Unit 5: Chemical Kinetics and Equilibria. Unit 5: CHEMICAL KINETICS AND EQUILIBRIA

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1 Chemistry A Unit 5: Chemical Kinetics and Equilibria Unit 5: CHEMICAL KINETICS AND EQUILIBRIA Chapter : Chemical Kinetics.: Reaction Rates Chemical Kinetics: - the branch of chemistry that studies the how fast reaction proceeds. Reaction Rates: - the speed of which the concentration of a reactant or product changes per unit of time. - unit is in mol/(l s) or mol/(l min) Rate t A Concentration A at t Concentration A at t t t Average Rate: - like average speed, it is the change in concentration per interval of time. Instantaneous Rate: - like instantaneous speed, it is the change in concentration at a particular time. Relationship between Molar Quantities and Reaction Rates For any chemical equation, aa bb, the Reaction Rate of B roduced is the Mole Ratio a b Reaction Rate of A Consumed. Example : For the reaction N (g) + Cl (g) NCl (g), what is the rate of consumption of N and Cl if the rate of production of NCl is 0.45 mol/(l s)? Rate of N Rate of NCl (0.45 mol/(l s)) Rate of N 0.7 mol/(l s) Rate of Cl Rate of NCl (0.45 mol/(l s)) Rate of Cl mol/(l s) Example : The concentrations of the reactant and product of the reaction, A B, are recorded below. a. Determine the average reaction rates for each time interval. b. Graph concentrations versus time for both reactant and product. c. Find the instantaneous reaction rates of reactant A at 0 seconds and product B at 40 seconds. Time (seconds) Reactant [A] roduct [B] Copyrighted by Gabriel Tang B.Ed., B.Sc. age 5.

2 Unit 5: Chemical Kinetics and Equilibria Chemistry A a. Average Reaction Rates (Note: Reactant Rate is Negative because A is consumed.) Time Intervals [ A ] (seconds) Reactant Rate (mol/(l s) roduct Rate [ B ] (mol/(l s) t t 0.00 to 0.00 ( ) mol/l (0. 0) mol/l ( ) s ( ) s to ( ) mol/l (0.5 0.) mol/l ( ) s ( ) s to ( ) mol/l ( ) mol/l ( ) s ( ) s to ( ) mol/l ( ) mol/l ( ) s ( ) s to ( ) mol/l ( ) mol/l ( ) s ( ) s b. Concentrations versus Time Graph. Concentrations over Time Reactant [A] roduct [B] Concentrations (mol/l 0. t seconds [B] 0.06 mol/l [A] 0.06 mol/l t 8 seconds Time (seconds) c. Instantaneous Rates Reactant Rate roduct Rate [ A ] t [ B ] t 0.06 mol/l 8 s 0.06 mol/l s Reactant [A] Rate mol/(l s) roduct [B] Rate mol/(l s) age 6. Copyrighted by Gabriel Tang B.Ed., B.Sc.

3 Chemistry A Unit 5: Chemical Kinetics and Equilibria.: Rate Laws: An Introduction Rate Law: - a reaction rate where it is depended solely on the concentrations of the reactants with the assumption that it is an irreversible reaction. Rate Constant (k): - the proportional coefficient of the rate law. (It is NOT the coefficient of the chemicals.) - the unit of the rate constant depends on the overall order of the reaction. Order (n): - the rational exponent of the reactant s concentration in the rate law as determined by experimentation. (It is NOT the coefficient of the chemicals.) Two Kinds of Rate Laws Differential Rate Law: - sometimes called rate law, which relates the reaction rate on reactants concentrations only. Integrated Rate Law: - the rate law that relates the concentration of the chemical species with time. - it is derived by integrating the differential rate law with respect of concentration and time..: Determining the Form of the Rate Law Overall Rate Order: - the sum of all orders of all reactants. Example : For the following rate laws, determine their overall rate order and the unit of the rate constant. a. Rate k[a] b. Rate k[a] k Rate Laws Single Reactant: Rate k[a] n Multiple Reactants: Rate k[a] n [B] m k rate constant n and m order of a particular reactant (Note: Both reaction rates are positive because they are net consumption rates.) n (First order) Rate [ A] m ol/(l mol/ L s) s k A n (Second order) Rate m ol/(l ( ) s) L/(mol s) mol/l c. Rate k[a][b] d. Rate k[a] [B] e. Rate k n + m + n + m + n 0 (Zero order) (Second order) (Third order) m ol/(l s) Rate k [ A][ B] ( ) mol/l k L/(mol s) Rate mol/(l s) mol/l k L /(mol s) k A [ B] ( ) ( mol/l) k mol/(l s) Copyrighted by Gabriel Tang B.Ed., B.Sc. age 7.

4 Unit 5: Chemical Kinetics and Equilibria Chemistry A Initial Rate ([A] 0 ): - the instantaneous reaction rate at the beginning instant (t 0) of the reaction. - for a single reactant, its order can be found by the instantaneous reaction rates and its concentrations at two different times. Determining Rate Law of a Single Reactant using Instantaneous Rates Rate Rate k k n [ A ] t [ A] t [ A] n [ A] t t Example : The decomposition of acetaldehyde (ethanal) at 98 K yields the following kinetic data. CH CHO (g) CH 4 (g) + CO (g) Time (seconds) [CH CHO (g)] (mol/l) Instantaneous Rate (mol/(l s)) a. Determine the rate law and the rate constant. b. Graph the concentration versus time plot for the decomposition of CH CHO (g). c. Calculate the reaction rate when the concentration of acetaldehyde is 0.50 mol/l. n a. To find Rate Law, pick any two concentrations and rates n n Rate k[ CHCHO] 0, t 0.79 k( 0.00) Rate k[ CH ] n n CHO.4 k(0.400) 0, t (0.75) n ( ) n log (0.75) log( ) n log 0.75 Rate k[ch CHO] ( ) n (Overall nd Order) b. Concentration versus Time Graph To calculate the Rate Constant, pick any one of the concentrations and rates. Rate k[ch CHO].4 k(0.400 mol/l).4 mol/ ( L s) k ( mol/l) k 8.75 L/(mol s) Rate (8.75 L/(mol s))[ch CHO] (The negative sign in k is dropped because we know reactant rate is one of consumption.) c. When [CH CHO (g) ] 0.50 mol/l, Concentrations (mol/l) Reaction Rate of Acetaldehyde Rate (8.75 L/(mol s))[ch CHO] Rate (8.75 L/(mol s))(0.50) mol /L Rate mol/(l s) Time (seconds) age 8. Copyrighted by Gabriel Tang B.Ed., B.Sc.

5 Chemistry A Unit 5: Chemical Kinetics and Equilibria Method of Initial Rates: - the use of initial rate(s) to determine the overall rate order of the reaction. - for x number of reactants, we will need at least (x + ) number of experiments with various initial rates and initial concentrations. - the order of a particular reactant is determined by dividing the initial rates of the two experiments with different initial concentrations of that particular reactant, while having the same initial concentrations of other reactants. Determining Rate Law of Multiple Reactants using Initial Rates Rate Rate Rate Rate k k k k n m [ A] 0, exp [ B] 0, [ A] n m 0, exp B 0, n m [ A] 0, exp [ B] 0, [ A] n [ B] m 0, exp exp exp exp 0, exp [ A] [ A] [ B] [ B] 0, exp 0, exp 0, exp 0, exp n m where [B] 0, exp [B] 0, exp where [A] 0, exp [A] 0, exp Example : The formation of nitrogen dioxide from the combustion of nitrogen monoxide at 660 K gives the following kinetic data. Determine the rate law and the rate constant. NO (g) + O (g) NO (g) Experiment [NO (g)] (mol/l) [O (g) ] (mol/l) Measured Initial Rate (mol/(l s)) To find Rate Law, we first have to pick the two concentrations and rates where [O ] stays constant. n m Rate k[ NO] 0, exp [ O ] 4 n m 0, exp.0 0 k( 0.00 mol/l) ( 0.00 mol/l) Rate k NO n 5 O.5 0 k 0.00 mol/l n 0.00 mol/l m 0, exp 0, exp ( ) ( ) m 4 () n n Then, we have to pick the two concentrations and rates where [NO] stays constant. n m Rate k[ NO] 0, exp [ O ] 5 n 0, exp k 0.00 mol/l 0.00 mol/l Rate k NO n 5 O.5 0 k 0.00 mol/l n 0.00 mol/l m 0, exp 0, exp ( ) ( m ) ( ) ( ) m Rate k[no] [O ] () m m (Overall rd Order) To calculate the Rate Constant, we can pick the concentration and rate from any experiment. Rate k[no] [O ] mol/(l s) k(0.00 mol/l) (0.00 mol/l) mol/(l s) k k 5 L /(mol s) 0.00 mol/l 0.00 mol/l ( ) ( ) Rate (5 L /(mol s))[no] [O ] Copyrighted by Gabriel Tang B.Ed., B.Sc. age 9.

