Unit 5: Chemical Kinetics and Equilibrium UNIT 5: CHEMICAL KINETICS AND EQUILIBRIUM

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1 UNIT 5: CHEMICAL KINETICS AND EQUILIBRIUM Chapter 14: Chemical Kinetics 14.4 & 14.6: Activation Energy, Temperature Depenence on Reaction Rates & Catalysis Reaction Rates: - the spee of which the concentration of a reactant or prouct changes over time. Collision Moel: - a moel that state for a reaction to occur, molecules must collie with each other. Factors Affecting the Collision Moel: 1. Activation Energy (E a ): - the threshol energy molecules neee to overcome to cause a chemical reaction that was first propose by Svante Arrhenius. - E a is the highest energy (top of the hill - E max ) minus the sum of energy of the reactants (ΣH reactants ) on the potential energy iagram. - in general, the Lower the Activation Energy, the Faster is the Rate of Reaction. E a E a Activation Energy of an Enothermic Reaction Activation Energy of an Exothermic Reaction Temperature (T): - the effective number of collisions increases exponentially with temperature. Because T > T 1, there are more molecules colliing with enough Kinetic Energy causing a reaction to occur. In general, the Higher the Temperature, the Faster the Rate of Reaction. age 17. Copyrighte by Gabriel Tang B.E., B.Sc.

2 . article Size: - the Smaller the article Size (the Larger the Surface Area expose), the Faster the Reaction Rate. Example 1: Grain sugar issolves faster than equal mass of sugar cubes because of smaller particle size an therefore increase surface area of the grain sugar. 4. Concentration: - the higher the Concentration (the more molecules in an available space, the higher the chance of collision), the Faster the Reaction Rate. 5. Catalyst: - a substance that spees up the reaction without being consume in the reaction. - unlike intermeiates, catalyst is use an recycle in the reaction. - lowers activation energy by proviing an alternate reaction pathway. (ΔE a is lowere but ΔH rxn remains the same.) - in general, the Aition of a Catalyst INCREASES the Rate of Reaction. Example : Enzyme is a catalyst in the boy that spees up certain boily reaction. E a without Catalyst E a with Catalyst ΔH rxn 6. Inhibitor: - a substance that inhibit the function of a catalyst to spee up the reaction. - in general, the Aition of an Inhibitor DECREASES the Reaction Rate. Example : Ammonia is forme from its elements using heterogeneous catalyst (catalyst in ifferent phase compare to the reactants) such as t (s): N (g) + H (g) t NH (g) (Check out Vieo at Example 4: The catalytic converter converts NO (g) (result of burning nitrogen at high temperature) to N (g) an O (g). The O (g) along with the catalytic converter is use to prouce CO (g) from CO (g). catalytic NO (g) converter catalytic CO (g) + O (g) converter N (g) + O (g) CO (g) (Left): A catalytic converter for most moern vehicle. Leae gasoline, an inhibitor, eactivates catalytic converter. Hence, it is not legally use in vehicles Copyrighte by Gabriel Tang B.E., B.Sc. age 17.

3 Example 5: The estruction of ozone in upper atmosphere can be attribute to NO (g) an CFCs acting as homogeneous catalysts (catalysts in the same phases as the reactants). NO (g) is prouce from the combustion of N (g) at high temperature commonly foun in internal combustion engine (high-altitue aircraft prouces lots of NO gas). CFCs (Chloro-Fluoro-Carbon compouns) are foun in aerosol can propellants, refrigerators, an air conitioners. They break own to form Cl (g) with the presence of light. NO (g) + O (g) NO (g) + O (g) O (g) + NO (g) NO (g) + O (g) O (g) + O (g) O (g) Cl (g) + O (g) ClO (g) + O (g) O (g) + ClO (g) Cl (g) + O (g) O (g) + O (g) O (g) (Above) The Ozone Hole over the South ole (Sept 000). A similar hole is present over the Arctic. (Left) rocess of Ozone Depletion. Ozone blocks harmful UV rays that can otherwise cause skin cancer. Factors Affecting Reaction Rate 1. Activation Energy (E a ) Reaction Rate. Temperature (T) Reaction Rate. article Size (Surface Area ) Reaction Rate 4. Concentration Reaction Rate 5. Catalyst Reaction Rate 6. Inhibitor Reaction Rate age 174. Assignment 14.4 & 14.6: pg #7, 0, 1, 49 to 5, 75a, 75c, 76 Copyrighte by Gabriel Tang B.E., B.Sc.

