 # Equilibrium L. Scheffler Lincoln High School 2009 1.

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Equilibrium L. Scheffler Lincoln High School 2009 1

Equilibrium Systems Many chemical reactions are reversible. Many chemical reactions are reversible. Such reactions do not go to completion. Such reactions do not go to completion. There is a state of balance between the products and the reactants There is a state of balance between the products and the reactants When the concentration of neither the reactants nor the products is changing, the system is in equilibrium. When the concentration of neither the reactants nor the products is changing, the system is in equilibrium. 2

Chemical equilibrium occurs in chemical reactions that are reversible. In a reaction such as: Chemical equilibrium occurs in chemical reactions that are reversible. In a reaction such as: CH 4 (g) + H 2 O(g)   CO(g) + 3H 2 (g) The reaction can proceed in both directions The reaction can proceed in both directions CO(g) + 3H 2 (g)   CH 4 (g) + H 2 O(g) Chemical Equilibrium 3

Equilibrium Conditions At equilibrium the rate of reaction in the forward direction and the rate in the reverse direction are equal At equilibrium the rate of reaction in the forward direction and the rate in the reverse direction are equal 4

An Equilibrium System CH 4 (g) + H 2 O(g)   CO(g) + 3H 2 (g)  After some of the products are created, the products begin to react to form the reactants.  At equilibrium there is no net change in the concentrations of the reactants and products.  The concentrations do not change but they are not necessarily equal. 5

At equilibrium two opposing processes are taking place at equal rates. At equilibrium two opposing processes are taking place at equal rates. In other words the rate in the forward direction = the rate in the reverse direction In other words the rate in the forward direction = the rate in the reverse direction Examples Examples H 2 O (l)   H 2 O (g) NaCl (s)   NaCl (aq) H2OH2O CO (g) + 2 H 2 (g)   CH 3 OH (g) 6 Dynamic Equilibrium

Equilibrium Conditions 7 H 2 O + CO   H 2 + CO 2

Law of Mass Action Given the reaction Given the reaction aA + bB   cC + dD aA + bB   cC + dD The rate in the forward direction is rate forward = k f [A] a [B] b rate forward = k f [A] a [B] b The rate in the reverse direction is rate reverse = k r [C] c [D] d rate reverse = k r [C] c [D] d 8

Law of Mass Action At equilibrium these rates are equal At equilibrium these rates are equal rate forward = rate reverse k f [A] a [B] b = k r [C] c [D] d The ratio of the rate constants is The ratio of the rate constants is Keq= [C] c [D] d [A] a [B] b 9

Writing Equilibrium Expressions Writing Equilibrium Expressions N 2 (g) + 3 H 2 (g)   2NH 3 (g) N 2 (g) + 3 H 2 (g)   2NH 3 (g) 2 SO 2 (g) + O 2 (g)   2SO 3 (g) 2 SO 2 (g) + O 2 (g)   2SO 3 (g) H 2 (g) + Br 2 (g)   2 HBr (g) H 2 (g) + Br 2 (g)   2 HBr (g) 2N 2 O (g)   2 N 2 (g) + O 2 (g) 2N 2 O (g)   2 N 2 (g) + O 2 (g) 10

Answers Check your work against the following: Check your work against the following: 4. K eq = [N 2 ] 2 [O 2 ] [N 2 O] 2 3. K eq 3. K eq= [HBr] 2 [H 2 ] [Br 2 ] 2. K eq 2. K eq= [SO 3 ] 2 [SO 2 ] 2 [O 2 ] 1. K eq 1. K eq= [NH 3 ] 2 [N 2 ] [H 2 ] 3 11

Reaction Quotient The equilibrium constant is a constant ratio only when the system is in equilibrium. The equilibrium constant is a constant ratio only when the system is in equilibrium. If the system it not at equilibrium the ratio is known as a Reaction Quotient If the system it not at equilibrium the ratio is known as a Reaction Quotient If the reaction quotient is equal to the equilibrium constant then the system is at equilibrium If the reaction quotient is equal to the equilibrium constant then the system is at equilibrium 12

