Equilibrium Chapter 15

A little Bit of History….

The Need for NH3 Plants need nitrogen fixation to grow the enzyme nitrogenase provides plants with fixed nitrogen through reaction N2(g) + 3H2(g) 2NH3(g) As demand for food became more, a way of producing NH3 was needed, but difficult, because to break the triple bond in N2, 550oC and 200atm pressure was needed

The Good and Bad of the Haber Process WWI – Germans also needed NH3 for explosives German Chemist Fritz Haber (1912) developed a way to synthesize NH3 from hydrogen and nitrogen gas. It prolonged WWI, but led to the development of nitrogen fixing fertilizers which helped increased world food supply

Haber process N2(g) + 3H2(g)↔2NH3(g)

Reversible Reactions Some reactions Most reactions aA + bB → cC + dD and cC + dD → aA + bB This can be written aA + bB ↔ cC + dD

Reversible reactions and equilibrium aA + bB ↔ cC + dD This reaction will continue in both directions until a CHEMICAL EQUILIBRIUM is reached this means that each of the opposite reactions aA + bB → cC + dD (forward rxn) cC + dD → aA + bB (reverse rxn) are occurring at the same time, at THE SAME RATE

Equilibrium When a system (chemical reaction) reaches equilibrium, the reaction does not stop – but because opposite reactions are happening at the same rate, the amount (concentration) of reactants and products at any given time is CONSTANT but NOT NECESSARILY EQUAL Neither reactants nor products can escape the system or equilibrium is disturbed Sometimes referred to as dynamic equilibrium If the concentrations of reactants and products are constant, the ratio of their concentrations also equals a constant

Equilibrium N2O4 (g) ↔ 2 NO2 (g)

Physical Equilibrium -occurs at phase changes Solid ↔ Liquid

Solution Equilibrium Occurs at saturation point

There are no units for Keq, or Kc. Types of Equilibrium Constants- relates to the composition of the mixture at equilibrium Keq = Kc = Kp = Equilibrium (basic equation) Concentrations (if molarity given) Pressures ( if pressures given) There are no units for Keq, or Kc. Kp is in atm

equilibrium aA + bB ↔ cC + dD For the reaction above, at any point in the reaction we can express the ratio of products to reactants (products over reactants) as a reactant quotient (Q) 𝑄= [𝐶] 𝑐 [𝐷] 𝑑 [𝐴] 𝑎 [𝐵] 𝑏

equilibrium expressions AT EQUILIBRIUM, the reaction quotient (Q) becomes the EQUILIBRIUM CONSTANT (Kc if concentration or Kp if pressure) - no units Concentrations of solids or pure liquids (not in solution) that are in the equilibrium expression (below) are assumed to be 1 since their concentrations do not change Dependent only on the stoichiometry of the reaction, not the mechanism 𝐾= [𝐶] 𝑐 [𝐷] 𝑑 [𝐴] 𝑎 [𝐵] 𝑏

Equilibrium expressions Example : Haber process N2(g) + 3 H2(g) ↔ 2 NH3(g) Equilibrium expression if concentration is molarity: Kc = [𝑁𝐻 3 ] 2 [ 𝑁 2 ] [𝐻 2 ] 3 Equilibrium expression if concentration is in partial pressure Kp = (𝑃 𝑁𝐻 3 ) 2 𝑃 𝑁 2 ∙ (𝑃 𝐻 2 ) 3

Example 1: Write the equilibrium expression, Kc, for the following reactions a. 2 O3 (g) ↔ 3 O2 (g) b. 2 NO (g) + Cl2 ↔ 2 NOCl (g) c. Ag + (aq) + 2 NH3 (aq) ↔ Ag(NH3)2 + (aq)

Evaluating (solving for) Kp There is a relationship between Kc and Kp remember the ideal gas law ? PV = nRT P = nRT → P = n (RT) → P = [A](RT) V V derivation…. 𝐾 𝑝 = 𝐾 𝑐 (𝑅𝑇) Δ𝑛

Example 2: N2(g) + 3 H2(g) ↔ 2 NH3(g) Kc = 9.60 at 300°C. Calculate Kp for this reaction at this temperature Answer: 4.34 x 10-3

Equilibrium Constants Magnitude (size) of equilibrium constants : If K >> 1 then equilibrium is product favored (lies to the right) If K << 1 then equilibrium is reactant favored (lies to the left) Ex: CO (g) + Cl2(g) → COCl2 (g) What is the equilibrium expression (Kc ) for this reaction? Kc = [COCl2] [CO] [Cl2] Kc = 4.56 x 109 Is the reaction product or reactant favored ?