4 Chapter 15: Chemical Equilibrium 15.1: The Concept of Equilibrium an the Equilibrium Constant Chemical Equilibrium: - the state at which the concentrations of all reactants an proucts remain constant with time (the Forwar Reaction Rate Reverse Reaction Rate). - the equilibrium state is ynamic (not static). Chemical species are continuously converting from reactants to proucts an vice versa. It appears that the reaction has stoppe only because the rate of consumption rate of prouction. - if an equilibrium state is isturbe (changing concentrations of species, pressure, volume an temperature change), the reaction will shift towars one sie in orer to re-establish the new equilibrium state. - like reaction rate, Equilibrium is affecte by Temperature. aa + bb cc + D a an b: Coefficients of reactant species A an B c an : Coefficients of prouct species C an D equilibrium symbol (inicating forwar an reverse reactions) Equilibrium State [D] Concentration [C] [B] [A] Time to reach Time Equilibrium (Check out Vieo at The N O 4 (g) NO (g) equilibrium. Initially (icture A), there were very little NO (g). As time proceee forwar, more NO (g) (brown color) is prouce (ictures B & C). However, there are still N O 4 (g) present at the equilibrium state (icture D). Copyrighte by Gabriel Tang B.E., B.Sc. age 175.

5 Law of Mass Action: - a reversible reaction at equilibrium an a constant temperature, a certain ratio of reactant an prouct concentrations (Equilibrium Expression) has a constant value, K equilibrium constant. Equilibrium Expression: - an expression relating the concentrations or pressures of the reactants an proucts when they are at the state of equilibrium. - it takes the form of the iniviual proucts raise to the power of their respective coefficients ivie by the iniviual reactants raise to the power of their respective coefficients. - the equilibrium expression is unique for each reaction, but it is the same for that particular reaction regarless of temperature. Equilibrium Constant (K): - the unitless numerical value of the equilibrium expression. - the equilibrium constant is the same for a particular reaction if it remains at the same temperature. Equilibrium Expression an Constant of a Reaction aa + bb cc + D K c C D [ A] a [ B] b Equilibrium Expressions a an b: Coefficients of reactant species A an B c an : Coefficients of prouct species C an D [A], [B], [C] & [D] Equilibrium Concentrations of Chemicals K Equilibrium Constant (Concentrations) Assignment 15.1 pg. 50 #1 to 4 15.: Ways of Expressing Equilibrium Constants Homogeneous Equilibria: - an equilibrium system where all chemical species are in the same phase. Equilibrium Constant (K): - the symbol for equilibrium constant when the expression eals with concentrations is simply K or K c. When the expression eals with pressures, it is symbolize as K. - for equilibrium of an ieal system, the activity of a substance is the ratio of its concentration or partial pressure to a stanar value of 1 M or 1 atm. The proceure eliminates all units but oes not alter the numerical value of the concentrations an pressures. Therefore, K has no units. age 176. Copyrighte by Gabriel Tang B.E., B.Sc.

6 Equilibrium Expression an Constant of a Reaction aa + bb cc + D K K c c C D [ A] a [ B] b Example 1: Write the equilibrium expression of the following reactions. a. Br (g) + 5 F (g) BrF 5 (g) b. N H 4 (g) + 6 H O (g) NO (g) + 8 H O (g) K a an b: Coefficients of reactant species A an B c an : Coefficients of prouct species C an D [A], [B], [C] & [D] [A, [B, [C & [D Equilibrium Concentrations of Chemicals (mol/l) A, B, C & D A, eq, B, eq, C, eq & D, eq Equilibrium ressures of Chemicals (atm) K or K c Equilibrium Constant (Concentrations) K [ BrF5 ] [ Br ][ F ] 5 BrF 5 K 5 Br Both K c an K are Unitless F c C a A D b B K Equilibrium Constant (ressures) 8 NO H O K [ N H ][ H O ] NO H K 6 N H 4 O H O Equilibrium osition: - the concentrations or pressures of all chemical species at equilibrium state. - epens strongly on the Initial Concentrations of the chemical species. (In contrast, K oes NOT epen on initial concentrations, only on temperature an the specific reaction.) - since there all many possible initial concentrations for any one reaction, there are infinite number of equilibrium position for a particular reaction. Note: Do NOT confuse initial concentrations [A] 0 with equilibrium concentration [A!! We only use Equilibrium Concentrations to calculate K by substituting them into the equilibrium expression. Example : The formation of HI (g) is an equilibrium reaction. Several experiments are performe at 710 K using ifferent initial concentrations. H (g) + I (g) HI (g) Experiment 1 Experiment Initial Equilibrium Initial Equilibrium [H ] M [H 0.0 M [H ] 0 0 M [H M [I ] M [I 0.0 M [I ] M [I M [HI] 0 0 M [HI M [HI] M [HI 0.80 M Experiment Experiment 4 Initial Equilibrium Initial Equilibrium [H ] M [H M [H ] 0 0 M [H M [I ] 0 0 M [I M [I ] 0 0 M [I M [HI] M [HI M [HI] M [HI 0.11 M a. Write the equilibrium expression for the formation of HI (g). b. Calculate the equilibrium constant for each experiment, an average them for an overall value. Copyrighte by Gabriel Tang B.E., B.Sc. age 177.