6 Unit 5: Chemical Kinetics and Equilibria Chemistry A Example 4: The reaction rate of the acetone iodination not only depends on the concentrations of the reactants, but it also relates to the concentration of the H + (aq) ion. The rate law has a format of Rate k[ch COCH ] n [I ] m [H + ] p. Several experiments were done at 98 K and the initial rates of the reaction are measured. Using the data below, find the final rate law and the rate constant. CH COCH (aq) + I (aq) CH COCH I (aq) + H + (aq) + I (aq) Experiment [CH COCH (aq) ] [I (aq) ] [H + (aq)] Measured Initial Rate (mol/(l s)) 0.80 mol/l 0.00 mol/l 0.0 mol/l mol/l 0.00 mol/l 0.0 mol/l mol/l 0.00 mol/l 0.40 mol/l mol/l mol/l 0.0 mol/l To find Rate Law, we first have to pick the two concentrations and rates where [I ] and [H + ] stay constant. Rate k Rate k n m + p [ CHCOCH] [ I ] [ H ] 0, exp 0, exp 0, n m + [ CH COCH ] [ I ] [ H ] p 0, exp 0, exp exp 0, exp ( ) n ( ) m ( ) p k.6 mol/l 0.00 mol/l 0.0 mol/l k( 0.80 mol/l) n ( 0.00 mol/l) m ( 0.0 mol/l) p () n n To find Rate Law, we first have to pick the two concentrations and rates where [CH COCH ] and [H + ] stay constant. Rate 4 k Rate k n m + p [ CHCOCH] [ I ] [ H ] 0, exp 4 0, exp 4 0, n m + [ CH COCH ] [ I ] [ H ] p 0, exp 0, exp exp 4 0, exp ( ) n ( ) m k 0.80 mol/l mol/l ( 0.0 mol/l ) k( 0.80 mol/l) n ( 0.00 mol/l) m ( 0.0 mol/l) p (0.5) m m 0 To find Rate Law, we first have to pick the two concentrations and rates where [I ] and [CH COCH ] stay constant. Rate k Rate k n m + p [ CHCOCH] [ I ] [ H ] 0, exp 0, exp 0, n m + [ CH COCH ] [ I ] [ H ] p 0, exp 0, exp 6 exp 6 0, exp ( ) n ( ) m ( ) p k 0.80 mol/l 0.00 mol/l 0.40 mol/l k( 0.80 mol/l) n ( 0.00 mol/l) m ( 0.0 mol/l) p () p p p Rate k[ch COCH ] [I ] 0 [H + ] Rate k[ch COCH ][H + ] (Overall nd Order) To calculate the Rate Constant, we can pick the concentration and rate from any experiment. Rate k[ch COCH ][H + ] mol/(l s) k(0.80 mol/l)(0.0 mol/l) mol/(l s) k k L/(mol s) ( 0.80 mol/l)( 0.0 mol/l) Rate ( L/(mol s))[ch COCH ][H + ] age 0. Assignment. pg. 598 #7 and 8. pg #9 to 6 Copyrighted by Gabriel Tang B.Ed., B.Sc.

7 Chemistry A Unit 5: Chemical Kinetics and Equilibria.4: The Integrated Rate Law. Integrated First-Order Rate Law for Single Reactant A Rate k[a] (Starting from First-Order Rate Law) t [ A] k t (Rearrange Equation for Integration) A A t [ A] [ A] d k A 0 0 dt ( t dt and [A] d[a]; Integrate Both Sides: dx x ln x) ln [A] ln [A] 0 kt A (Apply one of the Logarithm Laws: log A log B log ) B [ A ] ln [ A] kt 0 (Integrated First Order Rate Laws Two Versions) Graphing Integrated First-Order Rate Law for Single Reactant Initial Graph of [A] versus t [A] Re-Graph with ln[a] vs. t Modified Graph of ln[a] versus t ln[a] k t + ln[a] ln [A] 0 y m x + b Half-Life (t / ): - the amount of time it takes to half the original reactant s concentration. ln - at half-life, t /, the reactant s concentration is [A]0 [A]: [ A ] [ A] kt ln ( / [A] 0 ) 0 [A] 0 t If ln[a] versus t graph is a linear relation, then it is a First Order Reaction with k slope and ln[a] 0 y-intercept. kt / ln kt/ (Since ln(/) ln()) ln () kt / ln 0.69 t / k k Integrated First Order Rate Laws for Single Reactant ln[a] kt + ln[a] 0 ln [ A] 0 A ln 0.69 kt t/ k k [A] Concentration of Reactant at time t k Rate Constant [A] 0 Concentration of Reactant at time 0 t time t / half-life Copyrighted by Gabriel Tang B.Ed., B.Sc. age. t

8 Unit 5: Chemical Kinetics and Equilibria Chemistry A. Integrated Second-Order Rate Law for Single Reactant Rate t [ A] [ A] A A k[a] (Starting from Second-Order Rate Law) k t (Rearrange Equation for Integration) [ A] [ A] d k [ A] 0 A [ A] 0 t 0 dt ( t dt and [A] d[a]; Integrate Both Sides: kt (Integrated Second Order Rate Law) x dx ) x Graphing Integrated Second-Order Rate Law for Single Reactant Initial Graph of [A] versus t [A] Re-Graph with vs. t A [ A] Modified Graph of A A versus t k t + [ A] 0 t t y m x + b If A versus t graph is a linear relation, then it is a Second Order Reaction with k slope and [ A] 0 y-intercept. Half-Life for Second-Order Reaction with a Single Reactant - at half-life, t /, the reactant s concentration is [A]0 [A]: A [ A] 0 kt [ A] 0 [ A] 0 t / kt / k[ A] 0 [ A] 0 [ A ]0 kt / [ A ]0 kt / Integrated Second Order Rate Laws for Single Reactant kt + A [ A] t / 0 k[ A] 0 [A] Concentration of Reactant at time t k Rate Constant [A] 0 Concentration of Reactant at time 0 t time t / half-life age. Copyrighted by Gabriel Tang B.Ed., B.Sc.

9 Chemistry A Unit 5: Chemical Kinetics and Equilibria. Integrated Zero-Order Rate Law for Single Reactant Rate A k (Starting from Zero-Order Rate Law) t A k t (Rearrange Equation for Integration) [ A] A 0 [ A] d k dt [A] [A] 0 kt t 0 ( t dt and [A] d[a]; Integrate Both Sides: dx x) (Integrated Zero Order Rate Law) Graphing Integrated Zero-Order Rate Law for Single Reactant Initial Graph of [A] versus t [A] (No Need to Re-Graph) [A] kt + [A] 0 y m x + b t If the initial graph of [A] versus t graph is a linear relation, then it is a Zero Order Reaction with k slope and [A] 0 y-intercept. Half-Life for Zero-Order Reaction with a Single Reactant - at half-life, t /, the reactant s concentration is [A]0 [A]: [A] [A] 0 kt A 0 kt/ [A]0 [A] 0 kt / [ A ] t / 0 k Integrated Zero Order Rate Laws for Single Reactant [A] kt + [A] 0 A 0 t / k [A] Concentration of Reactant at time t k Rate Constant [A] 0 Concentration of Reactant at time 0 t time t / half-life Copyrighted by Gabriel Tang B.Ed., B.Sc. age.

10 Unit 5: Chemical Kinetics and Equilibria Chemistry A Example : The decomposition of HI (g) at 76 K yields the following data. HI (g) ½ H (g) + ½ I (g) Time (minutes) [HI] mol/l a. Complete the table below. b. Graph [HI] versus t, ln[hi] versus t, and /[HI] versus t. Assess the linearity to determine the overall rate order. c. State the integrated rate law for the decomposition of HI and find the rate constant. d. What is the half-life of this reaction? e. Calculate the time needed for the reactant to reached 0.0% of its original concentration. Time (minutes) [HI] mol/l ln[hi] /[HI] a. Using the TI-8 lus Calculator, we can enter the experimental results in the STATS Editor and quickly manipulate the data.. Enter Values in the L and L Columns of Stats Editor STAT ENTER. Enter ln(l ) in the heading of L. Enter /L in the heading of L 4. Cursor must be on the column heading before entering formula MEM must be used to enter formula ALHA + To access L, press nd L age 4. Copyrighted by Gabriel Tang B.Ed., B.Sc.

11 Chemistry A. Fill in the Table b. First, we Graph [HI] vs. t Unit 5: Chemical Kinetics and Equilibria Time (minutes) [HI] mol/l ln[hi] /[HI] Turn On STATS LOT. Set WINDOW. Graph nd STAT LOT ENTER WINDOW x: [0, 7, ] y: [0, 0.5, 0.005] GRAH Y To access L, press Next, we graph ln[hi] vs. t nd L t (min) [HI] vs. t Graph is a Curve. Hence, it is NOT a Zero-Order Reaction.. Turn On STATS LOT. Set WINDOW. Graph nd STAT LOT ENTER WINDOW Y To access L, press nd [HI] (mol/l) x: [0, 7, ] y: [ 5.5,.5, 0.5] ln[hi] GRAH L t (min) ln[hi] vs. t Graph is a Curve. Hence, it is NOT a First-Order Reaction. Copyrighted by Gabriel Tang B.Ed., B.Sc. age 5.

