Chemical Equilibrium “The speed of a chemical reaction is determined by kinetics. The extent of a chemical reaction is determined by thermodynamics.” If the absorption of O 2 by hemoglobin was not reversible we would suffocate, O 2 absorption illustrates the principle of chemical equilibrium

Hemoglobin  protein (Hb) found in red blood cells that reacts with O 2  enhances the amount of O 2 that can be carried through the blood stream Hb + O 2 HbO 2 the represents that the reaction is in dynamic equilibrium  protein (Hb) found in red blood cells that reacts with O 2  enhances the amount of O 2 that can be carried through the blood stream Hb + O 2 HbO 2 the represents that the reaction is in dynamic equilibrium

Hemoglobin

Hemoglobin Equilibrium System Hb + O 2 HbO 2 [Hb], [O 2 ], and [HbO 2 ] at equilibrium are related via the equilibrium constant, K  changing the concentration of any one of these necessitates changing the other concentrations to re-establish equilibrium  When [O 2 ] is big reaction shifts right  When [O 2 ] is small reaction shifts left  Equilibrium allows Hb to bring enhanced amount of O 2 into the body and release it into the bloodstream [Hb], [O 2 ], and [HbO 2 ] at equilibrium are related via the equilibrium constant, K  changing the concentration of any one of these necessitates changing the other concentrations to re-establish equilibrium  When [O 2 ] is big reaction shifts right  When [O 2 ] is small reaction shifts left  Equilibrium allows Hb to bring enhanced amount of O 2 into the body and release it into the bloodstream

Hypothetical Reaction: 2 Red Blue The reaction slows over time, But the Red molecules never run out! At some time between 100 and 110 sec, the concentrations of both the Red and the Blue molecules no longer change equilibrium has been established. Notice that equilibrium does not mean that the concentrations are equal! Once equilibrium is established, the rate of Red molecules turning into Blue is the same as the rate of Blue molecules turning into Red The reaction slows over time, But the Red molecules never run out! At some time between 100 and 110 sec, the concentrations of both the Red and the Blue molecules no longer change equilibrium has been established. Notice that equilibrium does not mean that the concentrations are equal! Once equilibrium is established, the rate of Red molecules turning into Blue is the same as the rate of Blue molecules turning into Red Time[Red][Blue]

Reaction Dynamics Time Rate Rate Forward Rate Reverse Initially, only the forward reaction takes place. As the forward reaction proceeds it makes products and uses reactants. Because the reactant concentration decreases, the forward reaction slows. As the products accumulate, the reverse reaction speeds up. Eventually, the reaction proceeds in the reverse direction as fast as it proceeds in the forward direction. At this time equilibrium is established. Once equilibrium is established, the forward and reverse reactions proceed at the same rate, so the concentrations of all materials stay constant.

Dynamic Equilibrium  as the forward reaction slows and the reverse reaction accelerates, eventually they reach the same rate  dynamic equilibrium is the condition where the rates of the forward and reverse reactions are equal  once the reaction reaches equilibrium, the concentrations of all the chemicals remain constant  as the forward reaction slows and the reverse reaction accelerates, eventually they reach the same rate  dynamic equilibrium is the condition where the rates of the forward and reverse reactions are equal  once the reaction reaches equilibrium, the concentrations of all the chemicals remain constant

Reaction Dynamics

Equilibrium: Multistep Reaction Overall

Equilibrium: Multistep Reaction Overall

Law of Mass Action The overall equilbrium constant is the same as you would get if you treated the overall reaction as occuring in one elementary reaction

Equilibrium Constant for a Reaction aA (aq) + bB (aq) cC (aq) + dD (aq) 2 N 2 O 5 4 NO 2 + O 2

Interpreting K eq K eq >> 1  more product molecules present than reactant molecules  the position of equilibrium favors products K eq << 1  more reactant molecules present than product molecules  the position of equilibrium favors reactants K eq >> 1  more product molecules present than reactant molecules  the position of equilibrium favors products K eq << 1  more reactant molecules present than product molecules  the position of equilibrium favors reactants

