Chemical Equilibrium

[H2][I2] = K (equilibrium constant) H2 + I2 2 HI At equilibrium, the rate of the forward and reverse reactions are equal. The concentration of reactants and products at equilibrium are related. [HI]2 [H2][I2] = K (equilibrium constant)

H2 + I2 2 HI Suppose the concentrations of H2 and I2 are each initially 0.0175 M at 425C and no HI is present. Over time, the concentrations of H2 and I2 will decrease and the concentration of HI will increase until equilibrium is reached. At equilibrium it was found the concentrations of [H2] = [I2] = 0.0037 M and [HI] = 0.0276 M. This data can be presented in an I.C.E. table.

I.C.E. I - initial concentration C - change in concentration E - equilibrium concentration

Equation H2 + I2 2 HI Initial conc.(I) 0.00175 0.00175 0 Change (C) - 0.0138 - 0.0138 + 0.0276 Equilibrium (E) 0.0037 0.0037 0.0276

The change in concentrations of reactants and products are always equal to the difference between the equilibrium and initial concentrations. Putting the equilibrium concentrations from the ICE table into the K expression gives: [HI]2 (0.0276)2 [H2][I2] = (0.0037)(0.0037) = 56

As long as the temperature is kept constant, you can change the concentrations of reactants and the value of K = 56.

equilibrium constant expression General Equation for Equilibrium: aA + bB cC + dD K = C c D d A a B b Where K = equilibrium constant equilibrium constant expression

Equilibrium Constant Expression When a reaction is at equilibrium: All concentrations are at equilibrium Products – numerator / reactants – denominator Powers are the coefficients from the balanced equation K depends on reaction temperature K has no units

Reactions involving solids: The concentrations of solids are not included in the equilibrium constant expression: Reactions in solution: Since water concentration is very high, the concentration of water is not included in the equilibrium constant expression.

Sometimes you will see K written as: Keq reaction is at equilibrium Ka reaction is at equilibrium for acids Kb reaction is at equilibrium for bases Kc reaction is at equilibrium for molar concentrations (M)

Write the equilibrium constant expression for: Problem 16.1 (page 725) Write the equilibrium constant expression for: N2 (gas) + 3H2 (gas) 2 NH3 (gas) H2 CO3 (aq) + H2O (liquid) HCO3- (aq) + H3O+ (aq)

Determining an Equilibrium Constant If the experimental values of concentrations and products are known, an equilibrium constant can be obtained by substituting the data into the equation: K = C c D d A a B b

Determining an Equilibrium Constant: 2SO2 (g) + O2 (g) 2SO3 (g) At equilibrium: [SO2] = 3.61 x 10-3 M [O2] = 6.11 x 10-4 M [SO3] = 1.01 x 10-2 M

K = SO3 2 SO2 2 O2 K = 1.01 x 10−2 2 3.61 x 10−3 2 6.11 x 10−4 K = 1.28 x 104

Determining an Equilibrium Constant More commonly, the concentrations of all reactants and products at equilibrium are not known. You must use an ICE table for these problems.

Determining an Equilibrium Constant: Suppose that 1.00 mol of SO2 and 1.00 mol of O2 are placed in a 1 L flask at 1000 K. When equilibrium is reached, 0.925 mol of SO3 has been formed. Calculate the equilibrium constant. 2SO2 (g) + O2 (g) 2SO3 (g) K = SO3 2 SO2 2 O2

Equation 2SO2 + O2 2 SO3 Initial conc.(I) 1.0 1.0 0 Change (C) Equilibrium (E) 0.925

Equation 2SO2 + O2 2 SO3 Initial conc.(I) 1.0 1.0 0 Change (C) -2x -X +2X Equilibrium (E) 0.925

Equation 2SO2 + O2 2 SO3 Initial conc.(I) 1.0 1.0 0 Change (C) -2x -X +2X Equilibrium (E) 1.0 - 2x 1.0 - x 2x = 0.925 Solve for X X = 0.463

Equation SO2 + O2 2 SO3 Initial conc.(I) 1.0 1.0 0 Change (C) -2(0.463) -0.463 +2X Equilibrium (E) 0.074 0.54 0.925