12 Unit 5: Chemical Kinetics and Equilibria Chemistry A Lastly, we graph /[HI] vs. t. Turn On STATS LOT. Set WINDOW. Graph nd STAT LOT ENTER WINDOW x: [0, 7, ] y: [0, 50, 50] GRAH Y To access L 4, press nd L4 4 /[HI] t (min) /[HI] vs. t Graph is a Straight Line. Hence, it is a Second-Order Reaction. c. Since the reaction is second order, the integrated rate law has a form of kt + HI can use the linear regression function of the calculator to find the value of the slope. HI 0. We. Turn Diagnostic On nd CATALOG 0 Select DiagnosticOn ENTER Again ENTER Note: After DiagnosticOn is selected; it will remain ON even when the calculator is turned Off. However, resetting the calculator will turn the Diagnostic Off (factory setting).. Obtain LinReg (ax + b) L, L 4, Y. Copy Equation to Y Screen, and Graph Least Square Line on Scatter lot. ENTER nd L To access Y, press Select Option STAT for Y Select CALC, use VARS ENTER, nd L4, 4 Choose Option 4 Linear Regression (ax + b) Select Option age 6. Copyrighted by Gabriel Tang B.Ed., B.Sc.

13 Chemistry A Unit 5: Chemical Kinetics and Equilibria ENTER Y Be Sure lot is On When r is extremely close to ±, it indicates the line has a linear relationship. GRAH /[HI] y x compares to kt + HI [ HI] 0 k L/(mol min) d. Half-Life t / k[ HI] 0 k e. Time needed to reach 0.0% of [HI] 0 HI 0 [HI] mol/l [HI] mol/l [HI] mol/l k L/(mol min) t? t (min) HI L/ HI ( ) mol min HI HI 0 kt + t kt HI L/mol or [HI] mol/l (49.85 L/mol) t /.656 min HI 0 k HI 0 t 4.88 min mol/l 0.00 mol/l ( L/ ( mol min) ) 4. Integrated Rate Law for Multiple Reactants seudo-first-order Rate Law: - when there are multiple reactants, if we can have excess amount for all reactants except one, we can simplify the rate law so it behaves like a first-order reaction. seudo-first-order Rate Law For reaction with Multiple Reactants, aa + bb roducts and Rate k[a][b] n, if [B] 0 >> [A] 0, then we can assume [B] [B] 0. We can simplify Rate k[a][b] n to Rate k [A] where k k[b] n Copyrighted by Gabriel Tang B.Ed., B.Sc. age 7.

14 Unit 5: Chemical Kinetics and Equilibria Chemistry A Example : A reaction, A + B + C D + E, has a rate law of Rate k[a][b] [C]. The concentrations at the beginning of the experiment are [A] mol/l, [B] 0.00 mol/l and [C] 0.00 mol/l. After 5.0 seconds, the concentration of A is 0.04 mol/l. a. Calculate k for this reaction. b. Determine the half-life for this reaction. c. Find the concentrations of A and C after 50.0 seconds. a. Since [B] 0 and [C] 0 >> [A] 0, we can assume that [B] [B] 0 and [C] [C] 0. Therefore, Rate k[a][b] [C] can be simplified to Rate k [A] where k k[b] [C]. Rate k [A] ln[a] k t + ln[a] 0 ln[a] ln[a] 0 k t ln A k t ln[ A] 0 Finally, we have to convert k back to k. k k[b] [C] k' B C k (seudo-first-order Rate Law) (seudo-integrated-first-order Rate Law) ln ( 0.05 s ) (.00 mol/l) (.00 mol/l) [A] mol/l t 5.0 s [A] 0.04 mol/l ( 0.04) ln( ) 5.0 s k 0.05 s k 0.00 L /(mol s) b. Because we used the seudo-integrated-first-order Rate Law, the corresponding half-life is: (Note: we are using k ) t / ( ) ln k' ln() 0.05 s t / 7.4 s c. At t 50.0 s, [A]? ln[a] k t + ln[a] 0 ln[a] (0.05 s )(50.0 s) + ln(0.0500) ln[a] [A] e [A] 0.04 mol/l Because [C] [C] 0.00 mol/l, at 50.0 s, [C].00 mol/l In reality, [A] reacted [A] 0 [A] [A] reacted mol/l 0.04 mol/l [A] reacted mol/l mol C [C] reacted mol/l of A mol/l mol A [C] [C] 0 [C] reacted.00 mol/l mol/l [C].9 mol/l Assignment.4 pg #7 to 44 age 8. Copyrighted by Gabriel Tang B.Ed., B.Sc.

15 Chemistry A Unit 5: Chemical Kinetics and Equilibria.5: Rate Laws: A Summary Summary of Rate Laws for Reactions with Single Reactant aa roducts Order First Second Zero Rate Law Rate k[a] Rate k[a] Rate k Unit of k s or min L/(mol s) or L/(mol min) mol/(l s) or mol/(l s) Integrated Rate Law ln[a] kt + ln[a] 0 A kt + [ A] 0 [A] kt + [A] 0 Relationship with k and Linear Slope Half-Life t /.6: Reaction Mechanisms Slope k Slope k Slope k ln k t / k[ A] 0 seudo-first-order Rate Law Reaction Mechanisms: - the series of steps involve in most chemical reactions. [ A ] t / 0 k For reaction with Multiple Reactants, aa + bb roducts and Rate k[a][b] n, if [B] 0 >> [A] 0, then we can assume [B] [B] 0. We can simplify Rate k[a][b] n to Rate k [A] where k k[b] n Intermediates: - a chemical species that is produced in the initial reaction but is consumed by the final reaction step. - an intermediate does NOT appear in the overall reaction. Molecularity: - the number of chemical species that must COLLIDE in a reaction step. Elementary Step: - a reaction step for which its rate can be written based on it molecularity.. Unimolecular: - an elementary step where one molecule dissociates. (A roducts where Rate k[a]). Bimolecular: - an elementary step where two molecule collides. (A + A roducts where Rate k[a] or A + B roducts where Rate k[a][b]). Termolecular: - an elementary step where three molecule collides. (A + A + B roducts where Rate k[a] [B] or A + B + C roducts where Rate k[a][b][c]) Copyrighted by Gabriel Tang B.Ed., B.Sc. age 9.

16 Unit 5: Chemical Kinetics and Equilibria Chemistry A Example : The reaction, SO (g) + H O (l) H SO (aq), has the following reaction mechanism. H O (l) H + (aq) + OH (aq) H + (aq)_+ SO (g) HSO + (aq) HSO + + OH (aq) H SO (aq) Determine the rate law for each of the elementary step above and state their molecularity. Identify all intermediates. H O (l) H + (aq) + OH (aq) Rate k [H O] (Unimolecular) H + + (aq) + SO (g) HSO (aq) Rate k [H + ][SO ] (Bimolecular) + HSO (aq) + OH (aq) H SO (aq) Rate k [HSO + ][OH ] (Bimolecular) Intermediates: H + (aq), OH (aq), and HSO + (aq) Rate-Determining Step: - the slowest step of the reaction mechanism for which the overall reaction rate is based on. Deduction of Reaction Mechanism: - possible reaction mechanisms are deduced by the overall reaction rate from experimentation and chemical experience. - a given rate law can give rise to many possible reaction mechanism. That is, a reaction mechanism can never be proven absolutely Two Conditions for a ossible Reaction Mechanism. All reaction steps should add up to the final balanced reaction.. The rate law as determined by experimentation must agree with the rate law of the rate-determining step. Example : The reaction, H O (aq) + H + (aq) + I (aq) I (aq) + H O (l), has a experimental rate law of Rate k[h O ][I ]. Are the following steps a possible reaction mechanism? H O (aq) + I (aq) H O (l) + IO (aq) (Slow) H + (aq) + IO (aq) HIO (aq) (Fast) HIO (aq) + H + (aq) + I (aq) I (aq) + H O (l) (Fast) H O (aq) + I (aq) H O (l) + IO (aq) (Slow) Rate-determining step: H + (aq) + IO (aq) HIO (aq) (Fast) Rate k [H O ][I ] HIO (aq) + H + (aq) + I (aq) I (aq) + H O (l) (Fast) Experimental Rate Law: H O (aq) + H + (aq) + I (aq) I (aq) + H O (l) Rate k[h O ][I ] Steps of the Reaction Mechanism added up to the final reaction ( st condition met). Rate-determining elementary step gives the same rate law compared to that from experiments ( nd condition met). Therefore, the proposed mechanism is a possible pathway for the overall final reaction. Assignment.6 pg #45 to 48 age 0. Copyrighted by Gabriel Tang B.Ed., B.Sc.

17 Chemistry A Unit 5: Chemical Kinetics and Equilibria.7: A Model for Chemical Kinetics Collision Model: - a model that state for a reaction to occur, molecules must collide with each other. Factors Affecting the Collision Model:. Activation Energy (E a ): - the threshold energy molecules needed to overcome to cause a chemical reaction that was first proposed 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 diagram. Activated Complex: - sometimes refer to as transition state. It is the transitional molecule found at the top hill of the activation energy. E a E a Activation Energy of an Endothermic Reaction Activation Energy of an Exothermic Reaction. Temperature (T): - the effective number of collisions increases exponentially with temperature. Collision Frequency (z): - measures how often effective collision occurs. Because T > T, there are more molecules colliding with enough Kinetic Energy causing a reaction to occur. Copyrighted by Gabriel Tang B.Ed., B.Sc. age.

18 Unit 5: Chemical Kinetics and Equilibria Chemistry A. Molecular Orientation: - the number of ways molecules collide that will lead to a reaction. Steric Factor (p): - a measure that reflects the fraction of collisions with the effective orientations (p < ). Arrhenius Equation: - developed by Arrehenius to relate rate constant (k) using the three factors that affect the collision model (Activation Energy, Temperature with Collision Frequency, and Steric Factor) k zp e Arrhenius Equation E a RT k A e k Rate Constant E a Activation Energy (J) T Temperature (K) z Collision Frequency p Steric Factor A zp Frequency Factor R Gas Constant (8.45 J/(K mol)) E a RT age. Copyrighted by Gabriel Tang B.Ed., B.Sc.