K eq >> 1

K eq << 1

 when the reaction is written backwards, the equilibrium constant is inverted for aA + bB cC + dDfor cC + dD aA + bB Relationships between K and Chemical Equations

 when the reaction is written backwards, the equilibrium constant is inverted for aA + bB cC + dDfor cC + dD aA + bB Relationships between K and Chemical Equations

when the coefficients of an equation are multiplied by a factor, the equilibrium constant is raised to that factor for aA + bB cC the equilibrium constant expression is: for 2aA + 2bB 2cC the equilibrium constant expression is:

Relationships between K and Chemical Equations when you add equations to get a new equation, the equilibrium constant of the new equation is the product of the equilibrium constants of the old equations for (1) aA bB (2) bB cC for aA cC

K backward = 1/K forward, K new = K old n for N 2 (g) + 3 H 2 (g) 2 NH 3 (g), K = 3.7 x 10 8 at 25°C K for NH 3 (g) 0.5N 2 (g) + 1.5H 2 (g), at 25°C Solution: Concept Plan: Relationships : Given: Find: N 2 (g) + 3 H 2 (g) 2 NH 3 (g)K 1 = 3.7 x 10 8 K’K’ K 2 NH 3 (g) N 2 (g) + 3 H 2 (g) NH 3 (g) 0.5 N 2 (g) H 2 (g)

Equilibrium Constants for Reactions Involving Gases  the concentration of a gas in a mixture is proportional to its partial pressure  therefore, the equilibrium constant can be expressed as the ratio of the partial pressures of the gases  for aA(g) + bB(g) cC(g) + dD(g)  the concentration of a gas in a mixture is proportional to its partial pressure  therefore, the equilibrium constant can be expressed as the ratio of the partial pressures of the gases  for aA(g) + bB(g) cC(g) + dD(g) or

K c and K p  in calculating K p, the partial pressures are always in atm  the values of K p and K c are not necessarily the same  because of the difference in units  K p = K c when Δn = 0  the relationship between them is:  in calculating K p, the partial pressures are always in atm  the values of K p and K c are not necessarily the same  because of the difference in units  K p = K c when Δn = 0  the relationship between them is: Δn is the difference between the number of moles of reactants and moles of products

Deriving the Relationship between K p and K c

Deriving the Relationship Between K p and K c for aA(g) + bB(g) cC(g) + dD(g) substituting

Find K c for the reaction 2 NO(g) + O 2 (g) 2 NO 2 (g) given K p = 2.2 x 10 25°C K is a unitless number since there are more moles of reactant than product, K c should be larger than K p, and it is K p = 2.2 x KcKc Check: Solution: Concept Plan: Relationships: Given: Find: KpKp KcKc 2 NO(g) + O 2 (g) 2 NO 2 (g) Δn = = -1

 Consider  The reaction occurs at the surface where the solid and water meet  Instead of the concentration of NaCl(s) the rate of dissolving is related to the surface area A NaCl  K is independent of the amount of solid!  Consider  The reaction occurs at the surface where the solid and water meet  Instead of the concentration of NaCl(s) the rate of dissolving is related to the surface area A NaCl  K is independent of the amount of solid! Heterogeneous Equilibria

 Consider  The Equilibrium constant would be  But what is [H 2 O]?  What is [H + ]?  The [H 2 O] is effectively a constant for changes in Equilibrium and an be folded into the equilibrium constant so the rate of the backward reaction doesn’t change  Consider  The Equilibrium constant would be  But what is [H 2 O]?  What is [H + ]?  The [H 2 O] is effectively a constant for changes in Equilibrium and an be folded into the equilibrium constant so the rate of the backward reaction doesn’t change Heterogeneous Equilibria

 pure liquids are materials whose concentration doesn’t change (much) during the course of a reaction and is essentially constant compared to other reagents  For solids, the forward and backward reactions depend on the surface area making the equilibrium constant independent of the amount of solid  for the reaction aA(s) + bB(aq) cC(l) + dD(aq) the equilibrium constant expression is:  pure liquids are materials whose concentration doesn’t change (much) during the course of a reaction and is essentially constant compared to other reagents  For solids, the forward and backward reactions depend on the surface area making the equilibrium constant independent of the amount of solid  for the reaction aA(s) + bB(aq) cC(l) + dD(aq) the equilibrium constant expression is:

