Chemical Equilibrium CHAPTER 15

Chemical Equilibrium CHAPTER 15
Chemistry: The Molecular Nature of Matter, 6th edition By Jesperson, Brady, & Hyslop

CHAPTER 15 Chemical Equilibrium
Learning Objectives: Reversible Reactions and Equilibrium Writing Equilibrium Expressions and the Equilibrium Constant (K) Reaction Quotient (Q) Kc vs Kp ICE Tables Quadratic Formula vs Simplifying Assumptions LeChatelier’s Principle Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Lecture Road Map: Dynamic Equilibrium Equilibrium Laws Equilibrium Constant Le Chatelier’s Principle Calculating Equilibrium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Dynamic Equilibrium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Chemical equilibrium exists when
Dynamic Eq Equilibrium Chemical equilibrium exists when Rates of forward and reverse reactions are equal Reaction appears to stop Concentration of reactants and products do not change over time Remain constant Both forward and reverse reaction never cease Equilibrium signified by double arrows ( ) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

N2O4 2 NO2 Dynamic Eq Equilibrium Initially have only N2O4
Only forward reaction As N2O4 reacts NO2 forms As NO2 forms Reverse reaction begins to occur NO2 collide more frequently as concentration of NO2 increases Eventually, equilibrium is reached Concentration of N2O4 does not change Concentration of NO2 does not change Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Dynamic Eq Equilibrium
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

N2O4 2NO2 Dynamic Eq Equilibrium Closed system
Equilibrium can be reached from either direction Independent of whether it starts with “reactants” or “products” Always have the same composition at equilibrium under same conditions Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Reactants Equilibrium Products N2O4 2NO2 Dynamic Eq Equilibrium

Mass Action Expression
Dynamic Eq Mass Action Expression Simple relationship among [reactants] and [products] for any chemical system at equilibrium Called the mass action expression Derived from thermodynamics Forward reaction: A  B Reverse reaction: A  B At equilibrium: A B Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Uses stoichiometric coefficients as exponent for each reactant
Dynamic Eq Reaction Quotient Uses stoichiometric coefficients as exponent for each reactant For reaction: aA + bB cC + dD Reaction quotient Numerical value of mass action expression Equals “Q ” at any time, and Equals “K ” only when reaction is known to be at equilibrium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Ex. 1 H2(g) + I2(g) 2HI(g) 440˚C Exp’t Initial Amts Equil’m Amts
Equil’m [M] I 1.00 mol H2 0.222 mol H2 M H2 10 L 1.00 mol I2 0.222 mol I2 M I2 0.00 mol HI 1.56 mol HI 0.156 M HI II 0.00 mol H2 0.350 mol H2 M H2 10 L 0.100 mol I2 0.450 mol I2 M I2 3.50 mol HI 2.80 mol HI 0.280 M HI

Ex. 1 H2(g) + I2(g) 2HI(g) 440 ˚C Exp’t Initial Amts Equil’m Amts
Equil’m [M] III mol H2 0.150 mol H2 M H2 10 L 0.00 mol I2 0.135 mol I2 M I2 1.27 mol HI 1.00 mol HI 0.100 M HI IV 0.00 mol H2 0.442 mol H2 M H2 10 L 0.00 mol I2 0.442 mol I2 M I2 4.00 mol HI 3.11 mol HI 0.311 M HI

Mass Action Expression
= same for all data sets at equilibrium Equilibrium Concentrations (M ) Exp’t [H2] [I2] [HI] I 0.0222 0.156 II 0.0350 0.0450 0.280 III 0.0150 0.0135 0.100 IV 0.0442 0.311 Average = 49.5

Write mass action expressions for the following: 2NO2(g) N2O4(g)
Group Problem Write mass action expressions for the following: 2NO2(g) N2O4(g) 2CO(g) + O2(g) CO2(g) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Cu2+(aq) + 4NH3(aq) [Cu(NH3)42+](aq)?
Group Problem Which of the following is the correct mass action expression for the reaction: Cu2+(aq) + 4NH3(aq) [Cu(NH3)42+](aq)? Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Equilibrium Laws Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