19 Chemistry A Unit 5: Chemical Kinetics and Equilibria Linear Form of Arrhenius Equation k A ln(k) ln(a E a e RT E a e RT E a RT (Arrhenius Equation) ) (Natural-log both sides) ln(k) ln( e ) + ln(a) (Applying Law of Logarithm: log (MN) log M + log N) E ln(k) a + ln(a) (Applying Law of Logarithm: ln e x x) RT E ln(k) a R T + ln(a) (We can graph ln(k) versus (/T) from an experiment to determine E a ) y m x + b We can also compare the rate constants and temperature from two different points on the graph. ln(k ) ln(k ) E a RT + ln ( A) E a RT + ln ( A) E ln(k ) ln(k ) a E + ln(a) + a k ln(a) ln RT RT k E a R T T Alternate Arrhenius Equation E ln(k) a k + ln(a) ln R T E a k R T T k Rate Constant E a Activation Energy (J) T Temperature (K) A Frequency Factor R Gas Constant (8.45 J/(K mol)) Example : The decomposition of N O 4 (g) NO (g) is a first order reaction. One experiment at 74 K, its rate constant is calculated to be s. At another experiment at 8 K, its rate constant was determined to be s. Find the activation energy of the decomposition of N O 4 (g). T 74 K k s T 8 K k s R 8.45 J/(K mol) E a? k ln k E a R T T k R ln E a k T T (.45 J/ ( K mol) ) K s ln.00 0 s 8 K 570 J/mol E a 57. kj/mol Copyrighted by Gabriel Tang B.Ed., B.Sc. age.

20 Unit 5: Chemical Kinetics and Equilibria Chemistry A Example : The decomposition of N O (g) N (g) + O (g) yields the following data. Temperature (K) k (L/(mol s) a. Complete the table below. b. Graph ln(k) versus (/T). Assess the linearity to determine the activation energy. c. Draw the potential energy diagram for the decomposition of N O (g) at 98 K. ( H f 8 kj/mol) Temperature (K) k (L/(mol s) /T (K ) ln(k) b. Using the TI-8 lus Calculator, we can enter the experimental results in the STATS Editor and quickly manipulate the data.. Enter Values in the L and L Columns of Stats Editor STAT ENTER. Enter /L in the heading of L. Enter ln(l ) in the heading of L 4. Cursor must be on the column heading before entering formula MEM must be used to enter formula ALHA + To access L, press nd L 4. Fill in the Table Temperature (K) k (L/(mol s) /T (K ) ln(k) age 4. Copyrighted by Gabriel Tang B.Ed., B.Sc.

21 Chemistry A Unit 5: Chemical Kinetics and Equilibria b. Graph ln(k) versus (/T). Assess the linearity to determine the activation energy.. Turn On STATS LOT. Set WINDOW. Graph nd STAT LOT ENTER WINDOW x: [0, 0.005, 0.000] y: [ 0, 0, 5] GRAH Y To access L, press ln(k) nd L /T (K ) To access L4 nd L 4, press 4 ln(k) versus (/T) Graph forms Straight Line. We can use the linear regression function of the calculator to find the value of the slope and y- intercept for calculating E a and A respectively.. Turn Diagnostic On nd ENTER Note: After DiagnosticOn is CATALOG Again selected; it will remain ON even when the 0 calculator is turned Off. However, resetting the calculator will turn the ENTER Select DiagnosticOn Diagnostic Off (factory setting).. Obtain LinReg (ax + b) L, L 4, Y. Copy Equation to Y Screen, and Graph Least Square Line on Scatter lot. ENTER nd L To access Y, press Select Option STAT Select CALC, use for Y VARS ENTER, nd L4, 4 Choose Option 4 Linear Regression (ax + b) Select Option Copyrighted by Gabriel Tang B.Ed., B.Sc. age 5.

22 Unit 5: Chemical Kinetics and Equilibria Chemistry A ENTER Y Be Sure lot is On GRAH When r is extremely close to ±, it indicates the line has a linear relationship. ln(k) /T (K ) y x compares to E ln(k) a + ln(a) R T slope K R E a slope Ea (slope)r ( K)(8. J/(mol K)) J/mol E a 5 kj/mol c. Draw the potential energy diagram for the decomposition of N O (g) at 98 K. ( H f 8 kj/mol) E 4 kj 8 kj N O (g) E a 5 kj 0 kj Reaction Coordinates N (g) + ½ O (g) H 8 kj age 6. Copyrighted by Gabriel Tang B.Ed., B.Sc.

23 Chemistry A Unit 5: Chemical Kinetics and Equilibria.8: Catalyst Catalyst: - a substance that speeds up the reaction without being consumed in the reaction. - unlike intermediates, catalyst is used and recycled in the reaction. - lowers activation energy by providing an alternate reaction pathway. ( E a is lowered but H rxn remains the same.) Enzyme: - a catalyst in the body that speeds up certain bodily reaction. E a without Catalyst E a with Catalyst H rxn Two Types of Catalyst. Heterogeneous Catalyst: - a catalyst that exists in a different phase compared to the reactants. - utilizes the concept of adsorption to hold down the reactants for faster reaction rate. Adsorption: - collection of a substance on the surface of another substance. - the opposite of absorption (passing a substance through another). In general, heterogeneous catalytic reaction involves four steps.. Adsorption of the reactants and activation of the heterogeneous catalyst on the surface.. Migration of the adsorbed reactants on the catalytic surface.. Reaction of the adsorbed substances. 4. Desorption or release of the products. Examples: a. Ammonia is formed from its elements using heterogeneous catalyst such as t (s): N (g) + H (g) t NH (g) (Check out Video at b. The catalytic converter converts NO (g) (result of burning nitrogen at high temperature) to N (g) and O (g). The O (g) along with the catalytic converter is used to produce CO (g) from CO (g). catalytic NO (g) converter N (g) + O (g) catalytic CO (g) + O (g) converter CO (g) A catalytic converter for most modern vehicle. Leaded gasoline deactivates catalytic converter. Therefore, they are not legally used in vehicles Copyrighted by Gabriel Tang B.Ed., B.Sc. age 7.

24 Unit 5: Chemical Kinetics and Equilibria Chemistry A. Homogeneous Catalyst: - a catalyst that exists in the same phase as the reactants. Example : The destruction of ozone in upper atmosphere can be attributed to NO (g) and CFCs acting as catalysts. NO (g) is produced from the combustion of N (g) at high temperature commonly found in internal combustion engine (high-altitude aircraft produces lots of NO (g)). CFCs (chlorofluoro-carbon compounds) are found in aerosol can propellants, refrigerators, and air conditioners. They break down to form Cl (g) with the presence of light. Use the following two reactions mechanisms, identify the intermediates and catalysts for the decomposition processes of ozone in the upper atmosphere. 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) 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) Intermediate: NO (g) Catalyst: NO (g) Intermediate: ClO (g) Catalyst: Cl (g) age 8. Assignment.7 pg. 60 #49 to 58.8 pg. 60 #6 and 6 (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. Copyrighted by Gabriel Tang B.Ed., B.Sc.

25 Chemistry A Unit 5: Chemical Kinetics and Equilibria Chapter : Chemical Equilibrium.: The Equilibrium Condition Chemical Equilibrium: - the state at which the concentrations of all reactants and products remain constant with time (the Forward Reaction Rate Reverse Reaction Rate). - the equilibrium state is dynamic (not static). Chemical species are continuously converting from reactants to products and vice versa. It appears that the reaction has stopped only because the rate of consumption rate of production. - if an equilibrium state is disturbed (changing concentrations of species, pressure, volume and temperature change), the reaction will shift towards one side in order to re-establish the new equilibrium state. - like reaction rate, Equilibrium is affected by Temperature. aa + bb cc + dd a and b: Coefficients of reactant species A and B c and d: Coefficients of product species C and D equilibrium symbol (indicating forward and reverse reactions) Equilibrium State [D] Concentration [C] [B] [A] Time to reach Time Equilibrium (Check out Video at The N O 4 (g) NO (g) equilibrium. Initially (icture A), there were very little NO (g). As time proceeded forward, more NO (g) (brown color) is produced (ictures B & C). However, there are still N O 4 (g) present at the equilibrium state (icture D). Copyrighted by Gabriel Tang B.Ed., B.Sc. age 9.

