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Equilibrium Constant (K)

To define the extent to which equilibrium favors one side of the reaction over the other, we can use the Equilibrium Constant.

Given the model reaction:
aA+bBcC+dD\large{aA +bB \rightleftharpoons cC + dD}
A & B are reactants and C & D are products.
a,b,c,d are their stoichiometric coefficients


The Equilibrium Expression (in terms of concentration) can be written as:


Kc=[ Products][Reactants]=[C]c[D]d[A]a[B]b\large\boxed{K_c=\frac{\left[\ Products\right]}{\left[Reactan ts\right]}=\frac{\left[C\right]^c\left[D\right]^d}{\left[A\right]^a\left[B\right]^b}}


The equilibrium constant (K) is a ratio used to describe the relative concentrations of products and reactants at equilibrium!
You may also see it called the Law of Mass Action.


Examples:
Keq (at equilibrium), Ka (for acids), Kb (for bases), Kw (for water), Ksp (solubility product constant)

Wize Concept
K can have different subscripts, but all K's follow the same rules!

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Example: What is the K expression for the reaction below?

N2(g) + 3H2(g) ⇌ 2NH3(g)


Wize Tip
Since coefficients become exponents in the K expression, we need to make sure that our equation is balanced!


K= [NH3]2[N2][H2]3\frac{\left[NH_3\right]^2}{\left[N_2\right]\left[H_2\right]^3}

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Additional Notes About K:

1) Above we wrote Kc. The c stands for concentration because we are considering the concentrations of the species in mol/L.

2) If instead we were considering gases and their pressures in atm, then we would write Kp!

Example: What is the Kp expression for the following reaction?

2SO2(g) + O2(g) ⇌ 2SO3(g)



Kp=(PSO3)2(PSO2)2(PO2)Kp=\frac{\left(PSO_3^{ }\right)^2}{\left(PSO_2\right)^2\left(PO_2\right)}


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3) If any species in the reaction are solids (s) or liquids (l), their concentrations do not affect equilibrium and therefore they do not appear in the equilibrium expression!

Example: What is the equilibrium expression for the following reaction?

PbCl2(s) ⇌ Pb2+ (aq) + 2Cl- (aq)

K= [Pb2+][Cl-]2


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4) We will see that changes in volume, pressure, temperature, and concentration can shift the equilibrium (according to Le Chatelier's Principle)

But the only factor that will change K is temperature!



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The Magnitude of the Equilibrium Constant, K

K tells us which side of the reaction (products or reactants) are favored and to what extent.

Example:
If we had the K value 1.8x1010 would products or reactants be favored at equilibrium?
Products!

Since there is an exponent of 10, then (products/reactants)
products
are favored to a (small/great)
great
extent!




Example:
If K was 1.2x10-8 then are products or reactants favored at equilibrium? And to a great or small extent?

Reactants
are favored to a
great
extent!

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Example: Are products or reactants favored at equilibrium according to the following reaction and concentration time graph?

2SO2(g) + O2(g) ⇌ 2SO3(g)

Photo by OpenStax / CC BY

Products are favored at equilibrium since at equilibrium (when the concentrations start to get flat in the graph), there is a higher concentration of SO3 (a product) than the reactants.

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Summary of What K Tells Us

Kc=[ Products][Reactants]\large{K_c=\frac{\left[\ Products\right]}{\left[Reactan ts\right]}}

Keq > 1

  • At equilibrium, products/reactants are favored:
    products
  • This is because products/reactants are more stable:
    products
  • As a result, at equilibrium there will be more reactants, more products, or equal amounts of products and reactants:
    more products

Keq = 1

  • At equilibrium, products/reactants are favored:
    neither
  • This is because products/reactants are more stable:
    equally stable
  • As a result, at equilibrium there will be more reactants, more products, or equal amounts of products and reactants:
    equal amounts

Keq < 1

  • At equilibrium, products/reactants are favoured:
    reactants
  • This is because products/reactants are more stable:
    reactants
  • As a result, at equilibrium there will be more reactants, more products, or equal amounts of products and reactants:
    more reactants

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Practice: Writing Equilibrium Constants

Which of the following expression is the correct equilibrium-constant expression for the reaction below?

HF(aq) + H2O(l) ⇌ H3O+(aq) + F-(aq)

1.
Kc=[HF][H2O][H3O+][F]K_c=\frac{[HF][H_2O]}{[H_3O^+][F^-]}
2.
Kc=1[HF]K_c=\frac{1}{[HF]}
3.
Kc=[H3O+][F][HF][H2O]K_c=\frac{[H_3O^+][F^-]}{[HF][H_2O]}
4.
Kc=[H3O+][F][HF]K_c=\frac{[H_3O^+][F^-]}{[HF]}
5.
Kc=[F][HF]K_c=\frac{[F^-]}{[HF]}




Practice: Calculating the Equilibrium Constant

For the process:
N2 (g) + 3 H2 (g) ⇌ 2 NH3 (g)

Determine the value of Kc when an equilibrium mixture contains 0.0420 mol N2, 0.516 mol H2, and 0.0357 mol NH3 in a 1.00 liter container at 400oC.

Extra Practice