Chapter 6: Chemical Equilibrium

Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6. The Equilibrium Constant 6.3 Equilibrium Expressions Involving Pressures 6.4 The Concept of Activity 6.5 Heterogeneous Equilibria 6.6 Applications of the Equilibrium Constant 6.7 Solving Equilibrium Problems 6.8 LeChatelier s Principle 6.9 OMIT: Equilibria Involving Real Gases HWs due Satd at 6 PM: HW #7: Chapter 5, part HW #8: Chapter 6, part 1

A new HW assignment (HW#8) over 1 st part of Ch. 6 is also due this Satd.. The nd part of Ch. 6 (HW#9) will be due 7 AM on Friday, Nov. 19. Exam on Nov.. ------------------- All Discussion Sessions after 1:00 on Nov. 18 have been rescheduled to: Wed. Nov. 17th from :45 to 3:45 in Bagley Hall room 154 (run by Emily Sprafka, a TA) --------------------------- She will also hold a Review for Exam on Fri. Nov. 19 th from 3:30 to 5:00 PM in BAG 131.

Molecular Picture of Establishment of Equilibrium CO(g) + H O(g) CO (g) + H (g) INITIAL 7 CO(g) + 7 H O(g) + 0 CO (g) + 0 H (g) AFTER IT STOPS CHANGING CO(g) + H O(g) + 5 CO (g) + 5 H (g)

Chemical Equilibrium Previously, we assumed that a chemical reaction goes to completion as written. H O(g) + CO (g) H (g) + CO (g) In general this is not correct. Instead, a stable state of the system in reached, which includes both reactants and products. It is called the equilibrium state, or simply equilibrium. H O(g) + CO (g) H (g) + CO (g) How far toward completion it goes depends on the specific reaction, and on temperature.

Concentration vs. Time CO(g) + H O(g) CO (g) + H (g) Ratio depends on temperature

Characteristics of Chemical Equilibrium States Reaching equilibrium requires reactions to occur. Once reached, they show no macroscopic evidence of further change. Reached through dynamic balance of forward and reverse reaction rates.

Figure 5.1: The collision rate of gas particles defines the maximum blue-pink reaction rate! Z = Collision rate (of one pink with blues)= # collisions with blues per s = 4 [N blue /V] d (πrt/m) 1/ = 4 [blues] d (πrt/m) 1/

Equilibrium arises through dynamic balance between forward and back reactions Forward rate = k 1 [NO ][NO ] = k 1 [NO ] Back rate = k -1 [NO 3 ][NO] The rate constants k 1 and k -1 reflect probabilities that one collision leads to a successful reaction. k -1 Reverse: NO 3 (g) + NO(g) NO (g)

Kinetics of Approach to Equilibrium At eqbm.: Forward rate = back rate k1 1 3 k k [ NO ] = k [ NO][ NO ] [ NO][ NO ] 1 3 = [ ] = K 1 NO Equilibrium constant

The Equilibrium Constant - Definition Consider the generalized chemical reaction: a A + b B c C + d D A, B, C and D represent chemical species and a, b, c, and d are their stoichiometric coefficients in the balanced chemical equation. At equilibrium, c d K [ C] [ D] = [ ] a [ ] b Note: The units for K are A B concentration units raised to some power = c+d (a+b) The square brackets indicate the concentrations of the species in equilibrium. K is a constant called the equilibrium constant. K depends only on T, and not on concentrations.

Reaching Equilibrium on the Macroscopic and Molecular Level N O 4 (g) NO (g) Colorless Brown K = [NO ] / [N O 4 ] Units are mol/l.

Equilibrium from Different Starting Points K = [ CH3OH ] [ CO][ H ] has same value at equil. for all 3 starting points CO(g) + H (g) CH 3 OH(g)

Fe N (g) + 3 H (g) NH 3 (g) K = [NH 3 ] [N ][H ] 3

Key Stages in the Haber Synthesis of Ammonia Gerhard Ertl, 007 Nobel Prize in Chemistry

Equilibrium Constants can have a wide range of values Small K N (g) + O (g) NO (g) K = 1 x 10 30 Essentially only reactants at eqbm. (10 15 x products) Large K CO (g) + O (g) CO (g) K =. x 10 Essentially only products at eqbm. Intermediate K BrCl (g) Br (g) + Cl (g) K = 5 Comparable amounts of products and reactants at eqbm.

