Entry Task: Jan 4 th Friday

Entry Task: Jan 4th Friday Define chemical equilibrium. You have 5 minutes!

Agenda: • Sign off on Ch. 15 sec. 1-2 notes • Use embedded notes for reinforcement • HW: Equilibrium constant ws.

I can…

Chapter 15Chemical Equilibrium

Warning!! Black text slides contains embedded notes to be done as a class

15.1- The Concept of Equilibrium Define Chemical Equilibrium The condition in which the concentration of all reactants and products in a closed system cease to change with time is called chemical equilibrium. OR The rate at which the products are formed front the reactants equals the rate at which the reactants are formed by the products

What are the 3 conditions necessary for equilibrium? • Must have a closed system • Must have a constant temperature • Ea must be low enough to allow a reaction.

15.1- The Concept of Equilibrium Explain kr[A] = kf[B]. As the concentration of A decreases the concentration increases no matter what direction the reaction is going.

The Concept of Equilibrium Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.

15.1- The Concept of Equilibrium Explain the equilibrium mixture. Once equilibrium is established, the concentrations of each (reactant/products) won’t change. They are a mixture. This does not mean they stopped reacting, the rate of the reaction is the same.

The Concept of Equilibrium • As a system approaches equilibrium, both the forward and reverse reactions are occurring. • At equilibrium, the forward and reverse reactions are proceeding at the same rate.

A System at Equilibrium Once equilibrium is achieved, the amount of each reactant and product remains constant.

The Equilibrium Constant • Forward reaction: N2O4 (g) 2 NO2 (g) • Rate law: Rate = kf [N2O4]

The Equilibrium Constant • Reverse reaction: 2 NO2 (g) N2O4 (g) • Rate law: Rate = kr [NO2]2

N2O4 (g) 2 NO2 (g) Depicting Equilibrium In a system at equilibrium, both the forward and reverse reactions are being carried out; as a result, we write its equation with a double arrow

15.2- The Equilibrium Constant What is the Haber process? Under high pressure and high temperature, nitrogen and hydrogen gas creates ammonia gas.

15.2- The Equilibrium Constant Read this section on the Haber process, how are the concentration of the reactants and the products change (or not) during the reaction. At equilibrium, the relative amounts concentrations of all 3 (H2, N2 and NH3) are all the same, regardless of whether the starting mixture was 3:1 molar ratio. The equilibrium conditions can be reached from either direction.

15.2- The Equilibrium Constant Explain the law of mass action. This law expresses the relationship between the concentrations of the reactants and products at equilibrium in reactions. aA + bBpP + qQ

15.2- The Equilibrium Constant Provide and explain the equilibrium-constant expression. Where K is the equilibrium constant, and is unitless

15.2- The Equilibrium Constant Provide and explain the equilibrium-constant expression. Kc is an equilibrium constant is a numerical value obtained when we substitute actual equilibrium concentration into the expression. The subscript indicates that concentrations (expressed in molarity).

15.2- The Equilibrium Constant Provide and explain the equilibrium-constant expression.  Concentration products  Concentration reactants

15.2- The Equilibrium Constant What variables affect the Kc constant value? Stoichiometry of the reaction and temperature.

15.2- The Equilibrium Constant What variables does not affect the Kc constant value? The mechanism and initial concentration.

15.2- The Equilibrium Constant When writing the equilibrium expression, what goes in the numerator and denominator? The products goes in the numerator. The reactants goes in the denominator

15.2- The Equilibrium Constant Go through the Sample Exercise 15.1, then belowSHOW how to do the practice exercise: Write the Kc expression for H2 + I2 2HI

Write the equilibrium expression for Keq or Kcthis reactions: 1. 2 O3(g) ⇋ 3 O2(g)

Write the equilibrium expression for Keq or Kcthis reactions: 2. 2 NO(g) + Cl2(g) ⇋ 2 NOCl(g)

Write the equilibrium expression for Keq or Kcthis reactions: 3. Ag+(aq) + 2 NH3(g) ⇋ Ag(NH3)2+(aq)

Write the equilibrium expression for Keq or Kcthis reactions: 4. Cd2+(aq) + 4 Br-(aq) ⇋ CdBr42-(aq)

Look at the phases, what do they all have in common? They are the same phase; all gases or aqueous solutions These are termed- homogeneous equilibrium.