7 a. Equilibrium Expression: K b. Equilibrium Constant: [ HI] eq Experiment 1: K 1 H I [ [ [ HI] [ H ][ I ] ( 0.156) ( 0.0)( 0.0) Experiment : K [ HI] eq [ H [ I K K 49.8 [ HI] eq ( 0.100) HI Experiment : K Experiment 4: K 4 H I H [ [ ( )( ) K 49. K eq [ [ I ( 0.80) ( 0.050)( ) ( 0.11) ( 0.044)( 0.044) Average K K avg 49.5 Converting K c to K : V nrt (Ieal Gas Law) V n (Solving for an let Concentration) V c C D K a b (Equilibrium Expression an Constant for aa + bb cc + D) A B Substituting pressures for each chemical species into the equilibrium expression: ( A [A]RT ; B [B]RT ; C [C]RT ; an D [D]RT) c ([ C] RT ) ([ D] RT ) c ( ) c ( ) C RT D RT K ([ A] RT ) a ([ B] RT ) b [ A] a ( RT ) a [ B] b ( RT ) b c ( ) c+ C D RT c C D [ c ] K (RT) a b a+ b [ A] [ B] ( RT ) [ A] a [ B] (c + ) (a + b) C D (Replace b [ A] a [ B] b K K c (RT) (c + ) (a + b) (Combine exponents on common base RT using Laws of Exponents) with K c ) Conversion between Concentration an ressure Equilibrium Constants aa + bb cc + D K K c (RT) Δn where Δn Σ Coefficients of roucts Σ Coefficients of Reactants (c + ) (a + b) R K 1 T Temperature in K R has a unit of K 1 to cancel out with the Kelvin from Temperature only, therefore K calculate will remain unitless age 178. Copyrighte by Gabriel Tang B.E., B.Sc.

8 Example : One possible way of removing NO (g) from the exhaust of an internal combustion engine is to cause it to react with CO (g) in the presence of suitable catalyst. NO (g) + CO (g) N (g) + CO (g) At 575 K, the reaction has K What is K of the same reaction at 575 K? K K c R K 1 T 575 K K K c (RT) Δn K ( )[( K 1 )(575 K)] 1 K ( )[ ] 1 Δn Σn proucts Σn reactants Δn (1 + ) ( + ) 4 Δn 1 K? K Example 4: The German s Haber-Bosch process evelope in 191 utilizes an iron surface that contains traces of aluminium an potassium oxie as a catalyst to manufacture ammonia from nitrogen an hyrogen. It is an important process as ammonia is commonly use in fertilizer an ammunition. In 1918, the scientist Fritz Haber won the Nobel rize in chemistry for his contribution. N (g) + H (g) NH (g) a. At 400 K, NH atm, N.8084 atm an H expression in terms of pressure an calculate K. b. Convert the calculate K above to K. NH a. Equilibrium Expression: K b. K.5 10 R K 1 T 400 K Δn Σn proucts Σn reactants Δn () (1 + ) 4 Δn K? N H K K c (RT) Δn K K c ( ) n RT Equilibrium Constant: K atm. Write the equilibrium K.5 10 ( ).5 10 Δ [( K )( )] K ( 0.104) (.8084)( 0.010) K c Heterogeneous Equilibria: - an equilibrium system where some chemical species are in ifferent phase compare to the others. - chemical species that are ure Soli or ure Liqui are NOT Inclue in the Equilibrium Expression. This is ue to the fact that pure solis an liquis o not have concentrations. Copyrighte by Gabriel Tang B.E., B.Sc. age 179.

9 Example 5: Write the equilibrium expression for the following systems. a. NaN (s) Na (s) + N (g) b. NH (g) + CO (g) H NCONH (s) + H O (g) K [N ] K N (NaN an Na are ure Solis) c. Ag SO 4 (s) Ag + (aq) + SO 4 (aq). HF (aq) + H O (l) H O + (aq) + F (aq) K [Ag + ] [SO 4 ] (Ag SO 4 is a ure Soli) (No K because there are no gases.) Multiple Equilibria: - the overall equilibrium constant of a multi-steps systems is the prouct of all the equilibrium constants of the iniviual steps involve. Consier the following multiple equilibria: K [ H O] [ NH ] [ CO] Step 1: A + B C + D K 1 [ C ][ D ] [ A][ B] Step : C + D E + F K [ E ][ F ] [ C][ D] A + B E + F K [ E ][ F ] [ C ][ D ] [ E ][ F ] [ A][ B] [ A][ B] [ C][ D] Multiple Equilibria K K 1 K K K K 1 K K (H NCONH is a ure Soli) [ O ][ F ] K H + [ HF] HO NH CO (H O is a ure Liqui) (No K because there are no gases.) Working with Equilibrium Constant an Equations: - reversing equilibrium reaction will cause the reciprocate the equilibrium constant (1/K) - multiplying the equation by a multiple n will result in raising the K by the power of n. Reversing Equilibrium Reaction aa + bb cc + D cc + D aa + bb K a b A B [ C] c [ D] c [ C] [ D] a [ A] [ B] b 1 K 1 K 1 K Reverse Equilibrium Constant Multiplying Equilibrium Reaction by a Factor of n n(aa + bb cc + D) naa + nbb ncc + nd) K nc n C D [ A] na [ B] nb c [ C] [ D] a [ A] [ B] b n K n K New Equilibrium Constant age 180. Copyrighte by Gabriel Tang B.E., B.Sc.