Heterogeneous Equilibria The amount of C is different, but the amounts of CO and CO 2 remains the same. Therefore the amount of C has no effect on the position of equilibrium.

Initial and Equilibrium Concentrations for H 2 (g) + I 2 (g) 445°C Initial Equilibrium concentrations Equilibrium Constant [H 2 ][I 2 ][HI][H 2 ][I 2 ][HI]

Calculating Equilibrium Concentrations  Stoichiometry can be used to determine the equilibrium concentrations of all reactants and products if you know initial concentrations and one equilibrium concentration  suppose you have a reaction 2 A (aq) + B (aq) 4 C (aq) with initial concentrations [A] = 1.00 M, [B] = 1.00 M, and [C] = 0. You then measure the equilibrium concentration of C as [C] = 0.50 M.  Stoichiometry can be used to determine the equilibrium concentrations of all reactants and products if you know initial concentrations and one equilibrium concentration  suppose you have a reaction 2 A (aq) + B (aq) 4 C (aq) with initial concentrations [A] = 1.00 M, [B] = 1.00 M, and [C] = 0. You then measure the equilibrium concentration of C as [C] = 0.50 M. [A][B][C] Initial molarity Change in concentration Equilibrium molarity ¼(0.50)-½(0.50)

Find the value of K c for the reaction 2 CH 4 (g) C 2 H 2 (g) + 3 H 2 (g) at 1700°C if the initial [CH 4 ] = M and the equilibrium [C 2 H 2 ] = M [CH 4 ][C 2 H 2 ][H 2 ] initial change equilibrium (0.035)+3(0.035) use the known change to determine the change in the other materials add the change to the initial concentration to get the equilibrium concentration in each column use the equilibrium concentrations to calculate K c

The following data were collected for the reaction 2 NO 2 (g) N 2 O 4 (g) at 100°C. Complete the table and determine values of K p and K c for each experiment [ ]/ [ ]

The following data were collected for the reaction 2 NO 2 (g) N 2 O 4 (g) at 100°C. Complete the table and determine values of K p and K c for each experiment.

The Reaction Quotient  if a reaction mixture, containing both reactants and products, is not at equilibrium; how can we determine which direction it will proceed?  the answer is to compare the current concentration ratios to the equilibrium constant  the concentration ratio of the products (raised to the power of their coefficients) to the reactants (raised to the power of their coefficients) is called the reaction quotient, Q  if a reaction mixture, containing both reactants and products, is not at equilibrium; how can we determine which direction it will proceed?  the answer is to compare the current concentration ratios to the equilibrium constant  the concentration ratio of the products (raised to the power of their coefficients) to the reactants (raised to the power of their coefficients) is called the reaction quotient, Q For the gas phase reaction aA + bB cC + dD the reaction quotient is:

The Reaction Quotient: Predicting the Direction of Change  if Q > K, the reaction will proceed fastest in the reverse direction  the [products] will decrease and [reactants] will increase  if Q < K, the reaction will proceed fastest in the forward direction  the [products] will increase and [reactants] will decrease  if Q = K, the reaction is at equilibrium  the [products] and [reactants] will not change  if a reaction mixture contains just reactants, Q = 0, and the reaction will proceed in the forward direction  if a reaction mixture contains just products, Q = ∞, and the reaction will proceed in the reverse direction  if Q > K, the reaction will proceed fastest in the reverse direction  the [products] will decrease and [reactants] will increase  if Q < K, the reaction will proceed fastest in the forward direction  the [products] will increase and [reactants] will decrease  if Q = K, the reaction is at equilibrium  the [products] and [reactants] will not change  if a reaction mixture contains just reactants, Q = 0, and the reaction will proceed in the forward direction  if a reaction mixture contains just products, Q = ∞, and the reaction will proceed in the reverse direction