(PC)c (PD)d (PA)a (PB)b Kp = Homogenous equilibrium applies to reactions in which all reacting species are in the same phase. 2 NO(g) + Cl2(g) ⇋ 2 NOCl(g) Since we are dealing with gases we need to factor in partial pressures of the gases. Provide the equilibrium expression for gases? 14.2

(PC)c (PD)d (PA)a (PB)b Kp = 15.2- The Equilibrium Constant Provide and explain the Kp expression. Because pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written

n V P = RT Relationship between Kc and Kp • From the ideal gas law we know that PV = nRT • Rearranging it, we get

Relationship between Kc and Kp Plugging this into the expression for Kp for each substance, the relationship between Kc and Kp becomes Kp= Kc(RT)n Where n = (moles of gaseous product) − (moles of gaseous reactant)

Write the equilibrium expression for Kpfor this reactions: 1. 2 O3(g) ⇋ 3 O2(g)

Write the equilibrium expression for Kpfor this reactions: 2. 2 NO(g) + Cl2(g) ⇋ 2 NOCl(g)

Using the Equilibrium Constant Exercise 15.2: In the synthesis of ammonia from nitrogen and hydrogen (The Haber process), N2 (g) + 3H2(g) 2NO3 (g) at temperature of 300 °C, the Kc value is 9.60. Calculate the Kp. Provide the Kp for this reaction

Using the Equilibrium Constant Exercise 15.2: In the synthesis of ammonia from nitrogen and hydrogen (The Haber process), N2 (g) + 3H2(g) 2NO3 (g) at temperature of 300 °C, the Kc value is 9.60. Calculate the Kp. Kp= Kc(RT)n Convert 300 °C to K = 573 Kp= 9.60 (0.0821)(573K)n

Using the Equilibrium Constant Provide the Kp for this reaction ∆n = moles of product – moles of reactant moles of product is 2 moles of reactant is 1+ 3 = 4 ∆n = 2– 4 = -2

Using the Equilibrium Constant Exercise 15.2: In the synthesis of ammonia from nitrogen and hydrogen (The Haber process), N2 (g) + 3H2(g) 2NO3 (g) at temperature of 300 °C, the Kc value is 9.60. Calculate the Kp. Kp= Kc(RT)n Use yx Kp= 9.60 ((0.0821)(573K))-2 Kp= 4.34 x 10-3

Using the Equilibrium Constant 2SO3 2SO2 + O2 at temperature of 1000 K, the Kc value is 4.08 x10-3. Calculate the Kp Provide the Kp expression for this reaction.

Using the Equilibrium Constant The ∆n superscript – use the stoichiometry. 2SO3 2SO2 + O2 2+1 (products) - 2 (reactants) = 1 Kp= 4.08 x10-3 (0.0821)(1000K)1 Kp= 0.335

CO (g) + Cl2(g) COCl2(g) [COCl2] [CO][Cl2] Calculating BOTH Kc and Kp The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 74°C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp. Provide the Kp expression Kc=

CO (g) + Cl2(g) COCl2(g) = 0.14 0.012 x 0.054 [COCl2] [CO][Cl2] Calculating BOTH Kc and Kp The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 74°C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp. Substitute the concentrations in the Kc expression. = 220 Kc= Kc= 220 Now we can plug n’ chug

CO (g) + Cl2(g) COCl2(g) = 0.14 0.012 x 0.054 [COCl2] [CO][Cl2] Calculating BOTH Kc and Kp The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 74°C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp. = 220 Kc= Kp = Kc(RT)Dn Dn = 1 – 2 = -1 T = 273 + 74 = 347 K R = 0.0821 Kp= 220 x (0.0821 x 347)-1 = • 7.7

[PNO] [PO2] 2 Kp = 2NO2 (g) 2NO (g) + O2 (g) [PNO2] 2 Calculating a concentration The equilibrium constant Kp for the reaction is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO = 0.400 atm and PNO2= 0.270 atm? Provide the Kp expression for this reaction.

[0.400]2 [0.400]2 [0.400] [PO2] 2 = 158 = Kp Kp = [0.270]2 [0.270]2 2NO2 (g) 2NO (g) + O2 (g) PO PO 2 2 [0.270] 2 Calculating a concentration The equilibrium constant Kp for the reaction is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO = 0.400 atm and PNO2= 0.270 atm? Substitute the concentrations in the Kpexpression. Rearrange to get PO2 alone and Kp= 158 14.2

[0.400]2 = 158 [0.270]2 2NO2 (g) 2NO (g) + O2 (g) PO PO = 158 x (0.400)2/(0.270)2 2 2 Calculating a concentration The equilibrium constant Kp for the reaction is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO = 0.400 atm and PNO2= 0.270 atm? Solve for the PO2 equilibrium pressure. = 347 atm 14.2

15.2- The Equilibrium Constant Explain the relationship between Kc value and the shift in the equilibrium. • If K >> 1, the reaction is product-favored; • When equilibrium is achieved, most of the reactanthas been converted to product

15.2- The Equilibrium Constant Explain the relationship between Kc value and the shift in the equilibrium. • If K << 1, the reaction is reactant-favored; • When equilibrium is achieved, very little reactant has been converted to product

Entry Task: Jan 4 th Friday