K>>1: The reaction is product-favored; equilibrium concentrations of products are greater than equilibrium concentrations of reactants. K>>1: The reaction is product-favored; equilibrium concentrations of products are greater than equilibrium concentrations of reactants. K<<1: The reaction is reactant-favored; equilibrium concentrations of reactants are greater than equilibrium concentrations of products. K<<1: The reaction is reactant-favored; equilibrium concentrations of reactants are greater than equilibrium concentrations of products. The Meaning of the Equilibrium Constant 13

Calculating Equilibrium Constants Nitrogen dioxide decomposes at high temperatures according to this equation: 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 ] = 0.080 M; calculate the equilibrium constant. If the equilibrium concentrations are as follows: [NO 2 ]= 1.20 M, [NO] = 0.160, and [O 2 ] = 0.080 M; calculate the equilibrium constant. K eq = [NO] 2 [O 2 ] [NO 2 ] 2 So K eq = (0.160) 2 (0.080) = 0.00142 (1.20) 2 14

Calculating Equilibrium Constants The equilibrium equation for the oxidation of sulfur dioxide is as follows: 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 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 ] 15

Practice Problem 1 The equilibrium equation for the carbon monoxide with steam to produce hydrogen gas is as follows: 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 ] = 0.075 M and [H 2 ] = 0.060 M, calculate the equilibrium constant. K eq = K eq = [CO] [H 2 O] [CO 2 ] [H 2 ] So K eq = (1.00) (0.025) = 0.139 (0.075) (0.060) 16

Calculating Equilibrium Concentrations 17

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 Equilibrium constants and concentrations can often be deduced by carefully examining data about the initial and equilibrium concentrations 18 I nitial C hange E quilbrium

Equilibrium Calculations ICE Model problem 1 Hydrogen and iodine are in equilibrium with Hydrogen iodide to this reaction: Hydrogen and iodine are in equilibrium with Hydrogen iodide to this reaction: H 2 + I 2 2HI 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. 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. 19

Equilibrium Calculations ICE Model Problem 1 -- Solution Hydrogen and iodine are in equilibrium with Hydrogen iodide to this reaction: H 2 + I 2 2HI 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. 20 I C E Since 2x = 0.4, x = 0.2 I C E Since 2x = 0.4, x = 0.2 [H 2 ] 1.5 - x 1.5- x [H2 ] = 1.5 – 0.2 = 1.3 [ I 2 ] 1.2 - x 1.2 –x [I2 ] = 1.2 – 0.2 = 1.0 [HI ] 0 +2x 0.4 Keq = [HI] 2 = (0.4) 2 = 0.123 [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: Sulfur dioxide reacts with oxygen to produce sulfur trioxide according to this reaction: 2 SO 2 + O 2 2SO 3 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 dm 3 of SO 3. Calculate the equilibrium concentrations of [SO 2 ] and [O 2 ] and the equilibrium constant. 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 dm 3 of SO 3. Calculate the equilibrium concentrations of [SO 2 ] and [O 2 ] and the equilibrium constant. 21

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 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 dm 3 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 ] 1.4 -2x 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 = 0.281 Keq = [SO 3 ] 2 = (0.6) 2 = 0.281 [SO 2 ] 2 [O 2 ] (0.8) 2 (0.5) [SO 2 ] 2 [O 2 ] (0.8) 2 (0.5) 22

Calculating Equilibrium Concentrations from K eq K eq = [] [H 2 ] [C 2 H 4 ] [H 2 ] [][C2H6][][C2H6] 23 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 0.300 M? Initial Change Equilibrium C2H4C2H4C2H4C2H40.300 - x 0.300-x H2H2H2H20.300 0.300-x C2H6C2H6C2H6C2H60 + x x 1.01 = [0.300-x][0.300-x] [][x][][x] 1.01x = 0.09 - 0.6x +x 2 x 2 - 1.61x +0.09 = 0 x = 1.61+ (1.61) 2 -4(1)(0.09) 2(1) x = 0.0580 and 1.552. The second root is extraneous. C 2 H 6 so [C 2 H 6 ] = 0.0580 C 2 H 4 and [H 2 ] = [C 2 H 4 ] = 0.242

Le Chatelier’s Principle –Le Chatelier's Principle states: When a system in chemical equilibrium is disturbed by a change of temperature, pressure, or a concentration, the system shifts in equilibrium composition in a way that tends to counteract this change of variable. –A change imposed on an equilibrium system is called a stress –The equilibrium always responds in such a way so as to counteract the stress 24

Le Chatelier’s Principle 25 A stress is any sudden change in conditions that drives the system out of equilibrium. When a stress is placed on an equilibrium system, the system will shift in such a way so as to lessen or mitigate the stress and restore equilibrium. A stress usually involves a change in the temperature, pressure, or in the concentration of one or more of the substances that are in equilibrium. Le Chatelier's principle predicts the direction of the change in the equilibrium.