CS (g) + 3 O (g) CO (g) + SO (g) The equilibrium expression for a reaction written in reverse is the reciprocal of that for the original expression. CO (g) + SO (g) CS (g) + 3 O (g) K 1 = [ CO ][ ] SO [ CS ][ O ] 3 [ ][ ] 3 CS O 1 [ ][ ] CO SO K1 K = = If the original reaction is multiplied by a factor n, the new equilibrium constant is the original raised to the power n. CS (g) + 6 O (g) CO (g) + 4 SO (g) K 4 [ CO ] [ SO ] = = K ( ) 3 [ ] [ ] 6 1 CS O

Example: Calculation of the Equilibrium Constant from equilibrium amounts At 454 K, the following reaction takes place: 3 Al Cl 6 (g) Al 3 Cl 9 (g) At this temperature, the eqbm concentration of Al Cl 6 (g) is 1.00 M and the equilibrium concentration of Al 3 Cl 9 (g) is 1.0 x 10 - M. Compute the equilibrium constant at 454 K. Strategy: Substitute values into K K ( 1.0 10 M ) [ ] Al3Cl9 4 = = = 1.04 10 M [ ] 3 ( ) 3 AlCl6 1.00M 1

Like Example 6.1 (P 01) - I The following equilibrium concentrations were observed for the reaction between CO and H to form CH 4 and H O at 97 o C: CO (g) + 3 H (g) CH 4 (g) + H O (g) [CO] = 0.613 mol/l [CH 4 ] = 0.387 mol/l [H ] = 1.839 mol/l [H O] = 0.387 mol/l a) Calculate the value of K at 97 o C for this reaction. b) Calculate the value of the equilibrium constant at 97 o C for: H O (g) + CH 4 (g) CO (g) + 3 H (g) c) Calculate the value of the equilibrium constant at 97 o C for: 1/3 CO (g) + H (g) 1/3 CH 4 (g) + 1/3 H O (g) Solution: a) For the first balanced reaction above: [CH 4 ][H O] K = [CO] [H = ] 3

Like Example 6.1 (P 01) - I The following equilibrium concentrations were observed for the Reaction between CO and H to form CH 4 and H O at 97 o C. CO (g) + 3 H (g) CH 4 (g) + H O (g) [CO] = 0.613 mol/l [CH 4 ] = 0.387 mol/l [H ] = 1.839 mol/l [H O] = 0.387 mol/l a) Calculate the value of K at 97 o C for this reaction. b) Calculate the value of the equilibrium constant at 97 o C for: H O (g) + CH 4 (g) CO (g) + 3 H (g) c) Calculate the value of the equilibrium constant at 97 o C for: 1/3 CO (g) + H (g) 1/3 CH 4 (g) + 1/3 H O (g) Solution: a) For the first balanced reaction above: [CH 4 ][H O] (0.387 mol/l) (0.387 mol/l) K = [CO] [H = = 0.0393 L /mol ] 3 (0.613 mol/l) (1.839 mol/l) 3

Like Example 6.1 (P 01) - II b) Calculate the value of the equilibrium constant at 97 o C for: [CO] [H K = ] 3 = [H O] [CH 4 ] H O (g) + CH 4 (g) CO (g) + 3 H (g) Easier way: This K is just the reciprocal of K from part a: 1 1 K = K a c) Calculate the value of the equilibrium constant at 97 o C for: [H K = O] 1/3 [CH 4 ] 1/3 = [CO] 1/3 [H ] 1/3 CO (g) + H (g) 1/3 CH 4 (g) + 1/3 H O (g) Easier way: K = (K from part a) 1/3

Like Example 6.1 (P 01) - II b) Calculate the value of the equilibrium constant at 97 o C for: H O (g) + CH 4 (g) CO (g) + 3 H (g) [CO] [H K = ] 3 (0.613 mol/l) (1.839 mol/l) 3 = = 5.45 mol /L [H O] [CH 4 ] (0.387 mol/l) (0.387 mol/l) Easier way: This K is just the reciprocal of K from part a: K = 1 K = 1 = 5.45 mol /L a 0.0393 L /mol c) Calculate the value of the equilibrium constant at 97 o C for: 1/3 CO (g) + H (g) = 1/3 CH 4 (g) + 1/3 H O (g) [H K = O] 1/3 [CH 4 ] 1/3 (0.387mol/L) = 1/3 (0.387 mol/l) 1/3 [CO] 1/3 [H ] (0.613 mol/l) 1/3 (1.839 mol/l) Easier way: K = (K a ) 1/3 = (0.0393L /mol ) 1/3