Q, K, and the Direction of Reaction

If Q = K, equilibrium; If Q K, reverse For the reaction below, which direction will it proceed if P I2 = atm, P Cl2 = atm & P ICl = atm for I 2 (g) + Cl 2 (g) 2 ICl(g), K p = 81.9 direction reaction will proceed Solution: Concept Plan: Relationships: Given: Find: I 2 (g) + Cl 2 (g) 2 ICl(g)K p = 81.9 QP I2, P Cl2, P ICl since Q (10.8) < K (81.9), the reaction will proceed to the right

If [COF 2 ] eq = M and [CF 4 ] eq = M, and K c = 1000°C, find the [CO 2 ] eq for the reaction given. Units & Magnitude OK Check: Check: Round to 1 sig fig and substitute back in Solution: Solve: Solve the equilibrium constant expression for the unknown quantity by substituting in the given amounts Concept Plan: Relationships: Strategize: You can calculate the missing concentration by using the equilibrium constant expression 2 COF 2 CO 2 + CF 4 [COF 2 ] eq = M, [CF 4 ] eq = M [CO 2 ] eq Given: Find: Sort: You’re given the reaction and K c. You’re also given the [X] eq of all but one of the chemicals K, [COF 2 ], [CF 4 ][CO 2 ]

A sample of PCl 5 (g) is placed in a L container and heated to 160°C. The PCl 5 is decomposed into PCl 3 (g) and Cl 2 (g). At equilibrium, moles of PCl 3 and Cl 2 are formed. Determine the equilibrium concentration of PCl 5 if K c = PCl 5  PCl 3 + Cl 2 equilibrium concentration, M ?

Finding Equilibrium Concentrations: Given the Equilibrium Constant and Initial Concentrations or Pressures  first decide which direction the reaction will proceed  compare Q to K  define the changes of all materials in terms of x  use the coefficient from the chemical equation for the coefficient of x  the x change is + for materials on the side the reaction is proceeding toward  the x change is - for materials on the side the reaction is proceeding away from  solve for x  for 2 nd order equations, take square roots of both sides or use the quadratic formula  may be able to simplify and approximate answer for very large or small equilibrium constants  first decide which direction the reaction will proceed  compare Q to K  define the changes of all materials in terms of x  use the coefficient from the chemical equation for the coefficient of x  the x change is + for materials on the side the reaction is proceeding toward  the x change is - for materials on the side the reaction is proceeding away from  solve for x  for 2 nd order equations, take square roots of both sides or use the quadratic formula  may be able to simplify and approximate answer for very large or small equilibrium constants

[I 2 ][Cl 2 ][ICl] Initial Change Equilibrium Construct an ICE table for the reaction determine the direction the reaction is proceeding since Q p (1) < K p (81.9), the reaction is proceeding forward For the reaction I 2 (g) + Cl 2 (g) 2 25°C, K p = If the initial partial pressures are all atm, find the equilibrium concentrations

[I 2 ][Cl 2 ][ICl] initial change equilibrium represent the change in the partial pressures in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression and solve for x +2x-x-x-x-x x x For the reaction I 2 (g) + Cl 2 (g) 2 25°C, K p = If the initial partial pressures are all atm, find the equilibrium concentrations

[I 2 ][Cl 2 ][ICl] initial change equilibrium substitute into the equilibrium constant expression and solve for x +2x-x-x-x-x x x For the reaction I 2 (g) + Cl 2 (g) 2 25°C, K p = If the initial partial pressures are all atm, find the equilibrium concentrations