Or, stated one more way… If Keq has a positive exponent then its mostly products at equilibrium If a negative exponent, then it’s mostly reactants

Example 3: H2 (g) + I2 (g) → 2 HI(g) At 298 K , Kp = 794 At 700 K , Kp = 55 Is the formation of HI favored more at the higher or lower temperature? Answer : lower

Direction of a reaction and k The value of the equilibrium constant is dependent on how you write the chemical reaction For example, The equilibrium-constant expression for a reaction written in one direction is the reciprocal of the expression for the reaction written in the reverse direction

Example 4: For the given reaction N2O4 (g) ↔ 2 NO2 (g) Kc = [NO2]2 = 0.212 at 100°C [N2O4 ] a. What is the equilibrium expression for the reverse reaction? Kc = [N2O4 ] [NO2]2 b. What is the value of the equilibrium constant for the reverse reaction? Kc = 4.72

Example 5: Kc = [𝑁𝐻 3 ] 2 [ 𝑁 2 ] [𝐻 2 ] 3 = 9.60 For the following reaction at 300 K, N2(g) + 3 H2(g) ↔ 2 NH3(g) Kc = [𝑁𝐻 3 ] 2 [ 𝑁 2 ] [𝐻 2 ] 3 = 9.60 What would the value of the equilibrium constant be for the following reaction? 2 N2(g) + 6 H2(g) ↔ 4 NH3(g)

Equilibrium constant We can calculate equilibrium constant for a reaction if we know the equilibrium constants for the other reactions that add up to give us the target equation For example :

Equilibrium constant cont. Reaction 1 : 2 NOBr(g) ↔ 2 NO(g) + Br2(g) Kc (1) = [NO]2 [Br2] = 0.014 [NOBr] Reaction 2 : Br2(g) + Cl2(g) ↔ 2 BrCl (g) Kc (2) = [BrCl]2 = 7.2 [Br2] [Cl2] The sum of these equations Reaction 3: 2 NOBr(g) + Cl2(g) ↔ 2 NO(g) + 2 BrCl (g) If you write the equilibrium expression for this reaction, it is equal to the two individual equilibrium expressions multiplied by each other, so we can multiply the individual equilibrium constants Kc (3) = [NO]2 [BrCl]2 = Kc (1) x Kc (2) = Kc (3) = 0.10 [NOBr] [Cl2]

Example 6: Given the following reactions, HF (aq) ↔ H + (aq) + F - (aq) Kc = 6.8 x 10 -4 H2C2O4 (aq) ↔ H + (aq) + C2O4 -2 (aq) Kc = 3.8 x 10 -6 determine the value of Kc for the reaction 2 HF (aq) + C2O4 -2 (aq) ↔ 2 F - (aq) + H2C2O4 (aq) Answer = 0.12

Equilibrium constant manipulations summary 1. the equilibrium constant of a reaction in the revers direction is the inverse (reciprocal) of the equilibrium constant of the reaction in the forward direction 2. if the coefficients of a reaction have been multiplied by a number, the new equilibrium constant is equal to the original equilibrium constant raised to a power equal to that number 3. The equilibrium constant for a net reaction made up of two or more reactions is the product of the individual equilibrium constants

Heterogeneous equilibria If substances in a chemical reaction are all in the same phase at equilibrium (usually a gas or liquid) – homogeneous equilibria If substances in different phases – heterogeneous equilibria Substances that have constant composition are not included in equilibrium expression - Solids and pure liquids are not included Only substances that can be expressed in terms of concentration (molarity) or pressure can be included in the equilibrium expression If a solvent is a reactant or product it is not written in in the equilibrium expression As long as the concentration of the reactants and products are low, the solvent is considered a pure liquid

example PbCl2 (s) ↔ Pb 2+ (aq) + Cl – (aq) Kc = [Pb 2+] [Cl – ]

Example 7: Write the equilibrium expression, Kc, for the following reactions a. CO2 (g) + H2 (g) ↔ CO (g) + H2O (l) b. SnO2 (s) + 2 CO (g) ↔ Sn (s) + 2 CO2 (g)

Calculating equilibrium constants when all concentrations are known, Example 8: An aqueous solution of acetic acid is found to have the following equilibrium concentrations at 25°C: [CH3COOH] = 1.65 x 10-2 M [H+] = 5.44 x 10-4 M [CH3COO -] = 5.44 x 10-4 M Calculate the equilibrium constant, Kc, for the ionization of acetic acid according to the following reaction CH3COOH (aq) ↔ H+ (aq) + CH3COO – (aq) Answer : 1.79 x 10-5