Determining Equilibrium Concentrations from K Example: Methane can be made by reacting carbon disulfide with hydrogen gas. K for this reaction is 7.8 (L/mol) at 900 C. CS (g) + 4 H (g) CH 4 (g) + H S (g) At equilibrium the reaction mixture in a 4.70 L flask contains 0.50 mol CS, 1.10 mol of H, and 0.45 mol of H S. How much methane was formed? Strategy: (1) Calculate the equilibrium concentrations from the moles given and the volume of the container. ()Use the value of K to solve for the concentration of methane. (3) Calculate the number of moles of methane from M and V.

CS (g) + 4 H (g) CH 4 (g) + H S(g) 0.50 mol CS = = 0.053 M 4.70 L [ ] 0.450 mol HS = = 0.0957 M 4.70 L [ ] 1.10 mol H = = 0.34 M 4.70 L [ ] [ CH ][ H S] 4 K = = 4 [ CS ][ H ] 7.8 L/mol ( ) [ ] [ ][ ] [ ] 4 4 ( ) [ ] [ ] M M [ M ] CS H 0.053 0.34 CH 4 = K = 7.8 L/mol = 0.484M H S 0.0957 moles CH 4 = M V = -1 0.484 moles L 4.70 L=.7 moles

K P : Equilibrium Constant Involving Pressures For a reaction of the type a A + b B c C + d D where the species are gases, It is sometimes convenient to write the equilibrium expression in terms of partial pressures: K ( ) c ( ) d P P = C D P ( ) a ( ) b P P A B P indicates the partial pressures of the species in equilibrium and K P is a constant called the equilibrium constant in terms of partial pressures. K P depends only on T, and not on pressure.

Expressing K with Pressure Units For gases, P i V = n i RT can be rearranged to give: P i = n i RT V or: n i = P i n i = Molar concentration of gas i = [i] V RT V For an equilibrium between gaseous compounds, there is a relationship between the equilibrium constants K and K P. Example: NO (g) + O (g) NO (g) K = [NO ] [NO] [O ] and K p = P NO P NO P O Using concentrations Using pressures

How is K P related to K? Answer: Through the use of the ideal gas law. For all species i: K ni PV i = nirt or Pi = RT = CiRT V ( ) c ( ) d ( ) c ( ) d PC PD CC RT CD RT P = = a b ( ) a ( ) b CA RT CB RT ( P ) ( ) A PB c d ( ) c+ d C D c+ d ( a+ b) Δn K( RT) K( RT) a b a+ b ( ) ( ) C C RT = = = RT ( C ) ( ) A CB ( ) Where Δn = c+d-(a+b) Note that K =K P when Δn = 0 NOTE: Δn here refers ONLY to the gases in rxn.

For reaction NO(g) + O (g) NO (g) Sum of gas coefficients (+ for products, for reactants) Δn = 1 = 1 K = K [ NO ] [ NO] [ O ] P = P P NO NO P O K P 1 = K( RT) = K RT

Unitless method for expressing K and K P Units for K are concentration raised to power Δn, and those for K P are pressure units raised to power Δn. In later chapters, we will divide each concentration in K by a reference concentration of 1 molar, and each partial pressure in K P by a reference pressure of 1 atm. This renders all concentrations and pressures unitless, at which point we will call them both activities (a i ). Just replace all [ i ] or P i with a i in K or K P.

Use of Activities in Equilibrium Constant Expressions For Pressures: Activity ( ith component) where ref P i = partial pressure of the ith gaseous component and P = 1 atm (exactly) = a = i P P i ref : Reference state with unit activity For Concentration: Activity ( ith component) = a = where and M ref M = i = molarity of the ith component 1 M (exactly) i M M i ref : Reference state with unit activity

For pure solids and liquids, activity = 1.0 CaCO 3 (s) CaO (s) + CO (g) K P = a CaO(s) P CO /a CaCO3(s) = P CO (The solids do NOT appear in K P! ) 1.0 1.0 The position of an equilibrium involving a pure solid or liquid does not depend upon the amounts of pure solid or liquids present. The activity of a pure solid or liquid is always equal to 1 if present, so its concentration does not appear in K expression.