[I 2 ][Cl 2 ][ICl] initial change equilibrium check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K to the given K (0.0729) K p (calculated) = K p (given) within significant figures For the reaction I 2 (g) + Cl 2 (g) 2 25°C, K p = If the initial partial pressures are all atm, find the equilibrium concentrations

For the reaction I 2 (g) 2 I(g) K c = 3.76 x at 1000 K If 1.00 moles of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]? (Hint: you will need to use the quadratic formula to solve for x)

For the reaction I 2 (g) 2 I(g) the value of K c = 3.76 x at 1000 K. If 1.00 moles of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]? [I 2 ][I] initial change-x-x+2x equilibrium x2x2x since [I] initial = 0, Q = 0 and the reaction must proceed forward

[I 2 ][I] initial change-x-x+2x equilibrium x2x2x For the reaction I 2 (g) 2 I(g) the value of K c = 3.76 x at 1000 K. If 1.00 moles of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]?

[I 2 ][I] initial change-x-x+2x equilibrium x = = [I 2 ] = M 2( ) = [I] = M  For the reaction I 2 (g) 2 I(g) the value of K c = 3.76 x at 1000 K. If 1.00 moles of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]?

Approximations to Simplify the Math  when the equilibrium constant is very small, the position of equilibrium favors the reactants  for relatively large initial concentrations of reactants, the reactant concentration will not change significantly when it reaches equilibrium  the [X] equilibrium = ([X] initial ± ax) ≈ [X] initial  we are approximating the equilibrium concentration of reactant to be the same as the initial concentration  when the equilibrium constant is very small, the position of equilibrium favors the reactants  for relatively large initial concentrations of reactants, the reactant concentration will not change significantly when it reaches equilibrium  the [X] equilibrium = ([X] initial ± ax) ≈ [X] initial  we are approximating the equilibrium concentration of reactant to be the same as the initial concentration

Checking the Approximation Refining as Necessary  we can check our approximation afterwards by comparing the approximate value of x to the initial concentration  if the approximate value of x is less than 5% of the initial concentration, the approximation is valid  we can check our approximation afterwards by comparing the approximate value of x to the initial concentration  if the approximate value of x is less than 5% of the initial concentration, the approximation is valid

[H 2 S][H 2 ][S 2 ] initial 2.50E-400 change equilibrium For 2 H 2 S(g) 2 H 2 (g) + S 2 800°C, K c = 1.67 x If a L flask initially containing 1.25 x mol H 2 S is heated to 800°C, find the equilibrium concentrations. Construct an ICE table for the reaction determine the direction the reaction is proceeding since no products initially, Q c = 0, and the reaction is proceeding forward

[H 2 S][H 2 ][S 2 ] initial 2.50E-400 change equilibrium represent the change in the partial pressures in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression +x+x +2x-2x 2.50E-4 -2x 2x2x x For 2 H 2 S(g) 2 H 2 (g) + S 2 800°C, K c = 1.67 x If a L flask initially containing 1.25 x mol H 2 S is heated to 800°C, find the equilibrium concentrations.

[H 2 S][H 2 ][S 2 ] initial 2.50E-400 change equilibrium +x+x +2x-2x 2.50E-4 -2x 2x2xx since K c is very small, approximate the [H 2 S] eq = [H 2 S] init and solve for x 2.50E-4 For 2 H 2 S(g) 2 H 2 (g) + S 2 800°C, K c = 1.67 x If a L flask initially containing 1.25 x mol H 2 S is heated to 800°C, find the equilibrium concentrations.

[H 2 S][H 2 ][S 2 ] initial 2.50E-400 change equilibrium check if the approximation is valid by seeing if x < 5% of [H 2 S] init +x+x +2x-2x 2x2xx 2.50E-4 the approximation is not valid  For 2 H 2 S(g) 2 H 2 (g) + S 2 800°C, K c = 1.67 x If a L flask initially containing 1.25 x mol H 2 S is heated to 800°C, find the equilibrium concentrations.

