7.2 The Position of Equilibrium.

Assessment Statements Deduce the equilibrium constant expression (K c ) from the equation for a homogeneous reaction Deduce the extent of a reaction from the magnitude of the equilibrium constant Apply Le Chatelier’s principle to predict qualitative effects of changes of temperature, pressure and concentration on the position of equilibrium and on the value of the equilibrium constant.

Assessment Statements State and explain the effect of a catalyst on an equilibrium reaction Apply the concepts of kinetics and equilibrium to industrial process Suitable examples include the Haber and Contact processes

Assessment Statements Deduce the equilibrium constant expression (K c ) from the equation for a homogeneous reaction.

The Equilibrium Constant Changing the temperature affects the rates of forward and reverse reactions differently because they have different activation energies Kc = expressed in terms of molar concentration (mol/L)

Homogeneous Equilibria Haber process to manufacture ammonia 3H 2 (g) + N 2 (g) 2NH 3 (g) Kc = [NH 3 ] 2 [H 2 ] 3 [N 2 ] Contact process to manufacture sulfuric acid 2SO 2 (g) + O 2 (g) 2SO 3 (g) Kc = [SO 3 ] 2 [SO 2 ] 2 [O 2 ]

Calculating Kc N 2 + 3Cl 2 2NCl 3 Temp remains constant in 5 L flask. Equilbrium was reached and the following was found: mol N 2, mol Cl 2 and 0.95 mol of NCl 3. Calculate the Kc for this reaction. Find molar concentration of everything (mol/L) Use equilibrium expression based on chemical equation

Answer 1. [N 2 ] = mol / 5 L = M [Cl 2 ] = mol / 5 L = M [NCl 3 ] = 0.95 mol / 5 L = 0.19 M 2. Kc = [NCl 3 ] 2 = [0.19] 2 [N 2 ][Cl 2 ] 3 [0.0014][ ] 3 = 3.0 x 10 11

Question CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g) 1. Write an expression for K eq 2. Calculate K at a given temperature if [CH 4 ] = M, [O 2 ] = M, [CO 2 ] = M, and [H 2 O] = M at equilibrium. (include units)

Assessment Statements Deduce the extent of a reaction from the magnitude of the equilibrium constant.

The value of K The value or magnitude of K tells us the extent to which reactants have been converted into products. Remember in the ratio for K, the concentration of products divided by the concentration of reactants.

A small value for K means that very little of the reactants were converted into products before equilibrium was reached. This is stated as “reactants are favoured”. A large value of K means that most of the reactants were converted into products before equilibrium was reached. This is stated as “products are favoured”.

The Magnitude of Equilibrium Constants If K >1, then products dominate at equilibrium and equilibrium lies to the right. If K <1, then reactants dominate at equilibrium and the equilibrium lies to the left.

If K = 1 neither reactants nor products are favoured. The value of K does not indicate how long it takes for equilibrium to be reached. The value of K varies with temperature and that’s why its usually mentioned with K

Calculating Equilibrium Constants Nitrogen dioxide decomposes at high temperatures according to this equation: 2 NO 2 (g)   2 NO (g) + O 2 (g) If the equilibrium concentrations are as follows: [NO 2 ]= 1.20 M, [NO] = 0.160, and [O 2 ] = M; calculate the equilibrium constant. K eq = [NO] 2 [O 2 ] [NO 2 ] 2 So K eq = (0.160) 2 (0.080) = (1.20) 2 19

Calculating Equilibrium Constants The equilibrium equation for the oxidation of sulfur dioxide is as follows: 2SO 2 (g) + O 2 (g)   2 SO 3 (g) If the equilibrium concentrations are as follows: [SO 2 ]= 0.44 M, [O 2 ] = 0.22, and [SO 3 ] = 0.78 M, Calculate the equilibrium constant So K eq = (0.78) 2 = 14.3 (0.44) 2 (0.22) K eq = [SO 3 ] 2 [SO 2 ] 2 [O 2 ] 20

Practice Problem 1 The equilibrium equation for the carbon monoxide with steam to produce hydrogen gas is as follows: CO 2 (g) + H 2 (g)   CO (g) + H 2 O (g) If the equilibrium concentrations are as follows: [CO] = 1.00 M, [H 2 O] = 0.025, [CO 2 ] = M and [H 2 ] = M, calculate the equilibrium constant. K eq = [CO] [H 2 O] [CO 2 ] [H 2 ] So K eq = (1.00) (0.025) = (0.075) (0.060) 21