26 Unit 5: Chemical Kinetics and Equilibria Chemistry A.: The Equilibrium Constant Law of Mass Action: - the general description of the equilibrium condition that is expressed in a form of an equilibrium expression. Equilibrium Expression: - an expression relating the concentrations or pressures of the reactants and products when they are at the state of equilibrium. - it takes the form of the individual products raised to the power of their respective coefficients divided by the individual reactants raised to the power of their respective coefficients. - the equilibrium expression is unique for each reaction, but it is the same for that particular reaction regardless 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. - the symbol for equilibrium constant when the expression deals with concentrations is simply K or K c. When the expression deals with pressures, it is symbolized as K. - reversing equilibrium reaction will cause the reciprocate the equilibrium constant (/K) - multiplying the equation by a multiple n will result in raising the K by the power of n. Equilibrium Expression and Constant of a Reaction aa + bb cc + dd K K c c d C D [ A] a [ B] b K c C a A d D b B Equilibrium Expressions a and b: Coefficients of reactant species A and B c and d: Coefficients of product species C and D [A], [B], [C] & [D] [A] eq, [B] eq, [C] eq & [D] eq 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) Both K c and K are Unitless K Equilibrium Constant (ressures) Reversing Equilibrium Reaction aa + bb cc + dd cc + dd aa + bb K a b A B [ C] c [ D] d c [ C] [ D] a [ A] [ B] K Reverse Equilibrium Constant age 40. d b K K Multiplying Equilibrium Reaction by a Factor of n n(aa + bb cc + dd) naa + nbb ncc + ndd) K nc nd C D [ A] na [ B] nb c [ C] [ D] a [ A] [ B] K New Equilibrium Constant Copyrighted by Gabriel Tang B.Ed., B.Sc. d b n K n

27 Chemistry A Unit 5: Chemical Kinetics and Equilibria Example : 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 [ BrF5 ] [ Br ][ F ] 5 BrF 5 K 5 Br F 8 NO H O K [ N H ][ H O ] NO H K 6 N H 4 O H O Example : For the following reaction and the equilibrium concentrations at 00 K. C H 6 (g) + NH (g) + O (g) C H N (g) + 6 H O (g) [C H 6 ] eq M [NH ] eq 0.50 M [O ] eq 0.50 M [C H N] eq.50 M [H O] eq.00 M a. Write the equilibrium expression and determine the equilibrium constant. b. Write the equilibrium expression and 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 and find 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)s 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 ] 6 H N H O CH6 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 K ( ) K [ CH N][ H O] [ C H ][ NH ][ O ] 6 [ CHN][ H O] [ C H ][ NH ][ O ] 6 K (.80 0 ) 6 (All Coefficients Multiplied by ½) 6 [ CHN] [ H O] [ C H ] [ NH ] [ O ] 6 K 6 K.6 0 Copyrighted by Gabriel Tang B.Ed., B.Sc. age 4.

28 Unit 5: Chemical Kinetics and Equilibria Chemistry A Equilibrium osition: - the concentrations or pressures of all chemical species at equilibrium state. - depends strongly on the Initial Concentrations of the chemical species. (In contrast, K does NOT depend on initial concentrations, only on temperature and 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] eq!! 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 performed at 70 K using different initial concentrations. H (g) + I (g) HI (g) Experiment Experiment Initial Equilibrium Initial Equilibrium [H ] M [H ] eq 0.0 M [H ] 0 0 M [H ] eq M [I ] M [I ] eq 0.0 M [I ] M [I ] eq M [HI] 0 0 M [HI] eq 0.56 M [HI] M [HI] eq 0.80 M Experiment Experiment 4 Initial Equilibrium Initial Equilibrium [H ] M [H ] eq M [H ] 0 0 M [H ] eq M [I ] 0 0 M [I ] eq 0.05 M [I ] 0 0 M [I ] eq M [HI] M [HI] eq 0.00 M [HI] M [HI] eq 0. M a. Write the equilibrium expression for the formation of HI (g). b. Calculate the equilibrium constant for each experiment, and average them for an overall value. a. Equilibrium Expression: K [ HI] [ H ][ I ] b. Equilibrium Constant: [ HI] eq ( 0.56) [ HI] eq Experiment : K Experiment : K [ H ] eq [ I ] eq ( 0.0)( 0.0) H eq I K 49.4 K 49.8 [ HI] eq ( 0.00) [ HI] eq Experiment : K Experiment 4: K [ H ] eq [ I ] eq ( 0.050)( 0.05) 4 H eq I K 49. K [ ] eq [ ] eq ( 0.80) ( 0.050)( ) ( 0.) ( 0.044)( 0.044) Average K K avg 49.5 age 4. Assignment. pg. 645 #5 and 7. pg #9 to 6 Copyrighted by Gabriel Tang B.Ed., B.Sc.

29 Chemistry A Unit 5: Chemical Kinetics and Equilibria.: Equilibrium Expressions Involving ressures Converting K c to K : V nrt (Ideal Gas Law) V n (Solving for and let Concentration) V c d C D K a b (Equilibrium Expression and Constant for aa + bb cc + dd) A B Substituting the pressures for each chemical species into the equilibrium expression. ( A [A]RT ; B [B]RT ; C [C]RT ; and D [D]RT) c d ([ C] RT ) ([ D] RT ) c K ( ) c d ( ) d C RT D RT ([ A] RT ) a ([ B] RT ) b [ A] a ( RT ) a [ B] b ( RT ) b c d K ( ) c+ d C D RT c d C D c d (RT) a b a+ b [ A] [ B] ( RT ) [ A] a [ B] (c + d) (a + b) C D (Replace b [ A] a [ B] b K K c (RT) (c + d) (a + b) (Combine exponents on common base RT using Laws of Exponents) with K c ) Conversion between Concentration and ressure Equilibrium Constants aa + bb cc + dd K K c (RT) n where n Σ Coefficients of roducts Σ Coefficients of Reactants (c + d) (a + b) R K T Temperature in K R has a unit of K to cancel out with the Kelvin from Temperature only, therefore K calculated will remain unitless 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 T 575 K K K c (RT) n K ( )[( K )(575 K)] K ( )[47.845] n Σn products Σn reactants n ( + ) ( + ) 4 n K K? Copyrighted by Gabriel Tang B.Ed., B.Sc. age 4.

30 Unit 5: Chemical Kinetics and Equilibria Chemistry A Example : The German s Haber-Bosch process developed in 9 utilizes an iron surface that contains traces of aluminum and potassium oxide as a catalyst to manufacture ammonia from nitrogen and hydrogen. It is an important process as ammonia is commonly used in fertilizer and ammunition. In 98, the scientist Fritz Haber won the Nobel rize in chemistry for his contribution. N (g) + H (g) NH (g) age 44. a. At 400 K, NH 0.04 atm, N.8084 atm and H expression in terms of pressure and calculate K. b. Convert the calculated K above to K. NH a. Equilibrium Expression: K b. K.5 0 R K T 400 K n Σn products Σn reactants n () ( + ) 4 n K?.4: Heterogeneous Equilibria 0.00 atm. Write the equilibrium Homogeneous Equilibria: - an equilibrium system where all chemical species are in the same phase. Heterogeneous Equilibria: - an equilibrium system where some chemical species are in different phase compare to the others. - chemical species that are ure Solid or ure Liquid are NOT Included in the Equilibrium Expression. This is due to the fact that pure solids and liquids do not have concentrations. Example : 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 and Na are ure Solids) c. Ag SO 4 (s) Ag + (aq) + SO 4 (aq) d. HF (aq) + H O (l) H O + (aq) + F (aq) K [Ag + ] [SO 4 ] (Ag SO 4 is a ure Solid) (No K because there are no gases.) Assignment. pg. 646 #7 and 0.4 pg. 646 # to 4 N H K K c (RT) n K K c ( ) n RT Equilibrium Constant: K K.5 0 ( ).5 0 [( K )( )] 400 K K [ H O] [ NH ] [ CO] ( 0.04) (.8084)( 0.00) K c K (H NCONH is a ure Solid) [ O ][ F ] K H + [ HF] HO NH CO (H O is a ure Liquid) (No K because there are no gases.) Copyrighted by Gabriel Tang B.Ed., B.Sc.

31 Chemistry A Unit 5: Chemical Kinetics and Equilibria.5: Applications of Equilibrium Constant Important Notes Regarding the Size of the Equilibrium Constant (K): aa + bb cc + dd c d C D K K c a eq eq A b eq B eq. When K >>, the equilibrium system favours the products. There are more products than reactants at the state of equilibrium. ([C] eq and [D] eq or C, eq and D, eq >> [A] eq and [B] eq or A, eq and B, eq ). When K <<, the equilibrium system favours the reactants. There are less products than reactants at the state of equilibrium. ([A] eq and [B] eq or A, eq and B, eq >> [C] eq and [D] eq or C, eq and D, eq ). When K, the equilibrium system favours neither the products nor the reactants. There are roughly the same amount of products and reactants at the state of equilibrium. ([C] eq and [D] eq or C, eq and D, eq [A] eq and [B] eq or A, eq and B, eq ) 4. The Size of K has NO Relationship with the Rate of Reaction to reach the state of equilibrium. Reaction Rate is dependent on Activation Energy and Temperature (T) (NOT K). K c C, eq a A, eq 5. The Size of K depends on Free Energy (G) and Temperature (T). d D, eq b B, eq Reaction Quotient (Q): - the mass action expression under any set of conditions (not necessarily equilibrium). - its magnitude relative to K determines the direction in which the reaction must occur to establish equilibrium. c d C D Q [ A] a [ B] b a. When Q > K, the system shifts to the reactants. At a specific condition, Q indicates that there are too much products. Therefore, the system has to shift back to the left. b. When Q < K, the system shifts to the products. At a specific condition, Q indicates that there are not enough products. Therefore, the system has to shift forward to the right. c. When Q K, the system is at equilibrium. At a specific condition, Q indicates that all concentrations are that of the equilibrium state. Therefore, there will be no shifting. Example : The reaction, NO (g) + O (g) NO (g) has an equilibrium constant of K 9.54 at 600 K. Indicate the direction in which the system will shift to reach equilibrium when the [NO] M, [O ] M, and [NO ] M. [NO] M [O ] M [NO ] M Q? Q [ NO ] 0 [ NO] 0[ O ] 0 ( 0.500) ( 0.00) ( 0.50) Q. Since Q < K (. < 9.54), the system will shift to the product (NO ). There are not enough NO at the initial conditions. Copyrighted by Gabriel Tang B.Ed., B.Sc. age 45.