[H 2 S][H 2 ][S 2 ] initial 2.50E-400 change equilibrium if approximation is invalid, substitute x current into K c where it is subtracted and re-solve for x new +x+x +2x-2x 2.50E-4 -2x 2x2xx x current = 1.38 x For 2 H 2 S(g) 2 H 2 (g) + S 2 800°C, K c = 1.67 x If a L flask initially containing 1.25 x mol H 2 S is heated to 800°C, find the equilibrium concentrations.

[H 2 S][H 2 ][S 2 ] initial 2.50E-400 change equilibrium Substitute x current into K c where it is subtracted and re-solve for x new. If x new is the same number, you have arrived at the best approximation +x+x +2x-2x 2.50E-4 -2x 2x2xx x current = 1.27 x since x current = x new, approx. OK For 2 H 2 S(g) 2 H 2 (g) + S 2 800°C, K c = 1.67 x If a L flask initially containing 1.25 x mol H 2 S is heated to 800°C, find the equilibrium concentrations.

[H 2 S][H 2 ][S 2 ] initial 2.50E-400 change equilibrium substitute x current into the equilibrium concentration definitions and solve +x+x +2x-2x 2.50E-4 -2x 2x2xx x current = 1.28 x E-42.56E-51.28E-5 For 2 H 2 S(g) 2 H 2 (g) + S 2 800°C, K c = 1.67 x If a L flask initially containing 1.25 x mol H 2 S is heated to 800°C, find the equilibrium concentrations.

[H 2 S][H 2 ][S 2 ] initial 2.50E-400 change equilibrium +x+x +2x-2x 2.24E E-51.28E-5 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K to the given K K c (calculated) = K c (given) within significant figures For 2 H 2 S(g) 2 H 2 (g) + S 2 800°C, K c = 1.67 x If a L flask initially containing 1.25 x mol H 2 S is heated to 800°C, find the equilibrium concentrations.

Summarize  Assume [H 2 S] eq =[H 2 S] o -2x ≈ [H 2 S] 0  Solve for x in K (call this x current )  Check ([H 2 S] o -2x current )/[H 2 S] o > 0.95 ?  If no replace [H 2 S] eq =[H 2 S] o -2x current in K and resolve for x new  replace [H 2 S] eq =[H 2 S] o -2x new in K and resolve for x newer  If x new ≈x newer then stop x=x newer  Assume [H 2 S] eq =[H 2 S] o -2x ≈ [H 2 S] 0  Solve for x in K (call this x current )  Check ([H 2 S] o -2x current )/[H 2 S] o > 0.95 ?  If no replace [H 2 S] eq =[H 2 S] o -2x current in K and resolve for x new  replace [H 2 S] eq =[H 2 S] o -2x new in K and resolve for x newer  If x new ≈x newer then stop x=x newer

For the reaction I 2 (g) Û 2 I(g) the value of K c = 3.76 x at 1000 K. If 1.00 mole of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]? (use the simplifying assumption to solve for x)

For the reaction I 2 (g) Û 2 I(g) the value of K c = 3.76 x at 1000 K. If 1.00 moles of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]? [I 2 ][I] initial change-x-x+2x equilibrium x2x2x since [I] initial = 0, Q = 0 and the reaction must proceed forward

For the reaction I 2 (g) Û 2 I(g) the value of K c = 3.76 x at 1000 K. If 1.00 moles of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]? [I 2 ][I] initial change-x-x+2x equilibrium x2x2x the approximation is valid!!