Equilibrium Calculations – Using I.C.E.Models Equilibrium constants and concentrations can often be deduced by carefully examining data about the initial and equilibrium concentrations 22 I nitial C hange E quilbrium

Equilibrium Calculations ICE Model problem 1 Hydrogen and iodine are in equilibrium with Hydrogen iodide to this reaction: H 2 + I 2   2HI Suppose that 1.5 mole of H 2 and 1.2 mole of I 2 are placed in a 1.0 dm 3 container. At equilibrium it was found that there were 0.4 mole of HI. Calculate the equilibrium concentrations of [H 2 ] and [I 2 ] and the equilibrium constant. 23

Equilibrium Calculations ICE Model Problem 1 -- Solution Hydrogen and iodine are in equilibrium with Hydrogen iodide to this reaction: H 2 + I 2   2HI Suppose that 1.5 mole of H 2 and 1.2 mole of I 2 are placed in a 1.0 dm 3 container. At equilibrium it was found that there were 0.4 mole of HI. Calculate the equilibrium concentrations of [H 2 ] and [I 2 ] and the equilibrium constant. 24 I C E Since 2x = 0.4, x = 0.2 I C E Since 2x = 0.4, x = 0.2 [H 2 ] x 1.5- x [H2 ] = 1.5 – 0.2 = 1.3 [ I 2 ] x 1.2 –x [I2 ] = 1.2 – 0.2 = 1.0 [HI ] 0 +2x 0.4 Keq = [HI] 2 = (0.4) 2 = [H 2 ] [ I 2 ] (1.3) (1.0) [H 2 ] [ I 2 ] (1.3) (1.0)

Equilibrium Calculations ICE Model Problem 2 Sulfur dioxide reacts with oxygen to produce sulfur trioxide according to this reaction: 2 SO 2 + O 2   2SO 3 Suppose that 1.4 mole of SO 2 and 0.8 mole of O 2 are placed in a 1.0 dm 3 container. At equilibrium it was found that there were 0.6 mole of SO 3. Calculate the equilibrium concentrations of [SO 2 ] and [O 2 ] and the equilibrium constant. 25

Equilibrium Calculations ICE Model Problem 2 - Solution Sulfur dioxide reacts with oxygen to produce sulfur trioxide according to this reaction: 2 SO 2 + O 2   2SO 3 Suppose that 1.4 mole of SO 2 and 0.8 mole of O 2 are placed in a 1.0 dm 3 container. At equilibrium it was found that there were 0.6 mole of SO 3. Calculate the equilibrium concentrations of [SO 2 ] and [O 2 ] and the equilibrium constant. ICE Since 2x = 0.6, x = 0.3 [SO 2 ] x 1.4-2x [SO 2 ] =1.4 – 2( 0.3) = 0.8 [O 2 ] 0.8 -x 0.8 –x [O 2 ] = 0.8 – 0.3 = 0.5 [SO 3 ] 0 +2x 0.6 Keq = [SO 3 ] 2 = (0.6) 2 = [SO 2 ] 2 [O 2 ] (0.8) 2 (0.5) 26

Calculating Equilibrium Concentrations from K eq K eq = [C 2 H 4 ] [H 2 ] [C2H6][C2H6] 27 What is the concentration for each substance at equilibrium for the following gaseous reaction C 2 H 6   C 2 H 4 + H 2 K C = 1.01 if the initial concentration of ethene, C 2 H 4 and that of hydrogen are both M? Initial Change Equilibrium C2H4C2H4C2H4C2H x x H2H2H2H x x C2H6C2H6C2H6C2H60 + x x 1.01 = [0.300-x][0.300-x] [x][x] 1.01x = x +x 2 x x = 0 x = (1.61) 2 -4(1)(0.09) 2(1) x = and The second root is extraneous. C 2 H 6 so [C 2 H 6 ] = C 2 H 4 and [H 2 ] = [C 2 H 4 ] = 0.242

2NO 2 (g) N 2 O 4 (g) Four experiments were performed. The initial concentrations of the two chemicals were different in each experiment and the concentration of each gas was measured once the system reached equilibrium. At Equilibrium Exp #[NO 2 ][N 2 O 4 ]

Calculate and Describe Using the data on the previous page, calculate the K for each experiment. What did you find? At Equilibrium Exp #[NO 2 ][N 2 O 4 ]

Starter CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g) 1. Write an expression for K eq 2. Calculate K at a given temperature if [CH 4 ] = M, [O 2 ] = M, [CO 2 ] = M, and [H 2 O] = M at equilibrium. (include units)

Objectives Apply Le Chatelier’s principle to predict qualitative effects of changes of temperature, pressure and concentration on the position of equilibrium and on the value of the equilibrium constant.