32 Unit 5: Chemical Kinetics and Equilibria Chemistry A Example : The reaction, Cl (g) + F (g) ClF (g) at 50 K has K 50., If the equilibrium concentrations of Cl (g) and ClF (g) are 0.49 M and 0.05 M respectively, what is the equilibrium concentration of F (g)? [Cl ] eq 0.49 M [ClF ] eq 0.05 M K 50. [F ] eq? K [ ClF ] eq [ Cl ] [ F ] eq eq 50. [F ] eq [F ] eq ( 0.05) ( 0.49)[ F ] eq ( 0.05) ( 50.)( 0.49) ( 0.05) ( 50.)( 0.49) [F ] eq 0.78 M ICE Box: - stands for Initial, Change, and Equilibrium. It is a table that organizes information to calculate final equilibrium concentrations given the equilibrium constant and 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 contained an initial concentration of.00 mol of HCl (g) and.85 mol of H (g). When the system reached equilibrium, it was found that there were mol of Cl (g). Determine the concentrations of H (g) and HCl (g) at equilibrium. K mol [H ] M 5.00 L.00 mol [HCl] M 5.00 L [Cl ] 0 0 M mol [Cl ] eq 0.7 M 5.00 L [H ] eq? [HCl] eq? 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, and the change to the HCl would be negative. Since there is 0.7 M of Cl at equilibrium, it means 0.7 M of H is added (: mol ratio between Cl and H ). It also means that there is (0.7 M) less HCl (: mol ratio between Cl and HCl). H (g) + Cl (g) HCl (g) Initial M 0 M M Change M M (0.7 M) Equilibrium 0.94 M 0.7 M 0.56 M Verify with K: K [ HCl] eq [ H ] eq [ Cl ] eq ( 0.56) ( 0.94)( 0.7) K This matches with K given in the question. Therefore, the equilibrium concentrations are: [H ] eq 0.94 M, [Cl ] eq 0.7 M and [HCl] eq 0.56 M age 46. Copyrighted by Gabriel Tang B.Ed., B.Sc.

33 Chemistry A Unit 5: Chemical Kinetics and Equilibria Example 4: The reaction, NO (g) NO (g) + NO (g) has K.7 at 700 K. A.50 L flask at 700 K has.00 mol of each species initially. Calculate the concentrations of all species at equilibrium. K.7.00 mol [NO ] M.50 L.00 mol [NO ] M.50 L.00 mol [NO] M.50 L [NO ] eq? [NO ] eq? [NO] eq? First, we must determine Q and the direction of the shift. [ NO ] NO 0 Q [ NO ] 0 0 ( )( ) ( 0.400) Q.00 Since Q < K (.00 <.7), the system will shift to the products (NO and NO). There are not enough products at the initial conditions. Hence, the change to the NO is negative, and the changes to the NO and NO would be positive. Let x amount of change per mole, since there are moles of NO reacted; NO will be lowered by x. Similarly, NO and NO will be increased by x 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 and solve for x. (Since we have a set of common exponents we can group it and square root both sides.) [ NO ] NO eq eq K.7 ( x )( x ) ( ) x ( x) [ NO ] ( 0.4 x) ( 0.4 x) ( 0.4 ) x eq ( x).7 ( 0.4 x) ( x ) ( 0.4 x) (0.4 x) x x x x x x Verify with K: NO NO eq K [ NO ] eq eq ( 0.55 )( 0.55 ) ( ) K.7 This matches with the K given in the question. Finally, substitute x back into the mathematical expressions. [NO ] eq x (0.506) [NO ] eq M [NO ] eq x (0.506) [NO ] eq 0.55 M [NO] eq x (0.506) [NO] eq 0.55 M Copyrighted by Gabriel Tang B.Ed., B.Sc. age 47.

34 Unit 5: Chemical Kinetics and Equilibria Chemistry A 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 0.0 L container at 500 K. K mol [SO ] M 0.0 L.00 mol [Cl ] M 0.0 L.00 mol [SO Cl ] M 0.0 L [SO ] eq? [Cl ] eq? [SO Cl ] eq? First, we must determine Q and the direction 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 and Cl ). There is too much product at the initial conditions. Hence, the changes to the SO and Cl are positive, and the change to the SO Cl would be negative. Let x amount of change per mole, since there is mole of SO reacted; SO will increase by x. Similarly, Cl will increase by x. SO Cl, on the other hand, will decrease by x. age 48. 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 and solve for x. (Since we don t have any common exponent, we have to expand the denominator.) [ SOCl ] eq ( 0. x) ( 0. x) K SO Cl 0. + x 0. + x x + x [ ] eq [ ] eq Assignment.5 pg. 646 #5 to 46 ( )( ) ( ) ( x + x ) (0. x) x x 0. x 0.486x +.944x (Quadratic Equation: Apply the Quadratic Formula!) b ± b 4ac a b.944 c x a.944 ± x ( ) (.944) 4( 0.486)( ) ( 0.486) x x.6005 (omit negative x) Verify with K: [ SOCl ] eq ( 0.057) K K [ SO ] eq [ Cl ] eq ( 0.4)( 0.4) This matches closely with the K given in the question. Finally, substitute x back into the mathematical expressions. [SO ] eq x (0.487) [SO ] eq 0.4 M [Cl ] eq x (0.487) [Cl ] eq 0.4 M [SO Cl ] eq 0.00 x 0.00 (0.487) [SO Cl ] eq M Copyrighted by Gabriel Tang B.Ed., B.Sc.

35 Chemistry A Unit 5: Chemical Kinetics and Equilibria.6: Solving Equilibrium roblems Steps to Solve Equilibrium roblems:. Write the Balanced Chemical Equation for the system. This includes all the correct states for all species.. Write the Equilibrium Expression and equate it to K value given.. List the Initial Concentrations. 4. Determine Q and compare it to K to determine the direction of the shift to equilibrium. 5. Construct the ICE Box and define the change amount of change per mole as x. Using the coefficients and the direction 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 methods of simplifying like common exponents if they exist. Otherwise, the quadratic equation will be needed to solve for x.) 7. Calculate the equilibrium concentrations of each species and verify that they indeed give the value of K. 8. For equilibrium involving pressures, the procedure is the same as above. Example : 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 had 6.00 mol of C H 6 (g), 6.00 mol of Cl (g) and 6.00 mol HCl (g) originally in a.00 L container at 8 K, determine 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 ] eq? [Cl ] eq? [HCl] eq? First, we must determine Q and the direction of the shift. (Be careful! This is a heterogeneous system. C H 5 Cl is a pure solid and it is not involved in the equilibrium expression. Q [ HCl] 0 [ C H 6 ] [ Cl ] 0 0 (.00) (.00)(.00) Q Since Q > K (0.500 > 0.00), the system will shift to the reactants (C H 6 and Cl ). There are too much products at the initial conditions. Hence, the changes to the C H 6 and Cl are positive, and the change to the HCl would be negative. Let x amount of change per mole, since there is mole of C H 6 reacted; C H 6 will increase by x. Similarly, Cl will increase by x. HCl, on the other hand, will decrease by x. 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 Copyrighted by Gabriel Tang B.Ed., B.Sc. age 49.

36 Unit 5: Chemical Kinetics and Equilibria Chemistry A Next, we set up the equilibrium expression, substitute the above mathematical expressions and solve for x. (Since we have a set of common exponents we can group it and square root both sides.) [ HCl] 0 ( x) ( x) K 0.00 [ C H 6 ] 0 [ Cl ] 0 ( + x)( + x) ( 4 + 4x + x ) 0.00 (4 + 4x + x ) ( x) x + 0.x x 0.x +.4x.6 0 b ± b 4ac x a 0. b.4 c.6 a (.4) ± (.4) 4( 0.)(.6) Finally, substitute x back into the x ( 0.) mathematical expressions. x.0658 x (omit negative x) [C H 6 ] eq.00 + x.00 + ( ) [C H 6 ] eq.06 M Verify with K: [ HCl] 0 K C H Cl [ 6 ] [ ] 0 0 ( 0.98) (.06)(.06) K 0.00 This matches closely with the K given in the question. [Cl ] eq.00 + x.00 + ( ) [Cl ] eq.06 M [HCl] eq.00 x.00 ( ) [HCl] eq 0.98 M Example : 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) and COCl (g) are atm, atm and 0.50 atm respectively. They are all mixed in a 7.50 L flask at 450 K. Find the equilibrium pressures. K 0.89 CO, atm atm Cl, 0 COCl, 0 CO, eq? Cl, eq? COCl, eq? 0.50 atm First, we must determine Q and the direction of the shift. Q COCl, 0 CO, 0 Cl, 0 ( 0.50) ( 0.800)( 0.900) Q 0.08 Since Q < K (0.08 < 0.89), the system will shift to the product (COCl ). There is too little product at the initial conditions. Hence, the changes to the CO and Cl are negative, and the change to the COCl would be positive. Let x amount of change per mole, since there is mole of CO reacted; CO will decrease by x. Similarly, Cl will decrease by x. COCl, on the other hand, will increase by x. CO (g) + Cl (g) COCl (g) Initial atm atm 0.50 atm Change x x + x Equilibrium (0.800 x) atm (0.900 x) atm ( x) M age 50. Copyrighted by Gabriel Tang B.Ed., B.Sc.