at equilibrium write the following equilibrium law
Equilibrium Laws For reaction H2(g) + I2(g) HI(g) at 440 ˚C at equilibrium write the following equilibrium law Equilibrium constant = Kc = constant at given T Use Kc since usually working with concentrations in mol/L For chemical equilibrium to exist in reaction mixture, reaction quotient Q must be equal to equilibrium constant, Kc Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Predicting Equilibrium Laws
For general chemical reaction: dD eE fF gG Where D, E, F, and G represent chemical formulas d, e, f, and g are coefficients Mass action expression is Note: Exponents in mass action expression are stoichiometric coefficients in balanced equation. Equilibrium law is: Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Predicting Equilibrium Laws
Only concentrations that satisfy this equation are equilibrium concentrations Numerator Multiply concentration of products raised to their stoichiometric coefficients Denominator Multiply concentration reactants raised to their stoichiometric coefficients is scientists’ convention Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

What is equilibrium law?
Example 3H2(g) N2(g) NH3(g) Kc = 4.26 × 108 at 25 °C What is equilibrium law? Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Various operations can be performed on equilibrium expressions
1. When direction of equation is reversed, new equilibrium constant is reciprocal of original A + B C + D C +D A + B Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Equilibrium Operations
1. When direction of equation is reversed, new equilibrium constant is reciprocal of original 3H2(g) + N2(g) NH3(g) at 25˚C 2NH3(g) H2(g) + N2(g) at 25 ˚C Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

2. When coefficients in equation are multiplied by a factor, equilibrium constant is raised to a power equal to that factor. A + B C + D 3A + 3B C + 3D Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

When coefficients in equation are multiplied by factor, equilibrium constant is raised to power equal to that factor 3H2(g) + N2(g) NH3(g) at 25 ˚C Multiply by 3 9H2(g) + 3N2(g) NH3(g) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

3. When chemical equilibria are added, their equilibrium constants are multiplied A + B C + D C + E F + G A + B + E D + F + G Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

NO3(g) + CO(g) NO2(g) + CO2(g) NO2(g) + CO(g) NO(g) + CO2(g)
Equilibrium Operations 3. When chemical equilibria are added, their equilibrium constants are multiplied 2 NO2(g) NO3(g) + NO(g) NO3(g) + CO(g) NO2(g) + CO2(g) NO2(g) + CO(g) NO(g) + CO2(g) Therefore Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

For: N2(g) + 3H2(g) 2NH3(g) Kc = 500 at a particular temperature.
Group Problem For: N2(g) + 3H2(g) NH3(g) Kc = 500 at a particular temperature. What would be Kc for following? 2NH3(g) N2(g) + 3H2(g) 1/2N2(g) + 3/2H2(g) NH3(g) 0.002 22.4 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Equilibrium Constant Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Equilibrium Constant Kc
Most often Kc is expressed in terms of a ratio of concentrations of products and reactants as shown on previous slides Sometimes partial pressures, in atmospheres, may be used in place of concentrations Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Based on reactions in which all substances are gaseous
Equilibrium Kp Based on reactions in which all substances are gaseous Gas quantities are expressed in atmospheres in mass action expression Use partial pressures for each gas in place of concentrations e.g. N2(g) + 3H2(g) NH3(g) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Relationship between Kp and Kc
Equilibrium Relationship between Kp and Kc Start with ideal gas law PV = nRT Rearranging gives Substituting P/RT for molar concentration into Kc results in pressure-based formula ∆n = moles of gas in product – moles of gas in reactant Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Kc = 11.7 Consider the reaction: 2NO2(g) N2O4(g)
Group Problem Consider the reaction: 2NO2(g) N2O4(g) If Kp = for the reaction at 25 ˚C, what is value of Kc at same temperature? n = nproducts – nreactants = 1 – 2 = –1 Kc = 11.7 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Δn = (4 – 3) = 1 Kp = Kc(RT)Δn Kp= 0.99 × (0.082057 × 298.15)1 Kp = 24
Group Problem Consider the reaction A(g) + 2B(g) C(g) If the Kc for the reaction is 0.99 at 25 ˚C, what would be the Kp? 0.99 2.0 24 2400 None of these Δn = (4 – 3) = 1 Kp = Kc(RT)Δn Kp= 0.99 × ( × )1 Kp = 24 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Homogeneous and Hetergeneous
Equilibrium Homogeneous and Hetergeneous Homogeneous reaction/equilibrium All reactants and products in same phase Can mix freely Heterogeneous reaction/equilibrium Reactants and products in different phases Can’t mix freely Solutions are expressed in M Gases are expressed in M Governed by Kc Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g) Equilibrium Law =
Heterogeneous 2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g) Equilibrium Law = Can write in simpler form For any pure liquid or solid, ratio of moles to volume of substance (M ) is constant e.g. 1 mol NaHCO3 occupies 38.9 cm mol NaHCO3 occupies 77.8 cm3 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g)
Equilibrium Heterogeneous 2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g) Ratio (n/V ) or M of NaHCO3 is constant (25.7 mol/L) regardless of sample size Likewise can show that molar concentration of Na2CO3 solid is constant regardless of sample size So concentrations of pure solids and liquids can be incorporated into equilibrium constant, Kc Equilibrium law for heterogeneous system written without concentrations of pure solids or liquids Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Write equilibrium laws for the following: Ag+(aq) + Cl–(aq) AgCl(s)
Heterogeneous Write equilibrium laws for the following: Ag+(aq) + Cl–(aq) AgCl(s) H3PO4(aq) + H2O H3O+(aq) + H2PO4–(aq) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Interpreting KC Large K (K >>1) Means product rich mixture
Reaction goes far toward completion e.g. 2SO2(g) + O2(g) SO3(g) Kc = 7.0  1025 at 25 °C