Le Chatelier’s Principle  When a system at equilibrium is placed under stress, the system will undergo a change in such a way as to relieve that stress. Henri-Louis Le Chatelier ( ) French industrial chemist

Le Chatelier Translated:  When you take something away from a system at equilibrium, the system shifts in such a way as to replace what you’ve taken away.  When you add something to a system at equilibrium, the system shifts in such a way as to use up what you’ve added.

Le Chatelier Example #1 A closed container of ice and water at equilibrium. The temperature is raised. Ice + Energy Water The equilibrium of the system shifts to the _______ to use up the added energy. right

Le Chatelier Example #2 A closed container of N 2 O 4 and NO 2 at equilibrium. NO 2 is added to the container. N 2 O 4 (g) + Energy 2 NO 2 (g) The equilibrium of the system shifts to the _______ to use up the added NO 2. left

Le Chatelier Example #3 A closed container of water and its vapor at equilibrium. Vapor is removed from the system. water + Energy  vapor The equilibrium of the system shifts to the _______ to replace the vapor. right

Le Chatelier Example #4 A closed container of N 2 O 4 and NO 2 at equilibrium. The pressure is increased. N 2 O 4 (g) + Energy 2 NO 2 (g) The equilibrium of the system shifts to the _______ to lower the pressure, because there are fewer moles of gas on that side of the equation. left

Pressure Changes to system If the volume decreases, the concentration increases, and there will be a shift to the side with the less amount of moles. If the volume increases, the concentration decreases, and there will be a shift to the side with the more amount of moles.

7.2.3Apply Le Chatelier’s principle to predict the qualitative effects of changes of temperature, pressure and concentration on the position of equilibrium and on the value of the equilibrium constant. N 2 (g) + 3H 2 (g) 2NH 3 (g) ΔH o = -92 kJ What effect does increasing the pressure have on the equilibrium? 39 Reaction will shift toward the side with the fewest gas molecules. The left side has 4 gas molecules (1N 2 & 3H 2 ). The right side has 2 gas molecules. Reaction will shift to the right. N 2 & H 2 will react and more NH 3 will be produced. Kc does not change

7.2.3Apply Le Chatelier’s principle to predict the qualitative effects of changes of temperature, pressure and concentration on the position of equilibrium and on the value of the equilibrium constant. N 2 (g) + 3H 2 (g) 2NH 3 (g) ΔH o = -92 kJ What effect does decreasing the pressure have on the equilibrium? 40 If a reaction has equal numbers of gas molecules on the left and on the right, changing the pressure has no effect. 2HI(g) H 2 (g) + I 2 (g)

Example; If I increase the pressure, where is the shift? (right) If I decrease the pressure, where is the shift? (left) 2SO 2 + O 2 2SO 3 (3moles)(2moles)

Effect of Concentration 1. If you add more reactant, it shifts to the right increasing the formation of product, using up the reactants. 2. If you add product, it shifts to the left 3. If you remove product, it shifts to the right, increasing the formation of product. 4. If you remove reactant, it shifts to the left

Le Chatelier’s Principle – The Concentration Effect 43 N 2 (g) +3 H 2 (g)   2NH 3 (g) DH = - 92 kJ An increase in the concentration of N 2 results in a decrease H 2 and an increase in NH 3 in such a way to keep the equilibrium constant the same K eq = K eq = [] 2 [ NH 3 ] 2 [N 2 ] [H 2 ] 3

Le Chatelier’s Principle – The Concentration Effect 44 N 2 (g) +3 H 2 (g)   2NH 3 (g) DH = - 92 kJ Likewise an increase in the concentration of H 2 results in a decrease in N 2 and an increase in NH 3 in such a way to keep the equilibrium constant the same K eq = K eq = [] 2 [ NH 3 ] 2 [N 2 ] [H 2 ] 3

Le Chatelier’s Principle – The Concentration Effect 45 N 2 (g) +3 H 2 (g)   2NH 3 (g)  H = - 92 kJ An increase in the concentration of NH 3 results in a increase in N 2 and an increase in H 2 in such a way to keep the equilibrium constant the same. K eq = K eq = [] 2 [ NH 3 ] 2 [N 2 ] [H 2 ] 3

Effect of temperature Energy is treated as a reactant if endothermic equation, and as a product if exothermic equation. If cooling a system, then it shifts so more heat is produced. If heating a system, then it shifts so extra heat is used up.