Applications of Le Chatelier’s Principle 26 N 2 (g) +3 H 2 (g)   2NH 3 (g)  H = -92 kJ Haber’s process for the production of ammonia is an example of an industrial equilibrium system. We will use this equilibrium as a model to explain the how Le Chatelier’s principle operates with the following stresses: Change in the concentration of one of the components Changes in pressure Changes in temperature Use of a catalyst

Applications of Le Chatelier’s Principle 27 N 2 (g) +3 H 2 (g)   2NH 3 (g)  H = - 92 kJ The equilibrium constant for this reaction is Any change in the concentration (or partial pressure) of any component will cause the equilibrium to shift in such a way so as to return the equilibrium constant to its original value. K eq = K eq = [] 2 [ NH 3 ] 2 [N 2 ] [H 2 ] 3

Le Chatelier’s Principle – The Concentration Effect Le Chatelier’s Principle – The Concentration Effect 28 N 2 (g) +3 H 2 (g)   2NH 3 (g)  H = - 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 29 N 2 (g) +3 H 2 (g)   2NH 3 (g)  H = - 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 30 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

Le Chatelier’s Principle – The Temperature Effect 31 N 2 (g) +3 H 2 (g)   2NH 3 (g)  H = - 92 kJ  The reaction is exothermic in the forward direction.  An increase in the temperature would trigger a response in the heat consuming (endothermic) direction.  An increase in temperature therefore causes the reaction to shift in the reverse direction. Some NH 3 decomposes to N 2 and H 2.

Le Chatelier’s Principle – The Pressure Effect 32 N 2 (g) +3 H 2 (g)   2NH 3 (g)  H = - 92 kJ  All molecules in the equilibrium are gases  When the reaction proceeds in the forward direction the number of moles are reduced from 4 to 2.  Pressure is proportional to the number of moles of gas  An increase in pressure therefore causes the reaction to moved in the forward direction. Some N 2 and H 2 combine to form more NH 3

Le Chatelier’s Principle – The Effect of Catalysts 33 N 2 (g) +3 H 2 (g)   2NH 3 (g)  H = - 92 kJ  Catalysts lower the activation energy  A catalyst affects the forward and the reverse direction equally  There is no change in the equilibrium position from a catalyst  A catalyst decreases the time required for the system to achieve equilibrium

Heterogeneous Equilibria In a heterogeneous equilibrium, the components are in two different phases. 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 solution A common form of a heterogeneous equilibrium is the equilibrium which exists between a solid and an aqueous solution When a substance dissolves in water there is an equilibrium established between the solid an its dissolved ions When a substance dissolves in water there is an equilibrium established between the solid an its dissolved ions Example Example AgCl (s)   Ag + (aq) + Cl - (aq) AgCl (s)   Ag + (aq) + Cl - (aq) 34

Heterogeneous Equilibria AgCl (s)   Ag + (aq) + Cl - (aq) 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 In this example the concentration of the solid phase is essentially 1. The equilibrium constant then takes the form K = [Ag + ][Cl - ]. K = [Ag + ][Cl - ]. This form is known as a solubility product and is usually designated as K sp This form is known as a solubility product and is usually designated as K sp 35

Different designations of equilibrium constants K eq General designation for an equilibrium constant K eq General designation for an equilibrium constant Kc Equilibrium constant based on concentration Kc Equilibrium constant based on concentration Kp Equilibrium constant based on pressure (gases) Kp Equilibrium constant based on pressure (gases) Ksp Solubility product Ksp Solubility product Ka Acid equilibrium constant Ka Acid equilibrium constant Kb Base equilibrium constant Kb Base equilibrium constant Kw Ion product of water Kw Ion product of water 36