Interpreting KC Small K (K << 1) Means reactant rich mixture
Only very small amounts of product formed e.g. H2(g) + Br2(g) HBr(g) Kc = 1.4  10–21 at 25 °C

Interpreting KC K  1 Means product and reactant concentrations close to equal Reaction goes only about halfway

Size of K gives measure of how reaction proceeds
K >> 1 [products] >> [reactants] K = [products] = [reactants] K << 1 [products] << [reactants]

CHAPTER 15 Chemical Equilibrium Le Chatelier’s Principle

Equilibrium positions Combination of concentrations that allow Q = K
Le Chatelier Definition Equilibrium positions Combination of concentrations that allow Q = K Infinite number of possible equilibrium positions Le Châtelier’s principle System at equilibrium (Q = K) when upset by disturbance (Q ≠ K) will shift to offset stress System said to “shift to right” when forward reaction is dominant (Q < K) System said to “shift to left” when reverse direction is dominant (Q > K) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Q = K reaction at equilibrium Q < K reactants go to products
Le Chatelier Q & K Relationships Q = K reaction at equilibrium Q < K reactants go to products Too many reactants Must convert some reactant to product to move reaction toward equilibrium Q > K products go to reactants Too many products Must convert some product to reactant to move reaction toward equilibrium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Change in Concentration
Le Chatelier Change in Concentration Cu(H2O)42+(aq) + 4Cl–(aq) CuCl42–(aq) + 4H2O blue yellow Equilibrium mixture is blue-green Add excess Cl– (conc. HCl) Equilibrium shifts to products Makes more yellow CuCl42– Solution becomes green Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Le Chatelier Change in Concentration Cu(H2O)42+(aq) + 4Cl–(aq) CuCl42–(aq) + 4H2O blue yellow Add Ag+ Removes Cl–: Ag+(aq) + Cl–(aq)  AgCl(s) Equilibrium shifts to reactants Makes more blue Cu(H2O)42+ Solution becomes increasingly more blue Add H2O? Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Change in Concentration: Example
Le Chatelier Change in Concentration: Example For the reaction 2SO2(g) + O2(g) SO3(g) Kc = 2.4 × 10–3 at 700 °C Which direction will the reaction move if moles of O2 is added to an equilibrium mixture? Towards the products Towards the reactants No change will occur Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Le Chatelier Change in Concentration When changing concentrations of reactants or products Equilibrium shifts to remove reactants or products that have been added Equilibrium shifts to replace reactants or products that have been removed Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Change in Pressure or Volume
Le Chatelier Change in Pressure or Volume Consider gaseous system at constant T and n 3H2(g) + N2(g) NH3(g) If volume is reduced Expect pressure to increase To reduce pressure, look at each side of reaction Which has less moles of gas Reactants = 3 mol + 1 mol = 4 mol gas Products = 2 mol gas Reaction favors products (shifts to right) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Le Chatelier Change in Pressure or Volume Consider gaseous system at constant T and n H2(g) + I2(g) HI(g) If pressure is increased, what is the effect on equilibrium? nreactant = = 2 nproduct = 2 Predict no change or shift in equilibrium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Le Chatelier Change in Pressure or Volume 2NaHSO3(s) NaSO3(s) + H2O(g) + SO2(g) If you decrease volume of reaction, what is the effect on equilibrium? Reactants: All solids, no moles gas Products: 2 moles gas Decrease in V, causes an increase in P Reaction shifts to left (reactants), as this has fewer moles of gas Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Le Chatelier Change in Pressure or Volume Reducing volume of gaseous reaction mixture causes reaction to decrease number of molecules of gas, if it can Increasing pressure Moderate pressure changes have negligible effect on reactions involving only liquids and/or solids Substances are already almost incompressible Changes in V, P and [X ] effect position of equilibrium (Q), but not K Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Cu(H2O)42+(aq) + 4Cl–(aq) CuCl42–(aq) + 4H2O blue yellow