Equilibrium involving pure solids or liquids: set their activity = 1 Example: NH 4 NO (s) N (g) + H O(g) The equilibrium constant for this reaction would normally be expressed as: K ' = [ N ][ ] HO [ NH NO] 4 However, a pure solid or liquid retains the same activity during the reaction. Thus we set the activity of NH NO ( s) to one. 4 [ N ][ ] ( )( ) HO and PN PH O K = K p = 1.0000 1.0000

[ C] [ D] [ A] [ B] The Reaction Quotient, Q Consider the reaction: a A( g) + b B( g) = c C( g) + dm D( g) The reaction quotient, Q is defined as Q = c t a t d t b t where the subscripts t indicate momentary concentrations at some time t before (or after) equilibrium has established. Q has same form as K, but the concentrations are the actual concentrations at any time t rather than the concentrations after equilibrium is reached.

Reaching Equilibrium on the Macroscopic and Molecular Level N O 4 (g) NO (g) Colorless Brown Q = [NO ] / [N O 4 ] Units are mol/l.

NO(g) Q = 4 [ NO ] [ NO] 4 NO(g) TIME

Reaction Direction and the Relative Sizes of Q and K Excess reactants initially Excess products initially

Which Direction? If Q < K, the system will shift to the right by converting reactants to products. If Q = K, the system is at equilibrium; the concentrations will not change. If Q > K, the system will shift to the left by converting products back to reactants.

Example: Predict Reaction Direction using Q For the following reaction: CH 4 (g) + H S(g) CS (g) + 4 H (g) 1.00 mol CH 4, 1.00 mol CS,.00 mol H S and.00 mol H are mixed in a 50 ml vessel at 960 o C. At this temperature, K = 0.036. In which direction will the reaction proceed in order to reach equilibrium? Strategy: 1) Calculate the actual concentrations ) Calculate Q. 3) Compare Q and K

Solution: Predict Reaction Direction using Q 1) [CH 4 ] 0 = 1.00 mol/0.50 L = 4.00 M. [H S] 0 =.00 mol/0.50 L = 8.00 M, [CS ] 0 = 1.00 mol/0.50 L = 4.00 M, and [H ] 0 =.00 mol/0.50 L = 8.00 M. ) Calculate the value of Q: [ ] [ ] [ CH ] [ H S] ( ) 4 4 0 0 4.00 (8.00) 4 0 0 CS H 4.00 8.00 Q = = = 64.00 >> 0.036 (3) Comparing Q and K: Q > K, so the reaction goes to the left. Therefore reactants concentrations increase and products concentrations decrease.

Writing the Reaction Quotient from the Balanced Equation Problem: Write the reaction quotient for each of the following reactions: (a) The thermal decomposition of potassium chlorate: KClO 3 (s) KCl (s) + O (g) (b) The combustion of butane in oxygen: C 4 H 10 (g) + O (g) CO (g) + H O (g) Plan: We first balance the equations, then construct the reaction quotient Solution: (a) KClO KCl 3 (s) (s) + 3 O [KCl] [O (g) Q c = ] 3 = [O ] 3 [KClO 3 ] (b) C 4 H 10 (g) + 13 O (g) 8 CO (g) + 10 H O (g) Q c = [CO ]8 [H O] 10 [C 4 H 10 ] [O ] 13

Reaction Direction and the Relative Sizes of Q and K Excess reactants initially Excess products initially

Example: Calculating Equilibrium pressures and concentrations from K and initial conditions. Consider the equilibrium: CO(g) + H O(g) CO (g) + H (g) 0.50 mol CO and 0.50 mol H O are placed in a 15 ml flask at 900 K. What is the composition of the equilibrium mixture if K = 1.56? The original reactant concentrations are: [CO] 0 = [H O] 0 = 0.50 mol/ 0.15 L =.00 M Q = 0.Therefore, Q < K, so reactants are consumed and products made.