Calculating equilibrium constant when not all concentrations are known Process Make a table of all known initial and equilibrium concentrations Calculate the change in the known concentrations as the system reaches equilibrium using the coefficients of the balanced equation Use the initial concentrations and changes in concentrations to determine the equilibrium concentrations Plug in concentrations into equilibrium expressions and solve for the equilibrium constant This is also known as making an ICE chart

Change in [HI] = 1.87  103 M  0 = 1.87  103 M A closed system initially containing 1.000  103 M H2 and 2.000  103 M I2 at 448 C is allowed to reach equilibrium, and at equilibrium the HI concentration is 1.87  103 M. Calculate Kc at 448 C for the reaction taking place, which is the initial and equilibrium concentrations of as many species as we can. We also provide space in our table for listing the changes in concentrations. use the chemical equation as the heading for the table. calculate the change in HI concentration, which is the difference Between the equilibrium and initial values: Change in [HI] = 1.87  103 M  0 = 1.87  103 M 35

Notice that the entries for the changes are negative when a reactant Third, we use the coefficients in the balanced equation to relate the change in [HI] to the changes in [H2] and [I2]: Fourth, we calculate the equilibrium concentrations of H2 and I2, using initial concentrations and changes in concentration. The equilibrium concentration equals the initial concentration minus that consumed: [H2] = 1.000  103 M  0.935  103 M = 0.065  103 M [I2] = 2.000  103 M  0.935  103 M = 1.065  103 M Notice that the entries for the changes are negative when a reactant is consumed and positive when a product is formed. 36

Finally, we use the equilibrium-constant expression to calculate the equilibrium constant: 37

Example 10: Calculating concentrations from initial concentrations H2 + I2 ↔ 2HI Given: Initial concentration H2 = .2 M Initial concentation I2 = .2 M Keq = 64 What are the equilibrium constants for each substance? (in other words, what is their molarity at equilibrium)

Step 1: Write Keq expression for reaction Keq = [HI]2 [H2] [ I2]

Chapter 15, Unnumbered Table, Page 629

Step 2: Ice Chart/Box I initial conc. 2HI I initial conc. C Change in conc. (use balanced eq.) E equilibrium conc.

Step 2: Ice Box H2 + I2 2HI I C -x 2x E (.2M- x) (.2M - x) (0 + 2x) Initial conc. .2M C Change in conc. -x 2x E Equilibrium conc. (.2M- x) (.2M - x) (0 + 2x) Fill in value for change in conc. Based on stoich. shown in equation (+ toward substance being produced)

Chapter 15, Unnumbered Table 1, Page 630

Plug in values into Keq equation 64 = ____[2x]2____ [.2-x] [.2-x]

How to put Ice Box Problem into calculator………

64 = (2x)2 (.2-x) (.2-x) 64 = 4x2______ 1 (.2-x ) ( .2 – x) Multiply top first 64 = 4x 2_______ 1 .04 - .4x + x2 “Foil” bottom 4x2 = 64x2 – 25.6x + 2.56 Cross multiply 0 = 60 x2 – 25.6x + 2.56 Subtract 4x2 from both sides 60 x2 – 25.6x + 2.56 = 0 “a” “b” “c” Use quadratic equation in calculator to solve for x *Program ---quad---fill in values for “a”,”b” and “c” x = .2666 or x = .16 You get 2 possible answers. Choose value that makes sense . If you started with 0.2 M and the molarity of the reactants goes down as it approaches equilibrium, then the correct value for X = .166 Plug in the value of x for the equilibrium concentration in the ICE Box

Step 3. Ans. X=.16 After solving for x, use it to determine the equilibrium concentrations So H2 =.2-.16 = .04M at equilibrium I2 = .2-.16 = .04M at equilibrium HI = 2X = 2x .16= .32M at equilibrium

How does q fit into all of this? Remember that Q is the reaction quotient, the concentrations that are used in the expression are not equilibrium concentrations 𝑄= [𝐶] 𝑐 [𝐷] 𝑑 [𝐴] 𝑎 [𝐵] 𝑏 The reaction quotient can vary throughout the reaction

How does q fit into all of this? We can use the value of Q to determine if a reaction is at equilibrium or not You can experimentally determine concentrations of reactants and products throughout a reaction and calculate Q If you know what Kc is, then you can compare it to determine if you are at equilibrium or not

How does Q fit into all of this? Q = K then the reaction is at equilibrium Q > K then the concentration of products is too large and that of reactants too small, the reaction needs to proceed to from right to left in order to reach equilibrium – reverse reaction takes over Q < K then the concentration of products is too small and the reaction needs to proceed from right to left in order to reach equilibrium – forward reaction takes over

Figure 15.8 Predicting the direction of a reaction by comparing Q and K at a given temperature.

