Law of Chemical Equilibrium

Since a reaction at equilibrium has fixed concentrations of products and reactants, we can calculate a constant for a given reaction at constant temperature.

For the generalized equation

aB + cD eF + gH

,

where [ ] represents the equilibrium concentration in moles/L

Keep in mind: You only include concentrations of aqueous and gaseous reactants/ products, not those of liquids or solids.

Example 1

Given the following concentrations in moles/L at 748 C

CO₂(g)   +   H₂(g)   CO(g)   +   H₂O(g)  

0.00630 0.00630 0.00552 0.00552

Write the equilibrium law expression, and then calculate the constant at that temperature.

= 0.768

Example 2

Suppose 6.000 mol of F₂ and 3.000 mol of H₂ are mixed in a 3.000 L container to make HF . The equilibrium constant at a certain temperature is 1.15x10 ².

Calculate the equilibrium concentrations, given:

H_2(g) + F_2(g) 2 HF_(g)

In such problems, it is very important to distinguish between

(1) the number of moles/L initially found in the reaction vessel

(2) the number of moles/L that actually react or form

(3) the number of moles/L found at equilibrium

	1 H2(g)	1 F2(g)	2 HF(g)
Initial # of moles	3.000/3 = 1.000	6.000/3 = 2.000	0 (since it is not mentioned,we have to assume there was none introduced.
Moles/L reacting/forming	x, since we don’t know how much reacts	x, since the ratio of H2(g) to F2(g) in the equation is 1: 1	2x , since 2 moles of HF form for every 1 mole of reacting H2.
# of moles/L at equilibrium	1-x remain	2-x remain	0 + 2x now exist.

We write an equilibrium law expression, and then substitute the equilibrium number of moles into that expression. Don’t forget to divide by the number of litres, which in this example is 3.0 L.

If we had only 1.00 mole/L of hydrogen initially, is the only sensible answer.

Now we can wrap up the problem and calculate the equilibrium concentrations: