ we are also saying that $k_\mathrm{forward} > k_\mathrm{reverse}$ and inversely if $K_c<1$ then $k_\mathrm{reverse} > k_\mathrm{forward}.$ To favor either the reactants or the products in equilibrium is to say the formation of either the reactants or products is favored, as indicated by the rate constants.

Fundamentally, though it may seem intuitive, adding concentrations on either side of the reaction and comparing them to see which is larger isn't a measure of favourability, since $K_c$ is defined by the multiplication of the concentrations. It's like taking two rectangles and trying to measure which has a larger area by comparing their perimeters; it doesn't work in general unless you specify some additional constraint(s).

Under your definition of favourability by addition, you're weighing reagents and products unfairly, in some sense, because in the expression of $K_c$ there are two reagents and one product. This is a bit (if indirectly) reminiscent of trying to compare salt solubilities by comparing their $K_\mathrm{sp}$ directly instead of actually calculating their solubilities; it doesn't always work, because $K_\mathrm{sp}$ of salts are weighted by how many ions are released when dissolved.

It is not strictly true that the products would be favoured by your measure for any $K_c > 1$ even if the number of reagents and products were equal - just add one reagent in heavy excess. For example, consider the hypothetical reaction:

$$\ce{A + B -> C + D} \qquad \qquad K_c=\frac{[\ce{C}][\ce{D}]}{[\ce{A}][\ce{B}]}=1000$$

One possible solution for equilibrium concentrations is $[\ce{A}]=1\ \mathrm{M}$, $[\ce{B}]=10^{-5}\ \mathrm{M}$ and $[\ce{C}]=[\ce{D}]=10^{-1}\ \mathrm{M}$, and here $[\ce{A}]+[\ce{B}]>[\ce{C}]+[\ce{D}]$.