Le Chatelier Change in Temperature Ice water Boiling water Cu(H2O)42+(aq) + 4Cl–(aq) CuCl42–(aq) + 4H2O blue yellow Reaction endothermic Adding heat shifts equilibrium toward products Cooling shifts equilibrium toward reactants Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Add heat energy, shift reaction right
Le Chatelier Change in Temperature Hf°=+6 kJ (at 0 °C) Energy + H2O(s) H2O(l ) Energy is reactant Add heat energy, shift reaction right 3H2(g) + N2(g) NH3(g) Hrxn= –47.19 kJ 3 H2(g) + N2(g) NH3(g) + energy Energy is product Add heat, shift reaction left H2O(s) H2O(l) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Le Chatelier Change in Temperature
Increase in temperature shifts reaction in direction that produces endothermic (heat absorbing) change Decrease in temperature shifts reaction in direction that produces exothermic (heat releasing) change Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Increase in temperature of exothermic reaction makes K smaller
Le Chatelier Change in Temperature Changes in T change value of mass action expression at equilibrium, so K changed K depends on T Increase in temperature of exothermic reaction makes K smaller More heat (product) forces equilibrium to reactants Increase in temperature of endothermic reaction makes K larger More heat (reactant) forces equilibrium to products Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Catalyst lowers Ea for both forward and reverse reaction
Le Chatelier Change with Catalyst Catalyst lowers Ea for both forward and reverse reaction Change in Ea affects rates k r and k f equally Catalysts have no effect on equilibrium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Addition of Inert Gas at Constant Volume
Le Chatelier Addition of Inert Gas at Constant Volume Inert gas One that does not react with components of reaction e.g. argon, helium, neon, usually N2 Adding inert gas to reaction at fixed V (n and T), increase P of all reactants and products Since it doesn’t react with anything No change in concentrations of reactants or products No net effect on reaction Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

How To Use Le Chatelier’s Principle
Write mass action expression for reaction Examine relationship between affected concentration and Q (direct or indirect) Compare Q to K If change makes Q > K, shifts left If change makes Q < K, shifts right If change has no effect on Q, no shift expected Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

H3PO4(aq) + 3OH–(aq) 3H2O(l) + PO43–(aq)
Group Problem Consider: H3PO4(aq) + 3OH–(aq) H2O(l) + PO43–(aq) What will happen if PO43– is removed? Q is proportional to [PO43–] Decrease [PO43–], decrease in Q Q < K equilibrium shifts to right Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

H3PO4(aq) + 3OH–(aq) 3H2O(aq) + PO43–(aq) is exothermic.
Group Problem The reaction H3PO4(aq) + 3OH–(aq) H2O(aq) + PO43–(aq) is exothermic. What will happen if system is cooled? heat Since reaction is exothermic, heat is product Heat is directly proportional to Q Decrease in T, decrease in Q Q < K equilibrium shifts to right Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

[Co(H2O)6]2+(aq) + 4Cl–(aq) [Co(Cl)4]2–(aq) + 6H2O pink blue
Group Problem The equilibrium between aqueous cobalt ion and the chlorine ion is shown: [Co(H2O)6]2+(aq) + 4Cl–(aq) [Co(Cl)4]2–(aq) + 6H2O pink blue It is noted that heating a pink sample causes it to turn violet. The reaction is: endothermic exothermic cannot tell from the given information Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Group Problem The following are equilibrium constants for the reaction of acids in water, Ka. Which reaction proceeds the furthest to products? Ka = 2.2 × 10–3 Ka = 1.8 × 10–5 Ka = 4.0 × 10–10 Ka = 6.3 × 10–3 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