10 Example 6: For the following reaction an the equilibrium concentrations at 00 K. C H 6 (g) + NH (g) + O (g) C H N (g) + 6 H O (g) [C H M [NH 0.50 M [O 0.50 M [C H N.50 M [H O.00 M a. Write the equilibrium expression an etermine the equilibrium constant. b. Write the equilibrium expression an calculate the equilibrium constant for the following reaction with the same equilibrium concentrations. C H N (g) + 6 H O (g) C H 6 (g) + NH (g) + O (g) c. Write the equilibrium expression an fin the equilibrium constant for the following reaction with the same equilibrium concentrations. a. Equilibrium Expression: K Equilibrium Constant: K C H 6 (g) + NH (g) + O (g) C H N (g) + H O (g) 6 [ CH N] [ H O] [ C ] H 6 NH O 6 [ CHN] [ H O] [ C H ] [ NH ] [ O ] C H 6 NH O b. Equilibrium Expression: K [ C H N] [ H O] 6 C H6 NH O Equilibrium Constant: K [ C H N] [ H O] 6 c. Equilibrium Expression: K Equilibrium Constant: K 6 6 (.50) (.00) ( 0.500) ( 0.50) ( 0.50) (Reverse Equilibrium) K CHN H O [ CH 6 ] [ NH ] [ O ] K 1 K ( ) 1 K [ CH N][ H O] [ C H ][ NH ][ O ] 6 [ CHN][ H O] [ C H ][ NH ][ O ] 1 6 K ( ) 6 1 (All Coefficients Multiplie by ½) 6 [ CHN] [ H O] [ C H ] [ NH ] [ O ] K 6 K Assignment 15. pg. 50 #5 to 9 Copyrighte by Gabriel Tang B.E., B.Sc. age 181.

11 15.: What Does the Equilibrium Constant Tell Us? Important Notes Regaring the Size of the Equilibrium Constant (K): aa + bb cc + D c C D K K c a eq eq A b eq B eq 1. When K >> 1, the equilibrium system favours the proucts. There are more proucts than reactants at the state of equilibrium. ([C an [D or C, eq an D, eq >> [A an [B or A, eq an B, eq ). When K << 1, the equilibrium system favours the reactants. There are less proucts than reactants at the state of equilibrium. ([A an [B or A, eq an B, eq >> [C an [D or C, eq an D, eq ). When K 1, the equilibrium system favours neither the proucts nor the reactants. There are roughly the same amount of proucts an reactants at the state of equilibrium. ([C an [D or C, eq an D, eq [A an [B or A, eq an B, eq ) 4. The Size of K has NO Relationship with the Rate of Reaction to reach the state of equilibrium. Reaction Rate is epenent on Activation Energy an Temperature (T) (NOT K). 5. The Size of K epens mainly on Temperature (T). K c C, eq a A, eq D, eq b B, eq Reaction Quotient (Q): - the mass action expression uner any set of conitions (not necessarily equilibrium). - its magnitue relative to K etermines the irection in which the reaction must occur to establish equilibrium. c C D Q [ A] a [ B] b a. When Q > K, the system shifts to the reactants. At a specific conition, Q inicates that there are too much proucts. Therefore, the system has to shift back to the left. b. When Q < K, the system shifts to the proucts. At a specific conition, Q inicates that there are not enough proucts. Therefore, the system has to shift forwar to the right. c. When Q K, the system is at equilibrium. At a specific conition, Q inicates that all concentrations are that of the equilibrium state. Therefore, there will be no shifting. Example 1: The reaction, NO (g) + O (g) NO (g) has an equilibrium constant of K 9.54 at 600 K. Inicate the irection in which the system will shift to reach equilibrium when the [NO] M, [O ] M, an [NO ] M. [NO] M [O ] M [NO ] M Q? Q [ NO ] 0 [ NO] 0[ O ] 0 ( 0.500) ( 0.00) ( 0.50) Q 11.1 Since Q < K (11.1 < 9.54), the system will shift to the prouct (NO ). There are not enough NO at the initial conitions. age 18. Copyrighte by Gabriel Tang B.E., B.Sc.