37 Chemistry A Unit 5: Chemical Kinetics and Equilibria Next, we set up the equilibrium expression, substitute the above mathematical expressions and solve for x. (Since we don t have any common exponent, we have to expand the denominator.) COCl, eq ( x) ( x) K 0.89 CO, eqcl, eq ( 0.8 x)( 0.9 x) ( 0.7.7x + x ) 0.89 (0.7.7x + x ) (0.5 + x) x x x 0.89x.49x (Quadratic Equation: Apply the Quadratic Formula!) b ± b 4ac x a 0.89 b.49 c x a.49 ( ) ± (.49) 4( 0.89)( ) ( 0.89) x (omit larger value bigger than atm and atm) Finally, substitute x back into the mathematical expressions. CO, eq x ( ) CO, eq 0.76 atm x Verify with K: COCl K CO, eq K 0.88, eq Cl, eq ( 0.89) ( 0.76)( 0.86) (This matches closely with the K given in the question.) Cl, eq x ( ) Cl, eq 0.86 atm COCl, eq x ( ) COCl, eq 0.89 M.7: Le Châtelier s rinciple Le Châtelier s rinciple: - a qualitative method to predict the shift on an equilibrium system if it is disturbed by means of changing concentration, pressure and temperature. - the equilibrium will shift in the direction that minimizes the change imposed on the system.. Effects of a Change in Concentration: a. An ADDITION of a species on one side of the equilibrium will Drive the System TOWARDS the Opposite Side. (There is more concentration of the species being added. Hence, the system will shift towards the opposite side to reduce the increased amount of that particular species.) b. A REMOVAL of a species on one side of the equilibrium will Drive the system TOWARDS the Same Side. (There is less concentration of the species being removed. Hence, the system will shift towards the removal side to compensate.) Copyrighted by Gabriel Tang B.Ed., B.Sc. age 5.

38 Unit 5: Chemical Kinetics and Equilibria Concentration [D] [C] [B] [A] Changes in Concentrations on Equilibrium System aa + bb cc + dd Chemistry A indicates an Increase in [D]. As [D], equilibrium shifts to the left (aa + bb cc + dd). Hence, [A], [B], and [C]. indicates a Decrease in [A]. As [A], equilibrium shifts to the left (aa + bb cc + dd). Hence, [B], [C], and [D]. indicates an Increase in [B]. As [B], equilibrium shifts to the right (aa + bb cc + dd). Hence, [A], [C], and [D]. indicates a Decrease in [C]. As [C], equilibrium shifts to the right (aa + bb cc + dd). Hence, [A], [B], and [D]. Time. Effects of a Change in ressure: a. Adding an Inert Gas has NO CHANGE on the equilibrium system. This is because an inert gas does not participate in the forward or reverse reaction. b. Reducing the Volume will Drive the System TOWARDS the Side With LESS Gaseous Molecules. Since there are less space for the number of molecules, the system will have to shift to the side with lesser gaseous molecules to compensate. c. Conversely, Expanding the Volume will Drive the System TOWARDS the Side With MORE Gaseous Molecules. Now that there is more room for the molecules to move about, the system will shift to the side that has more gaseous molecules to adjust to the new condition. d. When there are Equal Number of Gaseous Molecules on Both Side of the Equilibrium, any Change in Volume will NOT Affect System. age 5. Copyrighted by Gabriel Tang B.Ed., B.Sc.

39 Chemistry A Unit 5: Chemical Kinetics and Equilibria Changes in ressures on a Gaseous Equilibrium System aa + bb cc + dd when (c + d) > (a + b) [D] [C] Concentration [B] [Inert Gas] [A] indicates an Increase in Volume. As V, equilibrium shifts to the right (aa + bb cc + dd) since there are more gaseous molecules on the product side [(c + d) > (a + b)]. Hence, [A], [B], [C], and [D]. indicates a Decrease in Volume. As V, equilibrium shifts to the left (aa + bb cc + dd) since there are LESS gaseous molecules on the reactant side [(c + d) > (a + b)]. Hence, [A], [B], [C], and [D]. indicates an Addition of an Inert Gas. There is no shifting of the equilibrium. (aa + bb cc + dd) as inert gas does not affect the system. Hence, [A], [B], [C], and [D] remain unchanged. (See the Video at Effects of a Change in Temperature: - look at the energy (written in the reactant or product side) as a chemical species. Then, the predictions will be the same as those found with changing the concentrations. a. For an Exothermic Equilibrium System: aa + bb cc + dd + Energy Time - an Increase in Temperature will drive the system to the left (aa + bb cc + dd + Energy). There is more heat added and because energy is written on the product side, the system will shift to the opposite side to re-establish equilibrium. Hence, [A], [B], [C], and [D]. - a Decrease in Temperature will drive the system to the right (aa + bb cc + dd + Energy). There is less heat overall and because energy is written on the product side, the system will shift to the products to compensate. Hence, [A], [B], [C], and [D]. Copyrighted by Gabriel Tang B.Ed., B.Sc. age 5.

40 Unit 5: Chemical Kinetics and Equilibria Chemistry A b. For Endothermic Equilibrium System: aa + bb + Energy cc + dd - a Decrease in Temperature will drive the system to the left (aa + bb + Energy cc + dd). There is less heat overall and because energy is written on the reactant side, the system will shift to the reactants to compensate. Hence, [A], [B], [C], and [D]. - an Increase in Temperature will drive the system to the right (aa + bb + Energy cc + dd). There is more heat added and because energy is written on the reactant side, the system will shift to the opposite side to re-establish equilibrium. Hence, [A], [B], [C], and [D]. Changes in Temperature on an Exothermic Changes in Temperature on an Endothermic Equilibrium System aa + bb cc + dd + Energy Equilibrium System aa + bb + Energy cc + dd [D] [D] [C] [C] Concentration [B] [A] Concentration [B] [A] Time Time indicates a Decrease in Temperature. As T, equilibrium shifts to the right of an exothermic system (aa + bb cc + dd + Energy). Hence, [A], [B], [C], and [D]. indicates an Increase in Temperature. As T, equilibrium shifts to the left of an exothermic system (aa + bb cc + dd + Energy). Hence, [A], [B], [C], and [D]. indicates an Increase in Temperature. As T, equilibrium shifts to the right of an endothermic reaction (aa + bb + Energy cc + dd). Hence, [A], [B], [C], and [D]. indicates a Decrease in Temperature. As T, equilibrium shifts to the left of an endothermic reaction (aa + bb + Energy cc + dd). Hence, [A], [B], [C], and [D]. 4. Effects of Adding a Catalyst: - has NO CHANGE on the equilibrium system. (See Video at age 54. Copyrighted by Gabriel Tang B.Ed., B.Sc.

41 Chemistry A Unit 5: Chemical Kinetics and Equilibria Example : The equilibrium system, 4 F (g) 4 (s) + 6 F (g) kj, is put under the following changes. redict the shift of the system and the resulting concentrations of all species for each case. a. an increase in the concentration of F (g). b. a decrease 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 solid and does not involve with the equilibrium system) Effect: [F ] and [F ] remain the same. c. a decrease in the concentration of F (g). d. a decrease in Temperature. The system will shift to the LEFT. 4 F (g) 4 (s) + 6 F (g) kj Effect: [F ] (decrease) The system will shift to the RIGHT. 4 F (g) 4 (s) + 6 F (g) kj Effect: [F ] (decrease) and [F ] (increase) e. an addition 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 and does not involve with the equilibrium system) Effect: [F ] and [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 product side 6 moles of F (g) versus 4 moles of F (g) ) Effect: [F ] (decrease) and [F ] (increase) Example : The Haber-Bosch process, N (g) + H (g) NH (g) + 9 kj is essentially an equilibrium system. A chemical engineer would like the highest yield of ammonia. List all the possible method of production that will ensure maximum amount of NH (g) produced. N (g) + H (g) NH (g) + 9 kj (Desire Effect: [NH ], which means driving the system forward.). Increase the concentrations of N (g) or H (g) or both will drive the system forward.. Decrease the concentration of NH (g) as it is produced will shift the system forward.. Lower the Temperature will drive the system to the product side. 4. Decrease the Volume of the system will shift the system to the right due to smaller number of gaseous molecules on the product side. Assignment.6 pg #47 to 56.7 pg #57 to 64 Copyrighted by Gabriel Tang B.Ed., B.Sc. age 55.

