Equilibrium constant

In the reaction `a` A + `b` B ⇌ `c` C + `d` D where lowercase letters are coefficients and uppercase letters are chemical formulas, the equilibrium constant (`K`) can be calculated using the equilibrium law expression:

`K = ("[C]"^c"[D]"^d)/("[A]"^a"[B]"^b)`.

In words, the equilibrium constant is calculated by multiplying the E values of the products together, each with its coefficient as an exponent, doing the same thing with the reactants, and dividing. The products and reactions must be gaseous or aqueous—if they are solid or liquid, do not include them in the expression (make sure to leave a 1 in the numerator if there are no products left). The concentrations are all in mol/L, but `K` is unitless.

It is easy to calculate `K` given all the equilibrium concentrations, but you may also be asked to solve for a concentration given the other ones and the value of `K`. If there is an exponent on the concentration once you have isolated it, you will need to take the root of both sides.

The equilibrium constant depends on temperature. Its value indicates the relative amounts of reactants and products at equilibrium:

`K ~~ 1 => "[products]" ~~ "[reactants]"`
significant amounts of products and reactants
`K > 10^4 => "[products]" ≫ "[reactants]"`
very high yield; equilibrium favours products
`K < 10^-4 => "[products]" ≪ "[reactants]"`
very low yield; equilibrium favours reactants

The equilibrium constant for the reverse reaction is the reciprocal because the reactants and products swap places in the fraction:

`K_"reverse" = 1/K_"forward"`.

Example

At 2000 ºC, the equilibrium constant for 2 CO2(g) ⇌ 2 CO(g) + O2(g) is 6.4 × 10−7. There is initially 1.50 mol of carbon dioxide in a 3.0 L container. What will the equilibrium concentration of carbon monoxide be?

First we can calculate I for CO2(g):

`c = n/V = (1.50\ "mol")/(3.0\ "L") = 0.50\ "mol/L"`.

Let `x` represent the absolute value of the change in concentration of O2(g).

2 CO2(g) 2 CO(g) O2(g)
I `0.50` `0` `0`
C `-2x` `+2x` `+x`
E `0.50 − 2x` `2x` `x`

Now we set up the equilibrium law expression:

`K = (["CO"]^2["O"_2])/(["CO"_2]^2) = ((2x)^2(x))/((0.50 - 2x)^2)`.

Before going any further, we can check to see if `K` is small enough for us to use a time-saving approximation. This is called the hundred rule: if the smallest initial concentration divided by `K` is greater than 100, you can drop out `x` where it is added or subtracted. Using CO2(g) this gives us

`["CO"_2]_"initial"/K = 0.50/(6.4xx10^-7) = 781250`,

and `781250 > 100`, therefore `K` is very small and `0.50 - 2x ~~ 0.50`. Now we can continue solving for `x` (notice that we got rid of the `x` in the denominator, but we have to keep the exponent):

`K = ((2x)^2(x))/((0.50 - 2x)^2) ~~ ((2x)^2(x))/(0.50^2) = (4x^3)/0.25 = 16x^3`.

Since `K = 16x^3`,

`x = root3(K/16) = root3((6.4xx10^-7)/16) = 0.00342`.

Now that we know the value of `x`, we can finally find the equilibrium concentration of carbon monoxide (two significant digits):

`["CO"] = 2x\ "mol/L" = 0.0068\ "mol/L"`.

In problems where you can’t use the approximation and the fraction doesn’t work out to a perfect square, you must use the quadratic formula to solve for x (there are two solutions, so cross out the one that will result in negative concentrations):

`x = (-b +- sqrt(b^2 - 4ac))/(2a)`.