12 Example : The reaction, Cl (g) + F (g) ClF (g) at 50 K has K 50., If the equilibrium concentrations of Cl (g) an ClF (g) are M an 0.05 M respectively, what is the equilibrium concentration of F (g)? [Cl M [ClF 0.05 M K 50. [F? K [ ClF ] eq [ Cl ] [ F ] eq eq 50. [F ] eq [F ( 0.05) ( 0.149)[ F ] eq ( 0.05) ( 50.)( 0.149) ( 0.05) ( 50.)( 0.149) [F M ICE Box: - stans for Initial, Change, an Equilibrium. It is a table that organizes information to calculate final equilibrium concentrations given the equilibrium constant an initial concentration. Example : The formation of HCl (g) from its elements, H (g) + Cl (g) HCl (g) has K at 50 K. A 5.00 L flask at 50 K containe an initial concentration of.00 mol of HCl (g) an.85 mol of H (g). When the system reache equilibrium, it was foun that there were mol of Cl (g). Determine the concentrations of H (g) an HCl (g) at equilibrium. K mol [H ] M 5.00 L.00 mol [HCl] M 5.00 L [Cl ] 0 0 M mol [Cl 0.17 M 5.00 L [H? [HCl? The system must shift to the left because initially, we are missing one reactant ([Cl ] 0 0 M). Hence, the change to the H is positive, an the change to the HCl woul be negative. Since there is 0.17 M of Cl at equilibrium, it means 0.17 M of H is ae (1:1 mol ratio between Cl an H ). It also means that there is (0.17 M) less HCl (:1 mol ratio between Cl an HCl). H (g) + Cl (g) HCl (g) Initial M 0 M M Change M M (0.17 M) Equilibrium 0.94 M 0.17 M 0.56 M Verify with K: K [ HCl] eq [ H [ Cl ( 0.56) ( 0.94)( 0.17) K This matches with K given in the question. Therefore, the equilibrium concentrations are: [H 0.94 M, [Cl 0.17 M an [HCl 0.56 M Copyrighte by Gabriel Tang B.E., B.Sc. age 18.

13 Example 4: The reaction, NO (g) NO (g) + NO (g) has K 1.7 at 700 K. A.50 L flask at 700 K has 1.00 mol of each species initially. Calculate the concentrations of all species at equilibrium. K mol [NO ] M.50 L 1.00 mol [NO ] M.50 L 1.00 mol [NO] M.50 L [NO? [NO? [NO? First, we must etermine Q an the irection of the shift. [ NO ] NO 0 Q [ NO ] 0 0 ( )( ) ( 0.400) Q 1.00 Since Q < K (1.00 < 1.7), the system will shift to the proucts (NO an NO). There are not enough proucts at the initial conitions. Hence, the change to the NO is negative, an the changes to the NO an NO woul be positive. Let x amount of change per mole, since there are moles of NO reacte; NO will be lowere by x. Similarly, NO an NO will be increase by 1x each. NO (g) NO (g) + NO (g) Initial M M M Change x + x + x Equilibrium (0.400 x) M ( x) M ( x) M Next, we set up the equilibrium expression, substitute the above mathematical expressions an solve for x. (Since we have a set of common exponents we can group it an square root both sies.) [ NO ] NO eq eq K 1.7 ( x )( x ) ( ) x ( x) [ NO ] ( 0.4 x) ( 0.4 x) ( 0.4 ) x eq ( x) 1.7 ( 0.4 x) ( x ) ( 0.4 x) (0.4 x) x x x x x x Verify [ with K: NO ] NO eq eq Using K the SOLVE function ( of )( the TI-8 ) lus K Calculator: 1.7 NO eq ( ) This matches with the K given in the question. Finally, substitute x back into the mathematical expressions. [NO x ( ) [NO M [NO x ( ) [NO M [NO x ( ) [NO M age 184. Copyrighte by Gabriel Tang B.E., B.Sc.

14 Using the TI-Graphing Calculator to Solve Higher Degree Equations: Sometimes we may encounter equations with higher egrees when solving for equilibrium pressures or concentrations. In some cases, we may be able to use the quaratic formula, but for cubic an higher egrees equations, they can become quite ifficult. The TI-Graphing Calculator has a SOLVE function to aie the etermination of the roots. Example: Solve for the final concentrations of both proucts in the following equilibrium system when the initial concentration of the reactant is mol/l, an K A (g) B (g) + C (g) A (g) B (g) + C (g) Initial M 0 M 0 M Change x + x + x Equilibrium (0.400 x) M x M x M 1. Manipulate the Equation so one-sie is equal to 0. x 0.4 x Example:.4 10 ( ) becomes 0 ( 0.4 x) x Callout the solve function. solve (Equation Expression, X, guess, {lower bounary, upper bounary}) e x n CATALOG 0 LN S ENTER then, Scroll Down from S X,T,θ,n Enter 0 Enter 0 Enter arameters n { ( Enter Highest Initial Concentration / ressure Given ENTER n } ) Select solve ( Value of x Therefore, the concentrations of the proucts are: x [B [C mol/l Copyrighte by Gabriel Tang B.E., B.Sc. age 185.

