Chapter Fifteen Chemical Equilibrium 1
The Concept of Equilibrium
Dynamic Equilibrium Opposing processes occur at equal rates Forward and reverses reaction proceed at equal rates No outward change is observed Ratio of reactants to products is constant. A double sided arrow ( ) is used Often temperature dependent Physical equilibrium Equilibrium between phases H O(l) H O(g) Chemical equilibrium Equilibrium between reactants and products N O 4 (g) NO (g) colorless brown 3
Equilibrium Constant Expression, K c At equilibrium, [N O 4 ] and [NO ] are constant N O 4 NO Rate (forward) = Rate (reverse) k 1 [N O 4 ] = k -1 [NO ] k k 1 [ NO ] [ N O 1 4] K c 4
Initial vs. Equilibrium Concentrations Proof of K c Calculation 5
The Equilibrium Constant Expression, K c For a reaction of the type: aa bb... cc dd... c d [ C] [ D] The equilibrium expression is: Kc a b [ A] [ B] Equilibrium Constant Expression: Concentrations of the products appear in the numerator Concentrations of the reactants appear in the denominator. Exponents match coefficients in chemical equation. Units are not included! Also known as a Mass Action Expression Chemical Equation must be balanced to calculate K c K is based on HOW the equation is balanced 6
Chemical Reactions and K c a. SO (g) + ½O (g) SO 3 (g) b. O (g) + O (g) SO 3 (g) K c [ SO 3 ] [ SO ][ O ] 1/ [ SO K c [ SO ] ] [ O Equilibrium constants change if reaction balanced differently. Numerical values for K c are different K c (reaction b) = [K c (reaction a)] 3 ] You must know how the reaction was balanced 7
Rules for Modifying A Chemical Equation aa + bb cc + dd Reversing a chemical equation: Invert K c a [ A] [ B] cc dd aa bb Kc c [ C] [ D] Multiplying coefficients by n: Raise K c to n power na nb [ A] [ B] n n( cc dd aa bb) Kc K nc nd c [ C] [ D] Adding equations: Multiply K c s ee ff aa bb cc ee aa bb dd K ff cc dd c [ C ] c[ D ] a [ A] [ B] d b b d [ F ] [ E ] f e 1 K c K 1 K 8
Meaning of Equilibrium Constants aa bb cc dd At Equilibrium K c [ C] [ A] More products than reactants More reactants than products c a [ D] [ B] d b As K goes to infinity reaction goes to completion As K goes to zero, no reaction occurs 9
At a given temperature, 0.100 moles of NO were added to a.00 L vessel. At equilibrium, 0.044 moles of NO were remaining. What is the value of K c? Balanced Equation and Equilibrium Constant Expression: NO(g) N (g) + O (g) K c = [N ] [O ] [NO] Find equilibrium molarities of reactants and products molesn = moleso = ½ moles NO used up in reaction molesn = moleso = ½ (0.100-0.044)= 0.08moles [N ] = [O ] = 0.08moles/.00L = 0.014M [NO]= 0.044moles/.00L = 0.0M Calculate K c K c = [N ] [O ] = [0.014][0.014] = 0.40 [NO] [0.0] 10
Ways of Expressing Equilibrium Constants 11
Homogeneous Equilibria, K c and K p All products and reactants are in an aqueous or gas phase for ANY equilibrium constant K c K p All chemicals in units of molarity, moles/l Can be either gas or aqueous solutions CO(g) + O (g)co (g) All chemicals in gas phase Partial pressure in atmospheres Relationship between K c and K p K p = K c (RT) n Derived from ideal gas law R is the gas constant, T is temperature in kelvin n = moles gaseous products - moles gaseous reactants K p =K c if moles gas of product = moles gas of reactant Otherwise, a decrease in moles of gas decreases pressure 1
Calculate K p for the following reactions CO(g) + O (g)co (g) NO(g) N (g) + O (g) K c = 1800 K c = 1800 n gas = n gas = 0 K p = K c (RT) n gas K p K c = 1800 R = 0.081L atm/mol K T = 1500 o C= 1773K n gas 1800 = Moles products gas phase - Moles reactants gas phase 0.081 Latm x( molk x1773 K ) 1 1800 x(.00687 ) 87.9 If n gas = 0 : K c = K p 13
Heterogeneous Equilibrium and K eq Are the equilibrium constants for the reactions the same? SO 3 (g) + H O(g) H SO 4 (g) SO 3 (g) + H O(l) H SO 4 (aq) K 1 = [H SO 4 (g)] K = [H SO 4 (aq)] [SO 3 (g)][h O(g)] [SO 3 (g)][h O(l)] K 1 equals K if [H SO 4 (g)]=[h SO 4 (aq)] & [H O(g)]=[H O(l)] [H SO 4 (g)] variable dependent on partial pressure [H SO 4 (aq)] variable # moles dissolved in water [H O(g)] variable dependent on partial pressure [H O(l)] 56M Density =1 g/ml = 1000 g/l [H O(l)] =1000g/L x mol/18g= 56mol/L The two equilibrium constants are not the same The phase of the material matters! 14