Disturbing and Re- establishing Equilibrium  once a reaction is at equilibrium, the concentrations of all the reactants and products remain the same  however if the conditions are changed, the concentrations of all the chemicals will change until equilibrium is re- established  the new concentrations will be different, but the equilibrium constant will be the same  unless you change the temperature  once a reaction is at equilibrium, the concentrations of all the reactants and products remain the same  however if the conditions are changed, the concentrations of all the chemicals will change until equilibrium is re- established  the new concentrations will be different, but the equilibrium constant will be the same  unless you change the temperature

Le Ch âtelier’s Principle  Le Châtelier's Principle guides us in predicting the effect various changes in conditions have on the position of equilibrium  it says that if a system at equilibrium is disturbed, the position of equilibrium will shift to minimize the disturbance  disturbances all involve making the system open  Le Châtelier's Principle guides us in predicting the effect various changes in conditions have on the position of equilibrium  it says that if a system at equilibrium is disturbed, the position of equilibrium will shift to minimize the disturbance  disturbances all involve making the system open

An Analogy: Population Changes When the populations of Country A and Country B are in equilibrium, the emigration rates between the two states are equal so the populations stay constant. When an influx of population enters Country B from somewhere outside Country A, it disturbs the equilibrium established between Country A and Country B. The result will be people moving from Country B into Country A faster than people moving from Country A into Country B. This will continue until a new equilibrium between the populations is established, however the new populations will have different numbers of people than the old ones.

The Effect of Concentration Changes on Equilibrium  Adding a reactant will decrease the amounts of the other reactants and increase the amount of the products until a new position of equilibrium is found  that has the same K  Removing a product will increase the amounts of the other products and decrease the amounts of the reactants.  you can use this to drive a reaction to completion!  Equilibrium shifts away from side with added chemicals or toward side with removed chemicals  Remember, adding more of a solid or liquid does not change its concentration – and therefore has no effect on the equilibrium  Adding a reactant will decrease the amounts of the other reactants and increase the amount of the products until a new position of equilibrium is found  that has the same K  Removing a product will increase the amounts of the other products and decrease the amounts of the reactants.  you can use this to drive a reaction to completion!  Equilibrium shifts away from side with added chemicals or toward side with removed chemicals  Remember, adding more of a solid or liquid does not change its concentration – and therefore has no effect on the equilibrium

Disturbing Equilibrium: Adding Reactants  adding a reactant initially increases the rate of the forward reaction, but has no initial effect on the rate of the reverse reaction.  the reaction proceeds to the right until equilibrium is re-established.  at the new equilibrium position, you will have more of the products than before, less of the non- added reactants than before, and less of the added reactant  but not as little of the added reactant as you had before the addition  at the new equilibrium position, the concentrations of reactants and products will be such that the value of the equilibrium constant is the same  adding a reactant initially increases the rate of the forward reaction, but has no initial effect on the rate of the reverse reaction.  the reaction proceeds to the right until equilibrium is re-established.  at the new equilibrium position, you will have more of the products than before, less of the non- added reactants than before, and less of the added reactant  but not as little of the added reactant as you had before the addition  at the new equilibrium position, the concentrations of reactants and products will be such that the value of the equilibrium constant is the same

Disturbing Equilibrium: Removing Reactants  removing a reactant initially decreases the rate of the forward reaction, but has no initial effect on the rate of the reverse reaction.  so the reaction is going faster in reverse  the reaction proceeds to the left until equilibrium is re-established.  at the new equilibrium position, you will have less of the products than before, more of the non- removed reactants than before, and more of the removed reactant  but not as much of the removed reactant as you had before the removal  at the new equilibrium position, the concentrations of reactants and products will be such that the value of the equilibrium constant is the same  removing a reactant initially decreases the rate of the forward reaction, but has no initial effect on the rate of the reverse reaction.  so the reaction is going faster in reverse  the reaction proceeds to the left until equilibrium is re-established.  at the new equilibrium position, you will have less of the products than before, more of the non- removed reactants than before, and more of the removed reactant  but not as much of the removed reactant as you had before the removal  at the new equilibrium position, the concentrations of reactants and products will be such that the value of the equilibrium constant is the same

The Effect of Concentration Changes on Equilibrium When NO 2 is added, some of it combines to make more N 2 O 4