15 Example 5: The reaction, SO (g) + Cl (g) SO Cl (g), has an equilibrium constant of at 500 K. Determine the equilibrium concentrations of all species if there are.00 mol of each species in 10.0 L container at 500 K. K mol [SO ] M 10.0 L.00 mol [Cl ] M 10.0 L.00 mol [SO Cl ] M 10.0 L [SO? [Cl? [SO Cl? First, we must etermine Q an the irection of the shift. Q [ SOCl ] 0 [ SO ] 0 [ Cl ] 0 ( 0.00) ( 0.00)( 0.00) Q 5.00 Since Q > K (5.00 > 0.486), the system will shift to the reactants (SO an Cl ). There is too much prouct at the initial conitions. Hence, the changes to the SO an Cl are positive, an the change to the SO Cl woul be negative. Let x amount of change per mole, since there is 1 mole of SO reacte; SO will increase by 1x. Similarly, Cl will increase by 1x. SO Cl, on the other han, will ecrease by 1x. SO (g) + Cl (g) SO Cl (g) Initial 0.00 M 0.00 M 0.00 M Change + x + x x Equilibrium ( x) M ( x) M (0.00 x) M Next, we set up the equilibrium expression, substitute the above mathematical expressions an solve for x. (Since we on t have any common exponent, we have to expan the enominator.) [ SOCl ] eq ( 0. x) ( 0. x) K or 0 ( 0. x ) SO Cl 0. + x 0. + x x + x 0. + x [ [ x x ( )( ) ( ) ( ) ( x + x ) (0. x) x x 0. x x 0.486x x (Quaratic Equation: Apply the Quaratic Formula!) b ± b 4ac a ( ) ± ( ) 4( 0.486)( ) ( 0.486) a b c x x.6005 (omit negative x) Verify with K: K [ SOCl ] eq [ SO [ Cl ( ) ( 0.4)( 0.4) K This matches closely with the K given in the question. Finally, substitute x back into the mathematical expressions. [SO x (0.1487) [SO 0.4 M [Cl x (0.1487) [Cl 0.4 M [SO Cl 0.00 x 0.00 (0.1487) [SO Cl M age 186. Copyrighte by Gabriel Tang B.E., B.Sc.

16 Steps to Solve Equilibrium roblems: 1. Write the Balance Chemical Equation for the system. This inclues all the correct states for all species.. Write the Equilibrium Expression an equate it to K value given.. List the Initial Concentrations. 4. Determine Q an compare it to K to etermine the irection of the shift to equilibrium. 5. Construct the ICE Box an efine the change amount of change per mole as x. Using the coefficients an the irection of the shifts, state the mathematical expressions of each species at equilibrium. 6. Solve for x after substituting the mathematical expressions into the equilibrium expression. (Look for methos of simplifying like common exponents if they exist. Otherwise, the quaratic equation will be neee to solve for x.) 7. Calculate the equilibrium concentrations of each species an verify that they inee give the value of K. 8. For equilibrium involving pressures, the proceure is the same as above. Example 6: The equilibrium constant for the system, C H 6 (g) + Cl (g) C H 5 Cl (s) + HCl (g), at 8 K is If we ha 6.00 mol of C H 6 (g), 6.00 mol of Cl (g) an 6.00 mol HCl (g) originally in a.00 L container at 8 K, etermine the equilibrium concentrations for all species. K mol [C H 6 ] 0.00 M.00 L 6.00 mol [Cl ] 0.00 M.00 L 6.00 mol [HCl] 0.00 M.00 L [C H 6? [Cl? [HCl? First, we must etermine Q an the irection of the shift. (Be careful! This is a heterogeneous system. C H 5 Cl is a pure soli an it is not involve in the equilibrium expression. Q [ HCl] 0 [ C H 6 ] [ Cl ] 0 0 (.00) (.00)(.00) Q Since Q > K (0.500 > 0.100), the system will shift to the reactants (C H 6 an Cl ). There are too much proucts at the initial conitions. Hence, the changes to the C H 6 an Cl are positive, an the change to the HCl woul be negative. Let x amount of change per mole, since there is 1 mole of C H 6 reacte; C H 6 will increase by 1x. Similarly, Cl will increase by 1x. HCl, on the other han, will ecrease by 1x. C H 6 (g) + Cl (g) C H 5 Cl (s) + HCl (g) Initial.00 M.00 M M Change + x + x x Equilibrium (.00 + x) M (.00 + x) M (.00 x) M Copyrighte by Gabriel Tang B.E., B.Sc. age 187.