CHAPTER 15 Chemical Equilibrium Calculating Equilibrium

For gaseous reactions, use either KP or KC
Calculations Overview For gaseous reactions, use either KP or KC For solution reactions, must use KC Either way, two basic categories of calculations Calculate K from known equilibrium concentrations or partial pressures Calculate one or more equilibrium concentrations or partial pressures using known KP or KC Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Kc with Known Equilibrium Concentrations
Calculations Kc with Known Equilibrium Concentrations When all concentrations at equilibrium are known Use mass action expression to relate concentrations to KC Two common types of calculations Given equilibrium concentrations, calculate K Given initial concentrations and one final concentration Calculate equilibrium concentration of all other species Then calculate K Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Calculations Kc with Known Equilibrium Concentrations Ex. 3 N2O4(g) NO2(g) If you place mol N2O4 in 1 L flask at equilibrium, what is KC? [N2O4]eq = M [NO2]eq = M KC = 4.61  10–3 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Group Problem For the reaction: 2A(aq) + B(aq) C(aq) the equilibrium concentrations are: A = 2.0 M, B = 1.0 M and C = 3.0 M. What is the expected value of Kc at this temperature? 14 0.15 1.5 6.75 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Calculations Kc with Known Equilibrium Concentrations Ex SO2(g) + O2(g) SO3(g) At 1000 K, mol SO2 and mol O2 are placed in a L flask. At equilibrium mol SO3 has formed. Calculate K C for this reaction. First calculate concentrations of each Initial Equilibrium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Set up concentration table Based on the following:
Calculations Example Continued Set up concentration table Based on the following: Changes in concentration must be in same ratio as coefficients of balanced equation Set up table under balanced chemical equation Initial concentrations Controlled by person running experiment Changes in concentrations Controlled by stoichiometry of reaction Equilibrium concentrations Equilibrium Concentration Initial Concentration Change in Concentration = – Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

2SO2(g) + O2(g) 2SO3(g) Initial Conc. (M) 1.000 0.000
Calculations Example Continued 2SO2(g) + O2(g) 2SO3(g) Initial Conc. (M) 1.000 0.000 Changes in Conc. (M) Equilibrium Conc. (M) –0.925 –0.462 +0.925 0.075 0.538 0.925 [SO2] consumed = amount of SO3 formed = [SO3] at equilibrium = M [O2] consumed = ½ amount SO3 formed = 0.925/2 = M [SO2] at equilibrium = – = 0.075 [O2] at equilibrium = 1.00 – = M Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Kc = 2.8 × 102 = 280 Calculations Example Continued
Finally calculate KC at 1000 K Kc = 2.8 × 102 = 280 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

