The way we express the position of equilibrium is through the equilibrium constant `(K)`, which is unitless constant for a reaction at a given temperature.
The Law of Mass Action states that for any given reaction `aA + bB ↔ cC + dD` at equilibrium, the equilibrium constant `(K)` is defined as:
`K= ([A]^a[B]^b)/([C]^c[D]^d)`
Where `[A]` denotes the concentration of `A` . Note that the equilibrium constant takes into account the concentrations at equilibrium.
For example, the `K` for the reaction `2NO_((g)) + 2H_(2(g)) ↔ N_(2(g)) + 2H_2O_((g))` would be calculated as:
`K=([N_2][H_2O]^2)/([NO]^2[H_2]^2)`
`K` is of particular use when we want to know in which direction equilibrium is favored:
`K>1` , reaction is product-favored.
`K<1` , reaction is reactant-favored.
This makes sense mathematically. If `K>1`, that means that the products term `([C]^z[D]^t)` is greater than the reactants `([A]^x[B]^y)` . The reverse is true for `K<1` .
When writing equilibrium expressions, solids and liquids do not contribute towards the equilibrium constant. In other words, if we're given the reaction `A_((s)) + B_((g)) ↔ C_((g)) + D_((l))` at equilibrium , the equilibrium constant would be:
`K=[[C]]/[[B]]`
Because the solid `A` and liquid `D` do not contribute towards equilibrium.
For reactions that are not in equilibrium, we can express the current state of the reaction through the reaction quotient (`Q`) . The reaction quotient is calculated the same way as the equilibrium constant:
`Q= ([A]^a[B]^b)/([C]^c[D]^d)`
The only difference between the reaction quotient and the equilibrium constant is that the reaction quotient describes a reaction not yet in equilibrium. By comparing the reaction quotient and the equilibrium constant, we can determine the direction that a reaction's going to proceed:
`Q>K` , reaction will shift to the left.
`Q<K` , reaction will shift to the right.
`Q=K` , reaction is at equilibrium.
Let's illustrate this with a problem.
#1. In the following reaction:
`2A_((g)) + 3B_((g)) ↔ C_((g))`
The equilibrium constant is given as `K=0.8`. If the reaction is not yet in equilibrium and the concentrations are given as `[A]=1.4M, [B]=0.75M` and `[C]=0.2M`, determine which direction the reaction will shift.
We're told that the reaction is not yet in equilibrium. The first order of business is to calculate the reaction quotient.
`Q=[[C]]/([A]^2[B]^3)`
`Q=[[0.2]]/([1.4]^2[0.75]^3)=0.24`
Since `Q<K`, the reaction will shift towards the products. This means that more `C` will be produced.
Answer: The reaction will shift to the right.
So far, we've been writing the equilibrium constant in terms of concentrations, `K_c`. Most of the time however, the equilibrium constant for gaseous reactions is written in terms of partial pressures. When writing the equilibrium constant in terms of partial pressures, we denote the equilibrium constant as `K_p` .
For any given reaction of `aA_((g)) + bB_((g)) ↔ cC_((g)) + dD_((g))` , the equilibrium constant of pressure `(K_p)` is written as:
`K_p=((P_C)^c(P_D)^d)/((P_A)^a(P_B)^b)`
Where `(P_i)` is the partial pressure of reactant `i` at equilibrium.
All of the rules regarding `K_c` apply to `K_p`. The conversion between the two is through the following equation:
`K_p=K_c(RT)^(Deltan_(gas))`
`R` is the ideal gas constant.
`T` is the temperature in `K`
`Deltan_(gas)= ("moles gas")_"product"-("moles gas")_"reactants"`
1. The equilibrium constant `K` is a unitless constant that expresses the position of equilibrium.
2. If `K>1`, the reaction is product favored. If `K<1`, the reaction is reactant favored.
3. `Q` is the reaction quotient and can be used to determine the direction of reaction.
4. Solids and liquids do not contribute to the equilibrium expression.
5. The equilibrium constant of gases `K_p` can be expressed in terms of the partial pressures of the reactants.
1. Reactions going to completion
Whenever a reaction has `K ">>>" 1` , we can model the reaction as going to completion. This is because when `K` is significantly greater than `1`, the reaction is strongly product-favored.