K = SO3 2 SO2 2 O2 K = 0.925 2 0.074 2 0.54 K = 2.8 x 102

2 Fe+3 (aq) + 3 I- (aq) 2 Fe+2 (aq) + I3- (aq) Example 16.3, page 731 In aqueous solution iron (III) ions react with iodine ions to give iron (II) ions and triiodide ions, I3-. Suppose the initial concentration of Fe+3 is 0.20 M, the initial I- ion concentration is 0.30 M, and the equilibrium concentration of I3- ions is 0.0866 M. What is the value of K? 2 Fe+3 (aq) + 3 I- (aq) 2 Fe+2 (aq) + I3- (aq)

Equation 2 Fe+3 + 3 I- 2 Fe+2 + I3- Initial conc.(I) 0.20 0.30 0 0 Change (C) Equilibrium (E) 0.0866

Equation 2 Fe+3 + 3 I- 2 Fe+2 + I3- Initial conc.(I) 0.20 0.30 0 0 Change (C) -2x -3x +2x +x Equilibrium (E) 0.0866

Equation 2 Fe+3 + 3 I- 2 Fe+2 + I3- Initial conc.(I) 0.20 0.30 0 0 Change (C) -2x -3x +2x +x Equilibrium (E) 0.2 - 2x 0.3 - 3x 0 + 2x 0.0866 X = 0.0866

Equation 2 Fe+3 + 3 I- 2 Fe+2 + I3- Initial conc.(I) 0.20 0.30 0 0 Change (C) -2x -3x +2x +x Equilibrium (E) 0.0270 0.040 0.173 0.0866 X = 0.0866

[Fe+2]2 [I3-] [Fe+3]2 [I-]3 K = (0.173)2 (0.0866) (0.0270)2 (0.040)3 K = K = 5.6 x 104

Example 16.4, page 732 Suppose a tank initially contains H2S at a pressure of 10.0 atm and a temperature of 800 K. When the reaction has come to equilibrium, the partial pressure of S2 vapor is 0.020 atm. Calculate K. 2 H2S (g) 2 H2 (aq) + S2(g) Use equilibrium expression substituting gas pressure for concentrations

Equation 2 H2S 2 H2 + S2 Initial conc.(I) 10.0 0 0 Change (C) Equilibrium (E) 0.020

In many cases the value of K and the initial concentrations of reactants are known, and you want to know the concentrations at equilibrium. Use ICE tables to summarize the information given, and what you need to calculate.

Example 16.5, page 733: Suppose that 1.00 mol of H2 and 1.00 mol of I2 are placed in a 0.5 L flask at 425 K. What are the concentrations of H2, I2 and HI when equilibrium is reached? K = 55.64 H2 (g) + I2 (g) 2HI (g) K = HI 2 H2 I2 = 55.64

Equation H2 + I2 2HI Initial conc.(I) 2.0 M 2.0 M 0 Change (C) Equilibrium (E)

Equation H2 + I2 2HI Initial conc.(I) 2.0 M 2.0 M 0 Change (C) -x -x +2x Equilibrium (E) 2.0 – X 2.0 – x 2X

[HI]2 [H2] [I2] K = (2x)2 (2.0 – x) (2.0 – x) 55.64 = (2x)2 (2.0 – x)2 55.64 =

Take the square root of both sides: (2x)2 (2.0 – x)2 55.64 = 2x (2.0 – x) 7.459 = 7.459 (2.0 –x) = 2x 14.59 – 7.459x = 2x 14.9 = 9.459x X = 1.58

Equation H2 + I2 2HI [H2] = 0.42 M [I2] = 0.42 M [HI] = 3.16 M With x = 1.58, substitute in ICE Table: Equation H2 + I2 2HI Equilibrium (E) 2.0 – X 2.0 – x 2X [H2] = 0.42 M [I2] = 0.42 M [HI] = 3.16 M

Determining an Equilibrium Constant which involves a quadratic equation. Assumptions can be used to avoid this problem.

Consider the dissociation of I2 molecules to I atoms Consider the dissociation of I2 molecules to I atoms. The initial concentration of I2 is 0.45 M and the equilibrium constant K = 5.6 x 10-12 at 500C. Calculate the concentrations at equilibrium. I2 (g) 2 I (g) K = I 2 I2 = 5.6 x 10-12

Equation I2 2 I Initial conc.(I) 0.45 M 0 Change (C) -x +2x Equilibrium (E) 0.45 - x +2x

[ I ]2 [ I2 ] K = 5.6 x 10-12 = (2x)2 (0.45 – x) 5.6 x 10-12 = To avoid using the quadratic equation, assume x in the denominator is very small compared to 0.45.