ICE tables used for most equilibrium calculations:
ICE Table Summary ICE tables used for most equilibrium calculations: Equilibrium concentrations are only values used in mass action expression Values in last row of table Initial value in table must be in units of mol/L (M) [X]initial = those present when reaction prepared No reaction occurs until everything is mixed Changes in concentrations always occur in same ratio as coefficients in balanced equation In “change” row be sure all [reactants] change in same directions and all [products] change in opposite direction. If [reactant]initial = 0, its change must be an increase (+) because [reactant]final cannot be negative If [reactants] decreases, all entries for reactants in change row should have minus sign and all entries for products should be positive Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Calculate [X ]equilibrium from Kc and [X ]initial
Calculations Calculate [X ]equilibrium from Kc and [X ]initial When all concentrations but one are known Use mass action expression to relate Kc and known concentrations to obtain missing concentrations Ex. 5 CH4(g) + H2O(g) CO(g) + 3H2(g) At 1500 °C, Kc = An equilibrium mixture of gases had the following concentrations: [CH4] = M and [H2] = M and [CO] = M. What is [H2O] at equilibrium ? Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Calculate [X ]equilibrium from Kc and [X ]initial
Calculations Calculate [X ]equilibrium from Kc and [X ]initial Ex. 5 CH4(g) + H2O(g) CO(g) + 3H2(g) Kc = 5.67 [CH4] = M; [H2] = M; [CO] =0.300 M What is [H2O] at equilibrium? First, set up equilibrium Next, plug in equilibrium concentrations and Kc [H2O] = M Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Calculating [X ]Equilibrium from Kc
Calculations Calculating [X ]Equilibrium from Kc When Initial Concentrations Are Given Write equilibrium law/mass action expression Set up Concentration table Allow reaction to proceed as expected, using “x” to represent change in concentration Substitute equilibrium terms from table into mass action expression and solve Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Calculate [X]equilibrium from [X]initial and KC
Calculations Calculate [X]equilibrium from [X]initial and KC Ex. 6 H2(g) + I2(g) HI(g) at 425 ˚C KC = 55.64 If one mole each of H2 and I2 are placed in a L flask at 425 °C, what are the equilibrium concentrations of H2, I2 and HI? Step 1. Write Equilibrium Law Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Conc (M) H2(g) + I2(g) 2HI (g) Initial 2.00 0.000 Change Equilibrium
Calculations Calculate [X]equilibrium from [X]initial and KC Step 2: Construct an ICE table Initial [H2] = [I2] = 1.00 mol/0.500 L =2.00 M Amt of H2 consumed = Amt of I2 consumed = x Amount of HI formed = 2x Conc (M) H2(g) + I2(g) 2HI (g) Initial 2.00 0.000 Change Equilibrium – x – x +2x 2.00 – x 2.00 – x +2x Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Calculations Calculate [X]equilibrium from [X]initial and KC Step 3. Solve for x Both sides are squared so we can take square root of both sides to simplify Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Conc (M) H2(g) + I2(g) 2HI (g) Initial 2.00 0.00 Change Equilibrium
Calculations Calculate [X]equilibrium from [X]initial and KC Step 4. Equilibrium Concentrations [H2]equil = [I2]equil = 2.00 – 1.58 = 0.42 M [HI]equil = 2x = 2(1.58) = 3.16 Conc (M) H2(g) + I2(g) 2HI (g) Initial 2.00 0.00 Change Equilibrium – 1.58 – 1.58 +3.16 0.42 0.42 +3.16 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Calculations Calculate [X]equilibrium from [X]initial and KC Ex. 7 H2(g) + I2(g) HI(g) at 425 ˚C KC = 55.64 If one mole each of H2, I2 and HI are placed in a L flask at 425 ˚C, what are the equilibrium concentrations of H2, I2 and HI? Now have product as well as reactants initially Step 1. Write Equilibrium Law Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Conc (M) H2(g) + I2(g) 2HI (g) Initial 2.00 Change Equil’m
Calculations Calculate [X]equilibrium from [X]initial and KC Step 2. Concentration Table Conc (M) H2(g) + I2(g) 2HI (g) Initial 2.00 Change Equil’m – x – x +2x 2.00 – x 2.00 – x x Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Calculations Calculate [X]equilibrium from [X]initial and KC Step 3. Solve for x [H2]equil = [I2]equil = 2.00 – x = 2.00 – 1.37 = 0.63 M [HI]equil = x = (1.37) = = 4.74 M Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