Equilibria Involving Mixed Phases Use the subscript eq to designate general K K eq = K c = K=, etc. Value depends on equation Do not include solids and liquids in K eq expression SO 3 (g) + H O(l) H SO 4 (aq) K eq = [H SO 4 (aq)] [SO 3 (g)] K p for liquid-vapor equilibrium is vapor pressure of liquid H O(l) H O(g) K p = P(H O) Liquid will not be included Reactions must show phase for each reactant and product. 15
Equilibria Involving Pure Solids And Liquids CaCO 3 (s)cao(s) + CO (g) K p = P CO P CO does not depend on the amount of CaCO 3 or CaO 16
Exam 1 Feb. 3 rd 7:30AM-9:15AM Covers Chapters 14 and 15 8 questions, 3 parts per question (see practice exam) Will be assigning seats tomorrow morning by 10am Seat assignment in Sakai is random You cannot ask for a certain seat, just general preference Request preferences immediately Check seat before exam so you can get started Bring your ID to the exam TAs will check IDs once exam is started Exam will be on your seat with your name on it Sit in assigned seat and start exam 17
Alternate Exam: Feb. 4 th, 6:30AM-7:50AM If you do not take the exam on Tuesday. Ideally email me by 5pm on the nd so I know you are coming If you just miss the exam, email me as soon as possible No guarantee I ll have an extra exam if I don t know you are coming!!! If you do not take exam 1 on either day DSS Exams It is your dropped exam, no further makeups given I need DSS paperwork by next Tuesday (1 week before exam) I will try to get a time and date between Tues-Fri that works for all Once I know your schedules, I ll get back to you with options 18
What Does the Equilibrium Constant Tell Us? 19
Relative Meaning Of Equilibrium Constants c d [ C] [ D] aa bb... cc dd... Kc a b [ A] [ B] Large value of K c or K p [C] & [D] very large or [A] & [B] very small. Reaction goes to completion Equilibrium favors products. Small value of K c or K p [C] & [D] very small or [A] & [B] very large Reaction proceeds slowly or not at all (NR: no reaction) Equilibrium favors reactants Does not guarantee reaction will occur Thermodynamically favored, but may be kinetically controlled. 0
The Reaction Quotient, Q c or Q p Represents K c or K p at non-equilibrium conditions aa bb... cc dd... Qc Predicts direction of reaction to get to equilibrium. Q c < K Concentration of products is less than at equilibrium Reaction occurs in the forward direction Q c > K c Concentration of products is greater than at equilibrium Reaction proceeds in the reverse direction Q c = K c Concentration of products equals that at equilibrium No macroscopic changes are observed [ C] [ A] c a [ D] [ B] d b 1
For the reaction of hydrogen with nitrogen to form ammonia, the equilibrium concentrations were found to be 1.0 10-3 M in both hydrogen and nitrogen and 0.00 M in ammonia. Now add 0.010 M nitrogen. Which direction does the reaction shift? N (g) + 3H (g) NH 3 (g)
For the reaction of hydrogen with nitrogen to form ammonia, the equilibrium constant K c = 4.0 10 8 at 5 o C. The equilibrium concentrations were found to be 1.0 10-3 M in both hydrogen and nitrogen and 0.00 M in ammonia. Now add 0.010 M nitrogen. Which direction does the reaction shift? N (g) + 3H (g) NH 3 (g) Fill in Concentrations: [N ] = 1.0 10-3 M + 0.010M= 0.011 M [NH 3 ] = 0.00 M Q c [H ] = 1.0 10-3 M K c [0.00 ] [0.011][1.0 x10 [ NH 3] [ N ][ H 3 ] 3 ] 3 3.6 x10 7 Compare Q c to K c : 3.6 10 7 < 4.1 10 8 Q c <K c The reaction goes toward products Adding Nitrogen forces reaction to the right 3
Solving Equilibrium Problems 4
Use of Tabulated Equilibrium Constants 1. Write the properly balanced chemical reaction. This gives the stoichiometry and the species involved in the reaction. Set up a table of concentrations for all components Use reaction stoichiometry to express all unknown concentrations in terms of a single reactant or product, x. 3. Write out the equilibrium constant expression for the reaction This is the equation for K 4. Substitute these concentrations into the equation for K 5. Solve the equation for the unknown concentration, x. 6. Substitute the value you calculated for x into the expressions for the other equilibrium concentrations. 5
The decomposition of BrCl to bromine and chlorine has a K c, of 0.14 at 350 K. If the initial concentration of BrCl is 0.06 M, what are the equilibrium concentrations of all components? 1. Equation BrCl(g) Br (g) + Cl (g). Table BrCl Br Cl Initial.06 0 0 Change x +x +x Equilibrium 0.06-x +x +x 3. Equilibrium Constant Expression 5. Solve for x K c [ Br ][ Cl [ BrCl ] 4. Substitute equilibrium values 0.14 [ x][ x] [0.06 x] ] [ x] [0.06 x] [ x] 0.374 [0.06 x] x = 0.03-0.748x x = 0.013 6. Substitute back into table [BrCl] = 0.06 x = 0.06 (0.013) = 0.036 M [Br ] = [Cl ] = x = 0.013M 6 0.14 1/ 1/
