Equilibrium Constants (K):

         + are derived from experimental data

         + are temperature dependent

         + provide a measured of the equilibrium position

if K is large (> 102) products of the reaction are favored if K is small, reactants are favored. +  remain the same value for a constant temperature even if the equilibrium concentrations of the reactants or products are altered +  do NOT change in the presence of a catalyst (a catalyst changes the time taken to reach equilibrium but does not alter the equilibrium position or equilibrium constant) + For the reaction: aA + bB  cC + dD where A, B, C and D are the chemical species and a, b, c and d are the stoichiometric coefficients, K1 = [C]c[D]d [A]a[B]b For the same reaction written in reverse:cC + dD  aA + bB where A, B, C and D are the chemical species and a, b, c and d are the stoichiometric coefficients, K2 = [A]a[B]b [C]c[D]d K2 = 1/K1 and K1 = 1/K2 If the concentration of species in the reaction is used to calculate the equilibrium constant then it can be given the symbol Kc. If the equilibrium constant has no subscript, that is it is given as K, this refers to Kc and the concentration of each species in the reaction is used to calculate the equilibrium expression. If a solvent takes part in a reaction, its concentration is said to remain constant and is incorporated into the value of K. If a solid is present in a reaction, its concentration is said to remain constant and is incorporated into the value of K. Writing Equilibrium Expressions Reactions Involving Aqueous Specie

1. All species are present in aqueous solution eg:
   Ag⁺_(aq) +2NH_3(aq)  Ag(NH₃)₂⁺_(aq)
   

   K =	[Ag(NH3)2+]
   	[Ag+][NH3]2

2. Solvent takes part in reaction, eg:
   NH_3(aq) + H₂O_(l)  NH₄⁺_(aq) + OH^-_(aq)
   

   K =	[NH4+][OH-]
   	[NH3]

   
   The concentration of the water as solvent is said to be constant and is incorporated into the value of K.
3. A solid is present in the reaction, eg
   Pb(NO₃)_2(aq) + 2KI_(aq)  PbI_2(s)
   

   K =	1
   	[Pb(NO3)2][KI]2

   
   The concentration of the solid is said to be constant and is incorporated into the value of K.

          

            Reactions Involving Gases

1. All species are present as gases, eg:
   CO_(g) + 2H_2(g)  CH₃OH_(g)
   

   K =	[CH3OH]
   	[CO][H2]2

2. A solid is present in the reaction, eg:
   CaCO_3(s)  CaO_(s) + CO_2(g)
   K = [CO₂]The concentration of solids are said to be constant and are incorporated into the value of K.

Some Special Special Cases

• Ion Product for Water (K_w)
• Acid Dissociation Constants (K_a)
• Base Dissociation Constants (K_b)
• Solubility Products (K_sp)
• K_P

Calculating Equilibrium Constant : Concentration of All Species Known

For the reaction COCl_2(g)  CO_(g) + Cl_2(g) at 900^oC at equilibrium, 
the concentration of COCl_2(g) is 3.0 x 10^-6M while that of CO_(g) and Cl_2(g) is 4.97 x 10^-4M. 
Calculate the value of K.

1. Write the equilibrium expression using the balanced chemical equation: 

   K =	[CO][Cl2]
   	[COCl2]

2. Extract all the relevant data from the question:
   [CO] = 4.97 x 10^-4M
   [Cl₂] = 4.97 x 10^-4M
   [COCl₂] = 3.0 x 10^-6M
   K = ?
3. Substitute the values into the equilibrium expression and solve: 

   K =	[4.97 x 10-4][4.97 x 10-4]
   	[3.0 x 10-6]

   
   K = 0.082

Calculating Equilibrium Constant : Initial Concentrations and Equilibrium Concentration of One Species Known

A mixture of 5.0 mol H_2(g) and 10.0 mol I_2(g) are placed in a 5L container at 450^oC and allowed to come to equilibrium. At equilibrium the concentration of HI_(g) is 1.87M. 
Calculate the value for K for this reaction.

1. Write the balanced chemical equation for the reaction:H_2(g) + I_2(g)  2HI_(g)
2. Write the equilibrium expression for the reaction: 

   K =	[HI]2
   	[H2][I2]

3. Calculate the equilibrium concentrations for all reactants and products: 

   	[H2]	[I2]	[HI]
   initial concentration M = moles/Volume	5.0mol/5L = 1.0M	10.0/5L = 2.0M	0M
   equilibrium concentration expression Let x=amount used in reaction	1.0 - x	2.0 - x	1.87M
   calculate x using mole ratio from equation	x = ½ x 1.87 x = 0.935	x = ½ x 1.87 x = 0.935	1.87M
   calculate equilibrium concentrations	1.0 - 0.935 = 0.065M	2.0 - 0.935 = 1.065M	1.87M

4. Substitute the equilibrium concentrations into the equation and solve for K: 

   K =	[1.87]2
   	[0.065][1.065]

   
   K = 50.5
   

Calculating Equilibrium Constant : Initial Reactant Concentration and %Dissociation Known

4.0 moles of NOCl_(g) were placed in a 2L vessel at 460^oC. The following dissociation reaction occurred: 
2NOCl_(g)  2NO_(g) + Cl_2(g)
At equilibrium the NOCl_(g) was 33% dissociated. Calculate the equilibrium constant for this reaction.

1. Write the equilibrium expression for the reaction: 

   K =	[NO]2[Cl2]
   	[NOCl]2

2. Calculate the equilibrium concentrations for all reactants and products: 

   	[NOCl]	[NO]	[Cl2]
   initial concentration M = moles/Volume	4.0mol/2L = 2.0M	0M	0M
   equilibrium concentration expression Let x=amount used in reaction	2.0 - 2x	2x	x
   calculate x using 33% dissociation of NOCl	2x = 33/100 x 2.0 2x = 0.66M
   and using mole ratio from equation		2x = 0.66M	x = ½ x 0.66 x = 0.33M
   calculate equilibrium concentrations	2.0 - 0.66 = 1.34M	0.66M	0.33M

3. Substitute the equilibrium concentrations into the equation and solve for K: 

   K =	[0.66]2[0.33]
   	[1.34]2

   
   K = 0.08