(2x)2 (0.45) 5.6 x 10-12 = X = 7.9 x 10-7 From the value of x, we can determine [ I2 ] = 0.45 – x, or 0.45 M [ I ] = 2x = 2(7.9 x 10-7) = 1.60 x 10-6 M

Example 16.6, page 736: N2 (g) + O2 (g) 2 NO (g) The above reaction contributes to air pollution when gasoline is burned in air at a high temperature, such as a car engine. At 1500 K, K = 1.0 x 10-5. Suppose the sample of air has an initial [N2] = 0.80 M and [O2] = 0.20 M. Calculate the equilibrium concentrations.

Equation N2 + O2 2 NO K = NO 2 N2 O2 = 1.0 x 10-5 Initial conc.(I) 0.8 0.2 0 Change (C) -x -x +2x Equilibrium (E) 0.8 – X 0.2 – x 2X

(2x)2 (0.80 – x)(0.20 – x) 1.0 x 10-5 = To avoid using the quadratic equation, assume x in the denominator is very small. (2x)2 (0.80)(0.20) 1.0 x 10-5 = Solving for x: x = 6.3 x 10-4

Equation N2 + O2 2 NO Equilibrium (E) 0.8 – X 0.2 – x 2X [ N2 ] = 0.8 M [ O2 ] = 0.2 M [NO] = 1.3 x -3 M

Problem 16.7, Page 739 Equation 1: N2 + 3H2 2 NH3 K1 = 3.5 x 108 Equation 2 is ½ Equation 1 So K2 = 𝐾1 K2 = 3.5 𝑥 108 = 1.9 x 104

Problem 16.7, Page 739 Equation 1: N2 + 3H2 2 NH3 K1 = 3.5 x 108 Equation 3 is the reverse of Equation 1 so K3 = 1 / K1 K3 = 1 / 3.5 x 108 = 2.9 x 10-9

Disturbing Chemical Equilibrium The equilibrium between reactants and products can be disturbed in three ways: Change of temperature Change of concentrations of reactants and/or products Change in pressure (gases only)

LeChatelier’s Principle – a change in any of the factors that determine equilibrium conditions of a system (concentration, temperature, pressure) will cause the system to change to maintain equilibrium.

Change of concentrations of reactants and/or products N2 + 3H2 2 NH3 + heat N2 H2 NH3 heat

If you add more of a reactant (N2 or H2) N2 + 3H2 2 NH3 + heat NH3 heat N2 H2

To re-establish equilibrium, there is a shift to the right, more NH3 is formed N2 + 3H2 2 NH3 + heat NH3 heat N2 H2

If some reactant (N2 or H2) is removed N2 + 3H2 2 NH3 + heat

To re-establish equilibrium, there is a shift to the left, more N2 and H2 is formed N2 + 3H2 2 NH3 + heat H2 N2 NH3 heat

If you NH3 is removed form the reaction N2 + 3H2 2 NH3 + heat

To re-establish equilibrium, there is a shift to the right to from more NH3. N2 + 3H2 2 NH3 + heat heat NH3 N2 H2

2) Change of temperature (exothermic) N2 + 3 H2 2 NH3 + heat N2 H2 NH3 heat

If heat is increased for an exothermic reaction: N2 + 3 H2 2 NH3 + heat N2 H2 heat NH3

If heat is increased, equilibrium shifts to the left: N2 + 3 H2 2 NH3 + heat H2 N2 heat NH3

2) Change of temperature (exothermic) 2 NO2 N2O4 + heat NO2 N2O4 heat

If heat is decreased, equilibrium shifts to the right to form more product: 2 NO2 N2O4 + heat N2O4 heat NO2

2) Change of temperature (endothermic) N2 + O2 + heat 2 NO heat N2 O2 NO

If heat is increased, equilibrium shifts to the right to form more product: N2 + O2 + heat 2 NO NO heat N2 O2

concentration and temperature changes only General Rule concentration and temperature changes only TT AA Take away Addition Toward that side Away from that side