Carbon monoxide reacts with water at 1000 o C to give carbon dioxide and hydrogen with K c = 0.58. A reaction was started with the following composition: CO, 0.M M; H 1.0 M; H O, 0.50 M, and CO, 1.0 M. What are the equilibrium concentrations of all components? 1. Equation CO(g) + H O(g) CO (g) + H (g). Table CO H 0 CO H Initial 1.0 0.50 0. 1. Change x -x +x +x Equilibrium 1.0-x 0.50-x 0.+x 1..+x 3. Equilibrium Constant Expression 4. Substitute equilibrium values 5. Solve for x: K c 0.58 [ CO ][ H ] [ CO ][ H O ] [0.0 x][1. x] [1.00 x][0.50 x] Need quadratic equation x = 0.0 6. Substitute back into table [CO] = 1.00 0.0 = 0.98M [H O] = 0.50 0.0 = 0.48M [CO ] = 0.0+ 0.0 = 0.M [H ] = 1.0 + 0.0 = 1.M 7
Solving Quadratic Equations For equations: ax + bx + c = 0 From the previous problem: Solving for X x = 0.0 or 5.4 Only x = 0.0 makes sense x = 5.4 gives negative concentrations 8
Factors that Affect Chemical Equilibrium 9
Le Châtelier s Principle When stress is applied to a system at equilibrium, the system will shift to reduce the applied stress and reestablish equilibrium NO (g) 1 N O 4 (g) Addition of pressure causes reaction to shift towards products Number of moles decreases (decreases the stress). 30
General Rules for Le Châtelier s Principle 5 types of stresses will affect the equilibrium Change in Components or Concentrations Change in Partial Pressure Change in External Pressure Change in Total Volume Change in Temperature 31
Change in Components or Partial Pressures If the concentration of a reactant increases: Fe 3+ (aq) + SCN - (aq)fescn + (aq) yellow clear red [ FeSCN ] K c 3 [ Fe ][ SCN ] Denominator of the Q c increases Q c < K c To equalize Q c and K c, Concentration of other reactants decrease Concentration of the products increases Equilibrium is pushed right. 3
Exothermic Reaction Heat is released as a product reactant aa bb cc dd Change in Temperature Heat Endothermic Reaction Heat is needed as a Heat aa bb cc dd Increase T: K decreases Decrease T: K increases Increase T: K increases Decrease T: K Decreases NO (g) N O 4 (g) Exothermic Reaction red clear ΔH=-58kJ/mol Cold: more N O 4 K increases Hot: More NO K decreases 33
Change in External Pressure or Volume 3A B 1C 1D When pressure is applied (increased) Equilibrium shifts to produce smaller # of moles of gas When pressure is decreased Equilibrium shifts to produce larger # of moles of gas 3A B 3C D Reaction with same # of moles of gas on both sides Changes in external pressure do not affect equilibrium 34
Adding A Catalyst or An Inhibitor Catalysts The role of a catalyst is to lower activation energy. Catalysts and inhibitors do not affect the equilibrium Catalysts cause a reaction to reach equilibrium faster Inhibitors Inhibitors slow down the reaction to prevent equilibrium from being reached as quickly Inhibitors may slow the reaction rate to the point that there appears to be no reaction, but it is really just infinitely slow. 35
Which way does the reaction shift if N O 3 (g) NO(g) + NO (g) H o = +39.7 kj/mol NO is added? Shifts left to use up additional chemical Volume of reaction vessel reduced? Increased pressure: shifts left to reduce # moles The total internal pressure is increased by adding He gas? No shift: Total pressure on system increases, but no change in partial pressures of reactants or products. The temperature is increased? This an endothermic reaction, heat is absorbed. Increasing the temperature adds heat to the system. Reaction shifts right (endothermic) to remove heat. A catalyst is added? No shift: A catalyst does not affect the equilibrium composition 36
Summary Equilibrium is a balance between products and reactants Use stoichiometry to determine reactant or product ratios, but NOT reactant to product ratios. Capital K is used to represent the equilibrium constant Products over reactants raised to their stoichiometric coefficients Calculated from balanced equation, subscript designates units used, K c, K p No units used in final written K Equlibrium Calculations and Reaction Quotients ICE tables used to manipulate initial and equilibrium concentrations. Factors influencing K Concentration of chemicals and temperature affect all equilibria T, P, V changes affect K p in gases Catalysis Effect of catalysts and inhibitors, reasons they are used. 37