CO(g) + H O(g) CO (g) + H (g) Construct the reaction table: Conc. (M) CO(g) H O(g) CO (g) H (g) Init. [i].00.00 0 0

CO(g) + H O(g) CO (g) + H (g) Conc. (M) CO(g) H O(g) CO (g) H (g) Init. [i].00.00 0 0 Change = x times coefficient = neg. for reactants -x -x +x +x

CO(g) + H O(g) CO (g) + H (g) Conc. (M) CO(g) H O(g) CO (g) H (g) Init. [i].00.00 0 0 Change = x times coefficient = neg. for reactants Eqbm. [i] = Init.+Change -x -x +x +x.00 - x.00 - x 0+x = x 0+x = x Called an I.C.E. table.

Substitute [i] eqbm into the equilibrium expression: K [ CO][ H] [ ][ ] ( x)( x) x ( x)( x) ( x) = = = = 1.56 CO H O.00.00.00 Taking the square root of both sides: x = 1.56 =± 1.5.00 x Since only the positive root meaningful, ignore the negative root: x 1.5 =.00 x x = 1.11 M

Calculating equilibrium concentrations: [ ] [ ] CO = H O =.00 x =.00 M - 1.11 M = 0.89 M [ ] [ ] CO = H = x = 1.11 M Check results: K [ ][ ] [ ][ ] CO H 1.11 1.11 = = = 1.6 1.56 CO H O 0.89 0.89

Solving Equilibrium Problems Write the balanced equation for the reaction. Write the equilibrium expression K. List the initial concentrations, [i] initial. Calculate Q and determine the direction of shift to equilibrium. Define change in [i] to reach equilibrium: Change = x times the stoich. coefficient Express equilibrium concentrations in terms of x. Substitute the equilibrium concentrations into the equilibrium expression for K. Solve for x and calculate [i] eqbm for all i. Check the solution by calculating K and making sure it is identical to the original K.

Like Example 6.3 (P10) - I Calculate the equilibrium concentrations when Hydrogen Chloride gas is made from Hydrogen gas and Chlorine gas. Initially 4.000 mol of H, and 4.000 mol of Cl, are added to.000 mol of gaseous HCl in a.000 liter flask. H (g) + Cl (g) HCl (g) K =.76 x 10 [HCl] = [Cl ] = [H ] = 4.000mol/.000L =.000M [H ] [Cl ] [HCl] =.000 mol/.000l = 1.000M Initial Concentration Change Equilibrium Conc. (mol/l) (mol/l) (mol/l) [H ] o =.000M -x [H ] = [Cl ] o =.000M [Cl ] = [HCl] o = 1.000M [HCl] =

Like Example 6.3 (P10) - II [HCl] K =.76 x 10 = [H = ] [Cl ] Solve for x:

Like Example 6.3 (P10) - II [HCl] (1.000 + x) K =.76 x 10 = [H = = ] [Cl ] (.000 x)(.000 x) (1.000 + x) (.000 x) Take the square root of each side: (1.000 + x) 16.61 = (.000 x) 33. 16.61x = 1.000 + x Therefore: [H ] = 0.69 M 3. = 18.61x [Cl ] = 0.69 M x = 1.731 [HCl] = 4.46 M Check: [HCl] [H ] [Cl ] (4.46) = = 76 OK! (0.69)(0.69)

Example: Calculate equilibrium partial pressures using exact solution of Quadratic Equation The reaction between nitrogen and oxygen to form nitric acid proceeds according to the reaction: -4 N (g) + O (g) NO(g) K = 4.10 10 at 000. K 0.500 moles of N and 0.860 moles of O are put into a.00 L vessel initially. Calculate the partial pressure of all the species at equilibrium.