17 Next, we set up the equilibrium expression, substitute the above mathematical expressions an solve for x. (Since we have a set of common exponents we can group it an square root both sies.) [ HCl] 0 ( x) ( x) K or 0 ( x ) C H Cl ( + x)( + x) 4 + 4x + x + x [ 6 ] [ ] 0 x x 0 b ± b 4ac a (4 + 4x + x ) ( x) x + 0.1x x 0.1x + 1.4x ( 1.4) ± ( 1.4) 4( 0.1)( 1.6) ( 0.1) x x (omit negative x) Verify with K: [ HCl] 0 K C H Cl [ 6 ] [ ] 0 0 ( 0.98) (.06)(.06) ( ) a 0.1 b 1.4 c 1.6 K This matches closely with the K given in the question. x ( ) Finally, substitute x back into the mathematical expressions. [C H x.00 + ( ) [C H 6.06 M [Cl.00 + x.00 + ( ) [Cl.06 M [HCl.00 x.00 ( ) [HCl 0.98 M Example 7: The reaction of CO (g) + Cl (g) COCl (g) has an equilibrium constant of K 0.89 at 450 K. The initial pressures of CO (g), Cl (g) an COCl (g) are atm, atm an atm respectively. They are all mixe in a 7.50 L flask at 450 K. Fin the equilibrium pressures. K 0.89 CO, atm atm Cl, 0 COCl, 0 CO, eq? Cl, eq? COCl, eq? atm First, we must etermine Q an the irection of the shift. Q COCl, 0 CO, 0 Cl, 0 ( 0.150) ( 0.800)( 0.900) Q 0.08 Since Q < K (0.08 < 0.89), the system will shift to the prouct (COCl ). There is too little prouct at the initial conitions. Hence, the changes to the CO an Cl are negative, an the change to the COCl woul be positive. Let x amount of change per mole, since there is 1 mole of CO reacte; CO will ecrease by 1x. Similarly, Cl will ecrease by 1x. COCl, on the other han, will increase by 1x. CO (g) + Cl (g) COCl (g) Initial atm atm atm Change x x + x Equilibrium (0.800 x) atm (0.900 x) atm ( x) M age 188. Copyrighte by Gabriel Tang B.E., B.Sc.

18 Next, we set up the equilibrium expression, substitute the above mathematical expressions an solve for x. (Since we on t have any common exponent, we have to expan the enominator.) COCl, eq ( x) ( x) ( x) K 0.89 or x 0.9 x x + x 0.8 x 0.9 x x x CO, eq b ± Cl, eq ( )( ) ( ) ( )( ) 0.89 ( x + x ) ( x) x x x x 0.89x 1.491x (Quaratic Equation: Apply the Quaratic Formula!) b 4ac a a 0.89 b c ( 1.491) ± ( 1.491) 4( 0.89)( ) ( 0.89) x (omit larger value bigger than atm an atm) Finally, substitute x back into the mathematical expressions. CO, eq x ( ) CO, eq atm x Verify with K: K COCl, eq CO, eq K 0.88 Cl, eq ( 0.189) ( 0.761)( 0.861) (This matches closely with the K given in the question.) Cl, eq x ( ) Cl, eq atm COCl, eq x ( ) COCl, eq M 14.5: Factors That Affect Chemical Equilibrium Le Châtelier s rinciple: - a qualitative metho to preict the shift on an equilibrium system if it is isturbe by means of changing concentration, pressure an temperature. - the equilibrium will shift in the irection that minimizes the change impose on the system. 1. Effects of a Change in Concentration: Assignment 15. pg #11 to 6, 7 to 8, 54, 58, 60, 6, 78 a. An ADDITION of a species on one sie of the equilibrium will Drive the System TOWARDS the Opposite Sie. (There is more concentration of the species being ae. Hence, the system will shift towars the opposite sie to reuce the increase amount of that particular species.) b. A REMOVAL of a species on one sie of the equilibrium will Drive the system TOWARDS the Same Sie. (There is less concentration of the species being remove. Hence, the system will shift towars the removal sie to compensate.) Copyrighte by Gabriel Tang B.E., B.Sc. age 189.

19 Concentration [D] [C] [B] [A] Changes in Concentrations on Equilibrium System aa + bb cc + D Time inicates an Increase in [D]. As [D], equilibrium shifts to the left (aa + bb cc + D). Hence, [A], [B], an [C]. inicates a Decrease in [A]. As [A], equilibrium shifts to the left (aa + bb cc + D). Hence, [B], [C], an [D]. inicates an Increase in [B]. As [B], equilibrium shifts to the right (aa + bb cc + D). Hence, [A], [C], an [D]. inicates a Decrease in [C]. As [C], equilibrium shifts to the right (aa + bb cc + D). Hence, [A], [B], an [D].. Effects of a Change in ressure: a. Aing an Inert Gas has NO CHANGE on the equilibrium system. This is because an inert gas oes not participate in the forwar or reverse reaction. b. Reucing the Volume will Drive the System TOWARDS the Sie With LESS Gaseous Molecules. Since there are less space for the number of molecules, the system will have to shift to the sie with lesser gaseous molecules to compensate. c. Conversely, Expaning the Volume will Drive the System TOWARDS the Sie With MORE Gaseous Molecules. Now that there is more room for the molecules to move about, the system will shift to the sie that has more gaseous molecules to ajust to the new conition.. When there are Equal Number of Gaseous Molecules on Both Sie of the Equilibrium, any Change in Volume will NOT Affect System. age 190. Copyrighte by Gabriel Tang B.E., B.Sc.