N2(g) + O2(g) 2NO(g) Kc = 0.0123 at 3900 ˚C
Group Problem N2(g) + O2(g) NO(g) Kc = at 3900 ˚C If 0.25 moles of N2 and O2 are placed in a 250 mL container, what are the equilibrium concentrations of all species? A M, M, M B M, M, M C M, M, M D M, M, M Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Conc (M) N2(g) + O2(g) 2NO (g) Initial 1.00 1.00 0.00
Group Problem Conc (M) N2(g) + O2(g) NO (g) Initial Change – x – x x Equil – x – x x Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Calculations Calculate [X]equilibrium from [X]initial and KC Example: Quadratic Equation Ex. 8 CH3CO2H(aq) + C2H5OH(aq) CH3CO2C2H5(aq) + H2O(l) acetic acid ethanol ethyl acetate KC = 0.11 An aqueous solution of ethanol and acetic acid, each with initial concentration of M, is heated at 100 °C. What are the concentrations of acetic acid, ethanol and ethyl acetate at equilibrium? Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Calculations Calculate [X]equilibrium from [X]initial and KC Example: Quadratic Equation Step 1. Write equilibrium law Need to find equilibrium values that satisfy this Step 2: Set up concentration table using “x” for unknown Initial concentrations Change in concentrations Equilibrium concentrations Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Calculations Calculate [X]equilibrium from [X]initial and KC Example: Quadratic Equation Step 2 Concentration Table Amt of CH3CO2H consumed = Amt of C2H5OH consumed = – x Amt of CH3CO2C2H5 formed = + x [CH3CO2H]eq and [C2H5OH ] = – x [CH3CO2C2H5] = x (M) CH3CO2H(aq) + C2H5OH(aq) CH3CO2C2H5(aq) + H2O(l) I 0.810 0.000 C E –x – x +x 0.810 – x 0.810 – x +x Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Calculations Calculate [X]equilibrium from [X]initial and KC Example: Quadratic Equation Step 3. Solve for x Rearranging gives Then put in form of quadratic equation ax2 + bx + c = 0 Solve for the quadratic equation using Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Calculations Calculate [X]equilibrium from [X]initial and KC Example: Quadratic Equation Step 3. Solve for x This gives two roots: x = 10.6 and x = 0.064 Only x = is possible x = 10.6 is >> initial concentrations 0.810 – 10.6 = negative concentration, which is impossible Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Calculations Calculate [X]equilibrium from [X]initial and KC Example: Quadratic Equation Step 4. Equilibrium Concentrations [CH3CO2C2H5]equil = x = M [CH3CO2H]equil = [C2H5OH]equil = M – x = M – M = M CH3CO2H(aq) + C2H5OH(aq) CH3CO2C2H5(aq) + H2O I 0.810 0.000 C E –0.064 – 0.064 +0.064 0.746 0.746 +0.064 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

When KC is very small Calculations
Calculate [X]equilibrium from [X]initial and KC Example: Cubic When KC is very small Ex. 9 2H2O(g) H2(g) + O2(g) At 1000 °C, KC = 7.3  10–18 If the initial H2O concentration is M, what will the H2 concentration be at equilibrium? Step 1. Write Equilibrium Law Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Conc (M ) 2H2O(g) 2H2(g) + O2(g) Initial 0.100 0.00 Change Equil’m
Calculations Calculate [X]equilibrium from [X]initial and KC Example: Cubic Step 2. Concentration Table Cubic equation – tough to solve Make approximation KC very small, so x will be very small Assume we can neglect x Must prove valid later Conc (M ) 2H2O(g) 2H2(g) + O2(g) Initial 0.100 0.00 Change Equil’m – 2x +2x +x 0.100 – 2x +2x +x Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

= 7.3 × 10–20 Conc (M) 2H2O (g) 2H2 (g) + O2 (g) Initial 0.100 0.00
Calculations Calculate [X]equilibrium from [X]initial and KC Example: Cubic Step 3. Solve for x Assume (0.100 – 2x)  0.100 Now our equilibrium expression simplifies to Conc (M) 2H2O (g) 2H2 (g) + O2 (g) Initial 0.100 0.00 Change Equil’m – 2x +2x +x 0.100 +2x +x = 7.3 × 10–20 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Calculate [X]equilibrium from [X]initial and KC Example: Cubic
Calculations Calculate [X]equilibrium from [X]initial and KC Example: Cubic Step 3. Solve for x Now take cube root x is very small 0.100 – 2(2.6  10–7) = Which rounds to (3 decimal places) [H2] = 2x = 2(2.6  10–7) = 5.2  10–7 M Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Simplifications: When Can You Ignore x In Binomial (Ci – x)?
Calculations Simplifications: When Can You Ignore x In Binomial (Ci – x)? If equilibrium law gives very complicated mathematical problems and if K is small Then the change (x term) will also be small and we can assume it can be ignored when added or subtracted from the initial concentration, Ci. How do we check that the assumption is correct? If the calculated x is so small it does not change the initial concentration (e.g Minitial – Mx-calc = 0.10) Or if the answer achieved by using the assumption differs from the true value by less than five percent. This often occurs when Ci > 100 x Kc Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

≈0.2 For the reaction 2A(g) B(g)
Group Problem For the reaction 2A(g) B(g) given that Kp = 3.5 × 10–16 at 25 ˚C, and we place 0.2 atm A into the container, what will be the pressure of B at equilibrium? 2A  B I atm C –2x x E – 2x x ≈0.2 x = 1.4 × 10–17 [B]= 1.4 × 10–17 atm Proof: × 10–17 = 0.2 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Chemical Equilibrium CHAPTER 15

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