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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+bB⇌cC+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)



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

N2(g) + 3H2(g) ⇌ 2NH3(g)




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



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)}





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




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? ________


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





Example:

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


________ are favored to a ________ extent!




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


2SO2(g) + O2(g) ⇌ 2SO3(g)


Image illustrating The Magnitude of the Equilibrium Constant, K

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.


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


Keq = 1

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


Keq < 1

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



Image illustrating The Magnitude of the Equilibrium Constant, K


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]}





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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.


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Mark Yourself Question

  1. Grab a piece of paper and try this problem yourself.
  2. When you're done, check the "I have answered this question" box below.
  3. View the solution and report whether you got it right or wrong.

Earlier we learned about how only one factor changes K (the equilibrium constant) and that factor is temperature.


Given the reaction:

H2(g) + I2(g) ⇌ 2HI(g) + heat


At 448oC K=50.5. Would you predict the K value to be higher or lower at 348oC?

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Extra Practice