Example for temp. changes for Endothermic Reaction Heating the below reaction causes the system to shift to the right = more products, because you treat energy like a reactant. 2NaCl +H 2 SO 4 + energy 2HCl + Na 2 SO 4 Cooling the above reaction causes the system to shift to the left = less reactants, so need to make up more

7.2.3Apply Le Chatelier’s principle to predict the qualitative effects of changes of temperature, pressure and concentration on the position of equilibrium and on the value of the equilibrium constant. N 2 (g) + 3H 2 (g) 2NH 3 (g) ΔH o = -92 kJ Effects of Temperature Change Increasing temperature causes the equilibrium position to shift in the direction that absorbs heat (endothermic). Decreasing temperature causes the equilibrium position to shift in the direction that produces heat (exothermic). The value of Kc will change with a change in temp. If the reaction shifts right, the value of Kc increases. If the reaction shifts left, the Kc value decreases. 48

7.2.3Apply Le Chatelier’s principle to predict the qualitative effects of changes of temperature, pressure and concentration on the position of equilibrium and on the value of the equilibrium constant. N 2 (g) + 3H 2 (g) 2NH 3 (g) ΔH o = -92 kJ What effect does increasing the temperature have on the equilibrium? 49 Increasing the temperature causes the reaction to shift to use up some of the added heat (endothermic rx). The reaction as written is exothermic so the endothermic rx is from right to left. The rx will shift left. [N 2 ] increases, [H 2 ] increases, [NH 3 ] decreases. Kc value will decrease

Effect of temp change on exothermic reactions Heating the below reaction causes the system to shift to the left, to use up the extra heat. 2SO 2 + O 2 2SO 3 + energy Cooling the above reaction causes the system to shift to the right, to make up for the lost heat.

7.2.3Apply Le Chatelier’s principle to predict the qualitative effects of changes of temperature, pressure and concentration on the position of equilibrium and on the value of the equilibrium constant. N 2 (g) + 3H 2 (g) 2NH 3 (g) ΔH o = -92 kJ What effect does decreasing the temperature have on the equilibrium? 51

7.2.3Apply Le Chatelier’s principle to predict the qualitative effects of changes of temperature, pressure and concentration on the position of equilibrium and on the value of the equilibrium constant. N 2 O 4 (g) 2NO 2 (g) ΔH o = 58.0 kJ Both gases are present in a flask at equilibrium. N 2 O 4 is a colorless gas while NO 2 is brown. 1) What color will the contents of the flask be if the pressure is increased? Explain. 2) State and explain three (3) ways the amount of NO 2 production can be increased. 52

Assessment Statements State and explain the effect of a catalyst on an equilibrium reaction.

The effect of a catalyst on equilibrium Adding a catalyst speeds up a reaction by providing an alternative mechanism with a lower activation energy, thus speeding up both the forward and backward reaction rate. It shortens the time needed to attain equilibrium concentrations It has no effect on the position of equilibrium, however equilibrium will be attained more quickly.

7.2.4State and explain the effect of a catalyst on an equilibrium reaction. 1) A catalyst lowers the activation energy barrier for both the forward and the reverse reactions. 2) Therefore a catalyst increase the rates of both reactions by the same factor. 3) A catalyst increases the rate at which equilibrium is achieved, but does not change the final composition of the substances. 4) The Kc value is not affected by the presence of a catalyst. 55

Heterogeneous Equilibria In a heterogeneous equilibrium, the components are in two different phases. A common form of a heterogeneous equilibrium is the equilibrium which exists between a solid and an aqueous (water) solution. When a substance dissolves in water there is an equilibrium established between the solid an its dissolved ions Example: AgCl (s)   Ag + (aq) + Cl - (aq ) 56