42 Unit 5: Chemical Kinetics and Equilibria Chemistry A 6.7: The Dependence of Free Energy on ressure Relationship between Free Energy and ressure: - the free energy of a gas at any pressure, G, can be calculated when comparing it to the free energy at standard pressures ( atm), G, along with the stated temperature in Kelvin and the gas constant. Relationship between Free Energy and ressure G G + RT ln() G Free Energy (J) of the gas at ressure of atm R Gas Constant 8.45 J/K G Free Energy (J) of gas at Standard ressure of atm T Temperature in K ressure of Gas in atm When atm, then G G + RT ln() and G G because ln() 0 Calculating the G of a System with Gaseous roducts and Reactants at any ressure, : aa (g) + bb (g) cc (g) + dd (g) G cg C + dg D ag A bg B ( G ΣnG products ΣnG reactants ) G c[g C + RT ln( C )] + d[g D + RT ln( D )] a[g A + RT ln( A )] b[g B + RT ln( B )] (Expand each term using G G + RT ln()) G cg C + crt ln( C ) + dg D + drt ln( D ) ag A art ln( A ) bg B brt ln( B ) (Multiply moles into each brackets) G [c G C + dg D ag A bg B ] + crt ln( C ) + drt ln( D ) art ln( A ) brt ln( B ) (Collect G terms: [cg C + dg D ag A bg B ] G ) G G + RT [c ln( C ) + d ln( D ) a ln( A ) b ln( B )] (Take out common factor RT from other terms) G G + RT [ln( C c ) + ln( D d ) ln( A a ) ln( B b )] (Apply law of logarithm: n log x log x n ) c d C D G G + RT ln a b A B [Apply logarithm laws: logm + logn log(mn) and logm logn log(m/n)] G G + RT ln (Q) c d C D [Substitute Reaction Q as ] a b A B Free Energy and ressures of Gaseous System aa (g) + bb (g) cc (g) + dd (g) G G + RT ln(q) G Free Energy (J) of the gas at ressure of atm R Gas Constant 8.45 J/K G Free Energy (J) of gas at Standard ressure of atm T Temperature in K c d C D Q Reaction Quotient of Gaseous Species a b A B When Q, then G G + RT ln() and G G because ln() 0 age 56. Copyrighted by Gabriel Tang B.Ed., B.Sc.

43 Chemistry A Unit 5: Chemical Kinetics and Equilibria Example : At 5 C, the system, NO (g) + Cl (g) NOCl (g) has the following partial pressures of NO (g), Cl (g), and NOCl (g) are.40 atm, 5.0 atm and.5 atm respectively. Determine the change in free energy for the system at the partial pressures indicated. (G of NO (g) 87 kj/mol, G of Cl (g) 0 kj/mol and G of NOCl (g) 66 kj/mol) G NO 87 kj/mol o G Cl 0 kj/mol G NOCl 66 kj/mol NO.40 atm Cl 5.0 atm NOCl.5 atm T 5 C 98.5 K G? Q? G? NO (g) + Cl (g) NOCl (g) G ΣG products ΣG reactants G [( mol)(66 kj/mol)] [( mol)(87 kj/mol) + ( mol)(0 kj/mol)] G 4 kj J ( G < 0; Forward Reaction is Spontaneous) Q NOCl NOCl (.5) (.40) ( 5.0) Q G G + RT ln(q) G ( J) + (8.45 J/K)(98.5 K) ln( ) G J G 45 kj (It became more spontaneous towards the product side. This probably means that Q < K or K > 0.46) Example : At 5 C, the system, H O (g) + Cl O (g) HOCl (g) has the following partial pressures of H O (g), Cl O (g), and HOCl (g) are 0.50 atm, 0.0 atm and atm respectively. Determine the change in free energy for the system at the partial pressures indicated. (G of H O (g) 9 kj/mol, G of Cl O (g) 98 kj/mol and G of HOCl (g) 0 kj/mol) o G H O 9 kj/mol o G Cl O 98 kj/mol G HOCl 0 kj/mol H O 0.50 atm Cl O 0.0 atm HOCl atm T 5 C 98.5 K G? Q? G? H O (g) + Cl O (g) HOCl (g) G ΣG products ΣG reactants G [( mol)(0 kj/mol)] [( mol)( 9 kj/mol) + ( mol)(98 kj/mol)] G kj. 0 5 J ( G > 0; Forward Reaction is Non-spontaneous) Q HOCl HO ClO ( 0.860) ( 0.50)( 0.0) Q G G + RT ln(q) G (. 0 5 J) + (8.45 J/K)(98.5 K) ln( ) J G 7 kj (It became more spontaneous towards the reactants (even more nonspontaneous towards the product side). This probably means that Q > K or K < ) Copyrighted by Gabriel Tang B.Ed., B.Sc. age 57.

44 Unit 5: Chemical Kinetics and Equilibria Chemistry A 6.8: Free Energy and Equilibrium Equilibrium oint: - the point where the system attains thermodynamic equilibrium the lowest value of free energy available to the equilibrium system. - this occurs when Q K and G 0 at certain partial pressures. - we can use equilibrium point to calculate the value of the equilibrium constant K from G Equilibrium oint (Q K and G 0) of a Gaseous System aa (g) + bb (g) cc (g) + dd (g) 0 G + RT ln(k) or G RT ln(k) G Free Energy (J) of gas at Standard ressure of atm R Gas Constant 8.45 J/K c d C, eq D, eq K Equilibrium Constant of Gaseous System a b A, eq B, eq T Temperature in K Qualitative Relationship between G and K : G RT ln(k ) K > ln(k ) > 0 (ositive) G < 0 (Negative) roducts are favoured over the Reactants (Forward Reaction is Spontaneous) K ln(k ) 0 G 0 roducts and Reactants are Equally favoured K < ln(k ) < 0 (Negative) G > 0 (ositive) Reactants are favoured over the roducts (Reverse Reaction is Spontaneous) Example : Calculate the equilibrium constant of the formation of water at 5.0 C. (G of H (g) 0 kj/mol, G of O (g) 0 kj/mol and G of H O (l) 7 kj/mol) H (g) + O (g) H O (l) G ΣG products ΣG reactants G [( mol)( 7 kj/mol)] [( mol)(0 kj/mol) + ( mol)(0 kj/mol)] G 474 kj J ( G < 0; Forward Reaction is Spontaneous) G RT ln(k ) at thermodynamic equilibrium G o ln(k ) RT 5 ( J) (8.45 J/K) ( 98.5 K) K e K Since K >>, the equilibrium is extremely favourable towards the product at 5 C. This is in agreement with the conclusion made using the thermodynamic quantity of G << 0. age 58. Copyrighted by Gabriel Tang B.Ed., B.Sc.

45 Chemistry A Unit 5: Chemical Kinetics and Equilibria Example : Given that N (g) + H (g) NH (g) has the following thermodynamic values, calculate G and K of the given equilibrium at 5.0 C. Chemicals H (kj /mol) S [J/(K mol)] N (g) 0 9 H (g) 0 NH (g) 46 9 H ΣH products ΣH reactants [( mol)( 46 kj/mol)] 9 kj S ΣS products ΣS reactants S [( mol)(9 J/(K mol))] [( mol)(9 J/(K mol)) + ( mol)( J/(K mol))] 99 J/K H 9 kj 9000 J S 99 J/K T 5.0 C 98.5 K G? N (g) + H (g) NH (g) H 9 kj G H T S G ( 9000 kj) (98.5 K)( 99 J/K) kj G. 0 4 J at 5.0 C Since G < 0, the Forward Reaction is Spontaneous R 8.45 J/K G.6685 kj J K? G RT ln(k ) at thermodynamic equilibrium G o ln(k ) RT ( J) ( 98.5 K) (8.45 J/K) ln(k ).784 K e K As we can see K >>, which indicates the equilibrium is favouring the product at 5 C. This is in agreement with the conclusion made using the thermodynamic quantity of G < 0. Temperature Dependence on K : RT ln(k ) G (Start with Free Energy Formula with Equilibrium Constant) RT ln(k ) H T S (Substitute G with H T S ) H o S o ln(k ) + RT R (Divide both sides by RT and simplify) o Graphs of ln(k ) versus H S o T ln(k ) + R T R When ln(k ) is graphed against (/T): ln(k ) H slope R y m x + b S (y-intercept) R T m < 0 when H > 0 (negative slope if system is endothermic) ln(k ) m > 0 when H < 0 (positive slope if system is exothermic) b < 0 when S < 0 (negative y-int if system becomes less random) T b > 0 when S > 0 (positive y-int if system becomes more random) Copyrighted by Gabriel Tang B.Ed., B.Sc. age 59.

46 Unit 5: Chemical Kinetics and Equilibria Chemistry A Example : The equilibrium constant of the system, NO (g) N O 4 (g), is allowed to varied with the change in temperature. The experimental data are shown below. Temperature Equilibrium Constant (K ) 7 K K K K a. Complete the table below. b. Graph ln(k ) versus /T. Determine the slope and y-intercept of the graph using linear regression of your graphing calculator. c. Calculate the H and S of this equilibrium system. d. Comment on the thermodynamic spontaneity of the equilibrium system at various temperatures. Temperature Equilibrium Constant (K ) /T (K ) ln(k ) 7 K K K K a. Using the TI-8 lus Calculator, we can enter the experimental results in the STATS Editor and quickly manipulate the data.. Enter Values in the L and L Columns of Stats Editor STAT ENTER. Enter /L in the heading of L. Enter ln(l ) in the heading of L 4. Cursor must be on the column heading before entering formula must be used to enter formula To access L, press ALHA +. Fill in the Table Temperature Equilibrium Constant (K ) /T (K ) ln(k ) 7 K K K K nd L MEM age 60. Copyrighted by Gabriel Tang B.Ed., B.Sc.