Initial concentration of N = 0.500 mol.00 L = 0.50 M, Initial concentration of O = 0.860 mol.00 L = 0.430 M, Initial concentration of NO = 0 Q =0; therefore the reaction proceeds to the right Construct the reaction table: Molarity N (g) + O (g) NO(g) N (g) O(g) Initial 0.50 0.430 0 NO(g)

N (g) + O (g) NO(g) Molarity N (g) O (g) NO(g) Initial 0.50 0.430 0 change -x -x x

N (g) + O (g) NO(g) Molarity N (g) O (g) NO(g) Initial 0.50 0.430 0 change -x -x x At equib 0.50-x 0.430-x x

Substituting the equilibrium concentrations from the table into the equilibrium expression: K [ ] [ ][ ] ( x) ( x)( x) NO = = = 4.10 10 N O 0.50 0.430 This expression simplifies to 4 5 4.00x.78 10 x 4.41 10 0 This is a quadratic equation of the general form ax bx c + = + + = 0 where the (two) roots can be obtained from the quadratic formula: 4 x ± = b b ac ± a 4

We obtain two possible solutions x = 3.35 10 and x = 3.8 10 [ NO] [ ] [ ] -3-3 Since only the positive root leads to all positive concentrations, we ignore the negative root. Calculating equilibrium concentrations: = x= 6.56 10 N = 0.50M x= 0.47M O = 0.430M x= 0.47M Check: K [ NO] [ ][ ] -3 M ( 6.56 10 ) -3 = = = 4.09 10 4.10 10 N O 0.47 0.47 4 4

Example: Solving equilibrium problems with simplifying assumptions Phosgene decomposes into CO and Cl when heated according to the equation. COCl (g) -4 o CO(g) + Cl (g) K = 8.3 10 at 360 C Calculate the concentration of all species at equilibrium if 5.00 moles of phosgene is placed into a 10.0 L flask 5.00 moles COCl (g) = = 0.500M 10.0 L [ ]

COCl (g) -4 o CO(g) + Cl (g) K = 8.3 10 at 360 C Molarity COCl (g) CO(g) Cl (g) Init. 0.500 0 0 Change Equil.

COCl (g) -4 o CO(g) + Cl (g) K = 8.3 10 at 360 C Molarity COCl (g) CO(g) Cl (g) Init. 0.500 0 0 Change -x +x +x Equil.

COCl (g) -4 o CO(g) + Cl (g) K = 8.3 10 at 360 C Molarity COCl (g) CO(g) Cl (g) Init. 0.500 0 0 Change -x +x +x Equil. 0.500 - x x x

K [ CO][ Cl] [ ] x = = = 8.3 10 COCl 0.500 x For K to be so small, 0.500 and so 0.500 0.500 x x = 8.3 10 0.500 x 0.500 x =.037 10.04 10 Exact solution is x = 1.996 10.00 10 Approximation gives error of % 4 x >> x x 4

Calculating K from Concentration Data I Problem: Hydrogen iodide decomposes at moderate temperatures by the reaction below: HI (g) H (g) + I (g) When 4.00 mol HI was placed in a 5.00 L vessel at 458 C, the equilibrium mixture was found to contain 0.44 mol I. What is the value of K c? Plan: First we calculate the molar concentrations, and then put them into the equilibrium expression to find it s value. Solution: To calculate the concentrations of HI and I, we divide the amounts of these compounds by the volume of the vessel. Starting conc. of HI = 4.00 mol = 0.800 M 5.00 L Equilibrium conc. of I = 0.44 mol = 0.0884 M 5.00 L Conc. (M) HI (g) H (g) I (g) Starting 0.800 0 0 Change Equilibrium 0.0884

Calculating K from Concentration Data II [HI] = [H ] = K c = = = [H ] [I ] [HI]

Calculating K from Concentration Data II [HI] = M = (0.800 - x 0.0884) M = 0.63 M [H ] = x = 0.0884 M = [I ] [H K c = ] [I ] = ( 0.0884)(0.0884) = 0.001 [HI] (0.63) Therefore the equilibrium constant for the decomposition of Hydrogen Iodide at 458 C is only 0.001 meaning that the decomposition does not proceed very far under these temperature conditions. We were given the initial concentrations, and that of one at equilibrium, and found the others that were needed to calculate the equilibrium constant.