Heterogeneous Equilibria AgCl (s)   Ag + (aq) + Cl - (aq) In this example the concentration of the solid phase is essentially 1. The equilibrium constant then takes the form K = [Ag + ][Cl - ]. This form is known as a solubility product and is usually designated as K sp 57

Different designations of equilibrium constants K eq General designation for an equilibrium constant K c Equilibrium constant based on concentration K p Equilibrium constant based on pressure (gases) K sp Solubility product K a Acid equilibrium constant K b Base equilibrium constant K w Ion product of water 58

Assessment Statements Apply the concepts of kinetics and equilibrium to industrial process Suitable examples include the Haber and Contact processes

7.2.5Apply the concepts of kinetics and equilibrium to industrial processes. N 2 (g) + 3H 2 (g) 2NH 3 (g) ΔH o = -92 kJ The Haber process to manufacture ammonia Ammonia is an important starting point for the production of fertilizers, nitric acid, explosives and polymers (nylon). Under what conditions can will an industrial chemist run this reaction to increase the yield of ammonia? An optimum temperature must be found 60

Haber Process N 2 (g) + 3H 2 (g) 2NH 3 (g) ΔH= -92 kJ/mol Mixture’s volume is compressed and passed over a heated iron catalyst. Conditions for his equilibrium is critical. High pressure is favourable due to 4 moles on left and 2 moles on right. Increased pressure causes a shift to the left, favouring product formation. This is expensive to due and most production plants will resist compressing gases in terms of operating costs. Compromise will be met.

Compromise This is an exothermic reaction, so low temperatures would be favourable to produce product. Low temps mean low reaction rates, so we may get a higher yield but it will take a long time to get it. Not good for business. A compromise temp, as well as the use of a catalyst will aid in speeding up the reaction to a more acceptable standard.

Typical conditions Pressure between MPa ( atm) Temperatures around 700 K The reaction is not allowed to reach equilibrium, because reaction rate decreases as we approach equilibrium, and typically only 20% of N 2 and H 2 is converted. The gases are cooled and NH 3 is condensed and removed, leaving unused N 2 and H 2 available for further production.

action?quick=128# Animation of Haber process

Ammonia’s Uses Manufacture of fertilizers (ammonia salts and urea) Manufacturing nitrogen used in polymers for the fabrication of nylon Used in the production of explosives (TNT, dynamite)

Contact Process Production of sulfuric acid by the oxidation of sulfur. 1. Sulfur is burnt in air to form sulfur dioxide S(s) + O 2 (g) SO 2 (g) 2. Sulfur dioxide is mixed with air and passed over vanadium(V)oxide catalyst to produce sulfur trioxide. 2SO 2 (g) + O 2 (g) 2SO 3 (g) ΔH= -196 kJ/mol 3. Sulfur trioxide is reacted with water to produce sulfuric acid. SO 3 (g) + H 2 O(l)  H 2 SO 4 (l)

7.2.5Apply the concepts of kinetics and equilibrium to industrial processes. 2SO 2 (g) + O 2 (g) 2SO 3 (g) ΔH o = -192 kJ The contact process to manufacture sulfuric acid SO 3 (g) + H 2 O(l) H 2 SO 4 (l) Sulfuric acid is used in many chemical processes Under what conditions can an industrial chemist run this reaction to increase the yield of sulfur trioxide? An optimum temperature must be found. 68

More money, more SO 3 … High pressure would favour the formation of SO 3 in the 2 nd step, however its too expensive. Reactants are compressed to 2 atm to achieve the desired flow rate in the reactor. Pure O 2 would drive the equilibrium to the right, however its an unnecessary expense. Low temperatures, because its exothermic, would be best, but it slows the rate too much.

Compromised conditions Temp between K (fast initial reaction rate) The use of a finely divided V 2 O 5 catalyst Oxidation is done in converters at lower temperatures (slows reaction rate) Overall conversion is 90% to SO 3

uick=12b Contact process animation

Uses of H 2 SO 4 Fertilizers (converting insoluble phosphate rock into soluble phosphates) Polymers Detergents Paints Pigments Petrochemical industry Processing of metals Electrolyte in car batteries

Le Chatelier’s principle is a memory aid, it doesn’t explain why these changes occur. Listen carefully and read over text pages to help you develop further understanding of explanation. alchemistry/flash/lechv17.swf alchemistry/flash/lechv17.swf Haber process notes aber.html aber.html