Example 11: Predicting direction of equilibrium At 448 °C the equilibrium constant, Kc, for the reaction H2 + I2 ↔ 2HI Is 50.5. Predict which direction the reaction proceeds to reach equilibrium if we start with 2.0 x 10-2 mol of HI, 1.0 x 10-2 mol of H2 , and 3.0 mol of I2 in a 2.0 L container. Find the initial concentrations of reactants and products, then plug into Q expression

Because Qc < Kc, the concentration of HI must increase and the concentrations of H2 and I2 must decrease to Reach equilibrium; the reaction as written proceeds left to right to attain equilibrium.

Le Chatlier’s principle How can we increase the yield (amount of products) of a reaction? If we change the conditions of the reaction, can we change the yield?

Figure 15.9 Effect of temperature and pressure on NH3 yield in the Haber process.

Le Chatlier’s principle If a system at equilibrium is disturbed by a change in temperature, pressure, or a component concentration, the system will shift its equilibrium position in order to counteract the effect of the disturbance We can use this principle to predict how a system at equilibrium will respond when we add or remove a reactant or product, change the pressure by changing the volume, or changing the temperature

Chapter 15, Unnumbered Figure, Page 631

Le chatlier’s principle: concentration Shift means that the reactant or product concentration will change to accommodate the new situation THERE IS NO CHANGE IN EQUILIBRIUM CONSTANT If a chemical system in equilibrium and the concentration of any substance in the mixture is increased the system reacts to consume that substance OR if the concentration of a substance is decreased, the system will react to produce more of that substance

Le chatlier’s principle: concentration For example N2(g) + 3 H2(g) ↔ 2 NH3(g) if H2(g) is added, the reaction will shift to the right to use up the excess hydrogen If we remove NH3(g) what will happen? If we add NH3(g) what will happen? So in order to increase the yield of this reaction, we could remove NH3(g) as it is produced

Le chatlier’s principle: volume and pressure changes At constant temperature, reducing the volume of a gaseous equilibrium mixture causes the system to shift in the direction that reduces the number of moles of gas Increasing the volume causes a shift in the direction that produces more gas molecules

Le chatlier’s principle: volume and pressure changes For example N2(g) + 3 H2(g) ↔ 2 NH3(g) 4 molecules of reactants react to produce 2 moles of product If we increase the pressure, by reducing the volume, this will shift in the direction that produces fewer molecules of gases , which in this reaction is a shift to the right, formation of product How about this one? Which way will the reaction shift? H2(g) + I2(g) ↔ 2 HI (g)

Le chatlier’s principle: temperature When temperature changes, the equilibrium constant changes Treat temperature as a reactant or product Endothermic : reactants + heat ↔ products , increasing temp increases K Exothermic : reactants ↔ products + heat , decreasing temp decreases K When the temperature of a system at equilibrium is increased, the system reacts as if we added a reactant to an endothermic reaction or a product to an exothermic reaction. The equilibrium shifts in the direction that consumes the excess reactant (or product)

Le chatlier’s principle: temperature For example, if you cool an endothermic reaction the equilibrium shifts equilibrium to the left Endothermic : reactants + heat ↔ products If you cool an exothermic reaction the equilibrium will shift to the right Exothermic : reactants ↔ products + heat

Example 12: N2O4 (g) ↔ 2 NO2 (g) ΔH = 58.0 kJ In which direction will the equilibrium shift when N2O4 is added NO2 is removed Pressure is increased by addition of N2 (g) Volume is increased The temperature is decreased

A. Shift to the right to the products to decrease concentration of reactant B. Shift to the right to compensate for removal of NO2 C. Adding nitrogen gas will increase the pressure, but nitrogen is not involved in the reaction so no shift in equilibrium D. If the volume is increased, reaction shifts to the right to create more gas molecules E. The reaction is endothermic so a decrease in temperature will shift to the direction that produces heat, which will shift to the left

Catalyst effect on equilibrium Catalysts change the rate of a reaction, but not the value of K So catalysts increases the rate at which equilibrium is achieved but not the composition of the equilibrium mixture