Using the Quadratic Formula to Solve for the Unknown Given the Reaction between CO and H O: Concentration (M) CO (g) + H O (g) CO (g) + H (g) Initial.00 1.00 0 0 Change -x -x +x +x Equilibrium.00-x 1.00-x x x [CO Q c = ][H ] = (x) (x) = x = 1.56 [CO][H O] (.00-x)(1.00-x) x - 3.00x +.00 We rearrange the equation: 0.56x - 4.68x + 3.1 = 0 ax + bx + c = 0 quadratic equation: x = - b + b -4ac a [CO] = 1.7 M [H x = 4.68 + (-4.68) O] = 0.7 M - 4(0.56)(3.1) = 7.6 M [CO ] = 0.73 M (0.56) and 0.73 M [H ] = 0.73 M

Le Chatelier s Principle If a change in conditions (a stress such as change in P, T, or concentration) is imposed on a system at equilibrium, the equilibrium position will shift in a direction that tends to reduce that change in conditions. Henri Le Chatelier, 1884

The Effect of a Change in Concentration I Given an equilibrium equation such as : CH 4 (g) + NH 3 (g) HCN (g) + 3 H (g) If one adds ammonia to the reaction mixture at equilibrium, it will force the reaction to go to the right producing more product. Likewise, if one takes ammonia from the equilibrium mixture, it will force the reaction back to produce more reactants by recombining H and HCN to give more of the initial reactants, CH 4 and NH 3. CH 4 (g) + NH 3 (g) Add NH 3 HCN (g) + 3 H (g) Forces equilibrium to produce more product. CH 4 (g) + NH 3 (g) HCN (g) + 3 H (g) Remove NH 3 Forces the reaction equilibrium to go back to the left and produce more of the reactants.

The Effect of a Change in Concentration II CH 4 (g) + NH 3 (g) HCN (g) + 3 H (g) If to this same equilibrium mixture one decides to add one of the products to the equilibrium mixture, it will force the equilibrium back toward the reactant side and increase the concentrations of reactants. Likewise, if one takes away some of the hydrogen or hydrogen cyanide from the product side, it will force the equilibrium to replace it. CH 4 (g) + NH 3 (g) Forces equilibrium to go toward the reactant direction. HCN (g) + 3 H (g) Add H CH 4 (g) + NH 3 (g) Forces equilibrium to make more produce and replace the lost HCN. HCN (g) + 3 H (g) Remove HCN

The Effect of a Change in Concentration If a gaseous reactant or product is added to a system at equilibrium, the system will shift in a direction to reduce the concentration of the added component. If a gaseous reactant or product is removed from a system at equilibrium, the system will shift in a direction to increase the concentration of the removed component. NO (g) k 3 [ NO ] = k [ NO][ NO ] 1 1 3 k k 1 = [ NO][ NO ] [ NO ] 1 NO(g) + NO (g) 3

Examples: The Effect of a Change in Concentration Consider the following reaction: H S(g) + O (g) S(s) + H O(g) What happens to: (a) [H O] if O is added? The reaction proceeds to the right so H O increases. (b) [H S] if O is added? Some H S reacts with the added O to move the reaction to the right, so [H S] decreases.

H S(g) + O (g) S(s) + H O(g) (c) [O ] if H S is removed? The reaction proceeds to the left to re-form H S, more O is formed as well, O increases. (d) [H S] if S(s) is added? S is a solid, so its activity does not change. Thus, [H S] is unchanged.

The Effect of a Change in Pressure (Volume) Pressure changes are mainly involving gases as liquids and solids are nearly incompressible. For gases, pressure changes can occur in three ways: Changing the concentration of a gaseous component Adding an inert gas (one that does not take part in the reaction) Changing the volume of the reaction vessel When a system at equilibrium that contains a gas undergoes a change in pressure as a result of a change in volume, the equilibrium position shifts to reduce the effect of the change. If the volume is lower (pressure is higher), the total number of gas molecules decrease. If the volume is higher (pressure is lower), the total number of gas molecules increases.

Examples: The Effect of a Change in Pressure How would you change the total pressure or volume in the following reactions to increase the yield of the products: (a) CaCO 3 (s) CaO(s) + CO (g) The only gas is the product CO. To move the reaction to the right increase the volume. (b) S(s) + 3 F (g) SF 6 (g) With 3 moles of gas on the left and only one on the right, we increase the pressure to form more SF 6. (c) Cl (g) + I (g) ICl (g) The number of moles of gas is the same on both sides of the equation, so a change in pressure or volume will have no effect.