20 Changes in ressures on a Gaseous Equilibrium System aa + bb cc + D when (c + ) > (a + b) [D] [C] Concentration [B] [Inert Gas] [A] inicates an Increase in Volume. As V, equilibrium shifts to the right (aa + bb cc + D) since there are more gaseous molecules on the prouct sie [(c + ) > (a + b)]. Hence, [A], [B], [C], an [D]. inicates a Decrease in Volume. As V, equilibrium shifts to the left (aa + bb cc + D) since there are LESS gaseous molecules on the reactant sie [(c + ) > (a + b)]. Hence, [A], [B], [C], an [D]. inicates an Aition of an Inert Gas. There is no shifting of the equilibrium. (aa + bb cc + D) as inert gas oes not affect the system. Hence, [A], [B], [C], an [D] remain unchange. Time (See the Vieo at Effects of a Change in Temperature: - look at the energy (written in the reactant or prouct sie) as a chemical species. Then, the preictions will be the same as those foun with changing the concentrations. a. For an Exothermic Equilibrium System: aa + bb cc + D + Energy - an Increase in Temperature will rive the system to the left (aa + bb cc + D + Energy). There is more heat ae an because energy is written on the prouct sie, the system will shift to the opposite sie to re-establish equilibrium. Hence, [A], [B], [C], an [D]. - a Decrease in Temperature will rive the system to the right (aa + bb cc + D + Energy). There is less heat overall an because energy is written on the prouct sie, the system will shift to the proucts to compensate. Hence, [A], [B], [C], an [D]. Copyrighte by Gabriel Tang B.E., B.Sc. age 191.

21 b. For Enothermic Equilibrium System: aa + bb + Energy cc + D - a Decrease in Temperature will rive the system to the left (aa + bb + Energy cc + D). There is less heat overall an because energy is written on the reactant sie, the system will shift to the reactants to compensate. Hence, [A], [B], [C], an [D]. - an Increase in Temperature will rive the system to the right (aa + bb + Energy cc + D). There is more heat ae an because energy is written on the reactant sie, the system will shift to the opposite sie to re-establish equilibrium. Hence, [A], [B], [C], an [D]. Changes in Temperature on an Exothermic Equilibrium System aa + bb cc + D + Energy Changes in Temperature on an Enothermic Equilibrium System aa + bb + Energy cc + D [D] [D] [C] [C] Concentration [B] [A] Concentration [B] [A] Time Time inicates a Decrease in Temperature. As T, equilibrium shifts to the right of an exothermic system (aa + bb cc + D + Energy). Hence, [A], [B], [C], an [D]. inicates an Increase in Temperature. As T, equilibrium shifts to the left of an exothermic system (aa + bb cc + D + Energy). Hence, [A], [B], [C], an [D]. inicates an Increase in Temperature. As T, equilibrium shifts to the right of an enothermic reaction (aa + bb + Energy cc + D). Hence, [A], [B], [C], an [D]. inicates a Decrease in Temperature. As T, equilibrium shifts to the left of an enothermic reaction (aa + bb + Energy cc + D). Hence, [A], [B], [C], an [D]. 4. Effects of Aing a Catalyst: - has NO CHANGE on the equilibrium system. (See Vieo at age 19. Copyrighte by Gabriel Tang B.E., B.Sc.

22 Example 1: The equilibrium system, 4 F (g) 4 (s) + 6 F (g) kj, is put uner the following changes. reict the shift of the system an the resulting concentrations of all species for each case. a. an increase in the concentration of F (g). b. a ecrease in the concentration of 4 (s) The system will shift to the LEFT. There will be NO SHIFT on the system. 4 F (g) 4 (s) + 6 F (g) kj Effect: [F ] (increase) 4 F (g) 4 (s) + 6 F (g) kj ( 4 is a pure soli an oes not involve with the equilibrium system) Effect: [F ] an [F ] remain the same. c. a ecrease in the concentration of F (g).. a ecrease in Temperature. The system will shift to the LEFT. 4 F (g) 4 (s) + 6 F (g) kj Effect: [F ] (ecrease) The system will shift to the RIGHT. 4 F (g) 4 (s) + 6 F (g) kj Effect: [F ] (ecrease) an [F ] (increase) e. an aition of He (g). f. an increase in volume. There will be NO SHIFT on the system. 4 F (g) 4 (s) + 6 F (g) kj (He is an inert gas an oes not involve with the equilibrium system) Effect: [F ] an [F ] remain the same. The system will shift to the RIGHT. 4 F (g) 4 (s) + 6 F (g) kj (There are more gaseous molecules on the prouct sie 6 moles of F (g) versus 4 moles of F (g) ) Effect: [F ] (ecrease) an [F ] (increase) Example : The Haber-Bosch process, N (g) + H (g) NH (g) + 9 kj is essentially an equilibrium system. A chemical engineer woul like the highest yiel of ammonia. List all the possible metho of prouction that will ensure maximum amount of NH (g) prouce. N (g) + H (g) NH (g) + 9 kj (Desire Effect: [NH ], which means riving the system forwar.) 1. Increase the concentrations of N (g) or H (g) or both will rive the system forwar.. Decrease the concentration of NH (g) as it is prouce will shift the system forwar.. Lower the Temperature will rive the system to the prouct sie. 4. Decrease the Volume of the system will shift the system to the right ue to smaller number of gaseous molecules on the prouct sie. Assignment 15.4 pg. 5 5 #9 to 5 Copyrighte by Gabriel Tang B.E., B.Sc. age 19.