Figure 6.9: Brown NO (g) and colorless N O 4 (g) at equilibrium in a syringe NO (g) N O 4 (g) Brown Colorless Source: Ken O Donoghue

The Effect of a Change in Temperature Only temperature changes will alter the equilibrium constant, and that is why we always specify the temperature when giving the value of K c. The best way to look at temperature effects is to realize that energy is a component of the equation, the same as a reactant, or product. For example, if you have an exothermic reaction, heat (energy) is on the product side of the equation, but if it is an endothermic reaction, it will be on the reactant side of the equation. O (g) + H (g) H O (g) + Energy = Exothermic Electrical energy + H O (g) H (g) + O (g) = Endothermic A temperature increase favors the endothermic direction and a temperature decrease favors the exothermic direction.

Shifting the N O 4 (g) and NO (g) equilibrium by changing the temperature

Exothermic Reactions Release energy upon reaction. Treat energy as a reaction product. Use Le Chatelier s principle. Increasing T adds energy to system. The equilibrium shifts towards the reactants. (System absorbs energy) The value of K decreases in consequence. Note that change in concentration at constant T changes equilibrium position, but not value of K.

Endothermic Reactions Absorb energy upon reaction. Treat energy as a reactant. Use Le Chatelier s principle. Increasing T adds energy to system. The equilibrium shifts towards the products. (System absorbs energy) The value of K increases in consequence.

Example: How does an increase in temperature affect the equilibrium concentration of the indicated substance and K for the following reactions: (a) CaO(s) + H O (l) Ca(OH) (aq) + energy Increasing T adds thermal energy. This shifts the system to the left, where it absorbs energy. [Ca(OH) ] and K decrease. (b) CaCO 3 (s) + energy CaO(s) + CO (g) Increasing T adds thermal energy. This shifts the system to the right, where it absorbs energy.[co ] and K increase. (c) SO (g) + energy S(s) + O (g) Increasing T adds thermal energy. This shifts the system to the right, where it absorbs energy.[so ] will decrease and K increases.

Effect of Various Stresses on an Equilibrium System Stress Net Direction of Reaction Effect on Value of K Concentration Increase [reactant] Toward formation of products Decrease [reactant] Toward formation of reactants None None Volume (of closed container holding reaction at constant T) Increase V Toward formation of larger amount (mol) of gas unless Δn = 0 None Decrease V Toward formation of smaller amount (mol) of gas unless Δn = 0 None Temperature - Adding energy in form of heat (Opposite if T decrease ): Increase T Toward formation of products if energy K increases is a reactant (endothermic reaction) with T increase Toward formation of reactants if energy K decreases is a product (exothermic reaction) with T increase

CoCl - 4 (aq) + 6 H O(l) blue Co(H O) + 6 (aq) + 4 Cl - (aq) pink [Co(H O) + 6 ][Cl - ] 4 Q =, which initially = K [CoCl - 4 ] DEMO!

Ways of Expressing the Reaction Quotient, Q Form of Chemical Equation Form of Q Value of K Reference reaction: A B Q (ref) = [B] K (ref) = [A] Reverse reaction: B A Q = 1 = [A] K = Q (ref) [B] Reaction as sum of two steps: (1) A C Q [C] [B] 1 = ; Q [A] = [C] 1 K (ref) [B] eq [A] eq () C B Q overall = Q 1 x Q = Q (ref) K overall = K 1 x K [C] [B] [B] = x = = K [A] [C] [A] (ref) Coefficients multiplied by n Q = Q n (ref) K = K n (ref) Reaction with pure solid or Q = Q (ref) [A] = [B] K = K (ref) [A] = [B] liquid component, such as A (s)

Predicting the Effect of a Change in Concentration on the Position of the Equilibrium Problem: Carbon will react with water to yield carbon monoxide and and hydrogen, in a reaction called the water gas reaction that was used to convert coal into a fuel that can be used by industry. C (s) + H O (g) CO (g) + H (g) What happens to: (a) [CO] if C is added? (c) [H O] if H is added? (b) [CO] if H O is added? (d) [H O] if CO is removed? Plan: We either write the reaction quotient to see how equilibrium will be effected, or look at the equation, and predict the change in direction of the reaction, and the effect of the material desired. Solution: (a) No change, as carbon is a solid, and not involved in the equilibrium, as long as some carbon is present to allow the reaction. (b) The reaction moves to the product side, and [CO] increases. (c) The reaction moves to the reactant side, and [H O] increases. (d) The reaction moves to the product side, and [H O] decreases.