Effect of Volume Change on Equilibrium  decreasing the size of the container increases the concentration of all the gases in the container  increases their partial pressures  if their partial pressures increase, then the total pressure in the container will increase  according to Le Châtelier’s Principle, the equilibrium should shift to remove that pressure  the way the system reduces the pressure is to reduce the number of gas molecules in the container  when the volume decreases, the equilibrium shifts to the side with fewer gas molecules  decreasing the size of the container increases the concentration of all the gases in the container  increases their partial pressures  if their partial pressures increase, then the total pressure in the container will increase  according to Le Châtelier’s Principle, the equilibrium should shift to remove that pressure  the way the system reduces the pressure is to reduce the number of gas molecules in the container  when the volume decreases, the equilibrium shifts to the side with fewer gas molecules

Disturbing Equilibrium: Changing the Volume  after equilibrium is established, how will decreasing the container volume affect the total pressure of solids, liquid, and gases? How will it affect the concentration of solids, liquid, solutions, and gases? What will this cause? How will it affect the value of K?

Disturbing Equilibrium: Reducing the VolumeReducing the Volume Disturbing Equilibrium: Reducing the VolumeReducing the Volume  for solids, liquids, or solutions, changing the size of the container has no effect on the concentration, therefore no effect on the position of equilibrium  decreasing the container volume will increase the total pressure  Boyle’s Law  if the total pressure increases, the partial pressures of all the gases will increase  Dalton’s Law of Partial Pressures  decreasing the container volume increases the concentration of all gases  same number of moles, but different number of liters, resulting in a different molarity  since the total pressure increases, the position of equilibrium will shift to decrease the pressure by removing gas molecules  shift toward the side with fewer gas molecules  at the new equilibrium position, the partial pressures of gaseous reactants and products will be such that the value of the equilibrium constant is the same  for solids, liquids, or solutions, changing the size of the container has no effect on the concentration, therefore no effect on the position of equilibrium  decreasing the container volume will increase the total pressure  Boyle’s Law  if the total pressure increases, the partial pressures of all the gases will increase  Dalton’s Law of Partial Pressures  decreasing the container volume increases the concentration of all gases  same number of moles, but different number of liters, resulting in a different molarity  since the total pressure increases, the position of equilibrium will shift to decrease the pressure by removing gas molecules  shift toward the side with fewer gas molecules  at the new equilibrium position, the partial pressures of gaseous reactants and products will be such that the value of the equilibrium constant is the same

Since there are more gas molecules on the reactants side of the reaction, when the pressure is increased the position of equilibrium shifts toward the products. When the pressure is decreased by increasing the volume, the position of equilibrium shifts toward the side with the greater number of molecules – the reactant side. The Effect of Volume Changes on Equilibrium

The Effect of Temperature Changes on Equilibrium Position The Effect of Temperature Changes on Equilibrium Position  exothermic reactions release energy and endothermic reactions absorb energy  if we write Heat as a product in an exothermic reaction or as a reactant in an endothermic reaction, it will help us use Le Châtelier’s Principle to predict the effect of temperature changes  even though heat is not matter and not written in a proper equation  exothermic reactions release energy and endothermic reactions absorb energy  if we write Heat as a product in an exothermic reaction or as a reactant in an endothermic reaction, it will help us use Le Châtelier’s Principle to predict the effect of temperature changes  even though heat is not matter and not written in a proper equation

The Effect of Temperature Changes on Equilibrium for Exothermic Reactions  for an exothermic reaction, heat is a product  increasing the temperature is like adding heat  according to Le Châtelier’s Principle, the equilibrium will shift away from the added heat  adding heat to an exothermic reaction will decrease the concentrations of products and increase the concentrations of reactants  adding heat to an exothermic reaction will decrease the value of K  how will decreasing the temperature affect the system?  for an exothermic reaction, heat is a product  increasing the temperature is like adding heat  according to Le Châtelier’s Principle, the equilibrium will shift away from the added heat  adding heat to an exothermic reaction will decrease the concentrations of products and increase the concentrations of reactants  adding heat to an exothermic reaction will decrease the value of K  how will decreasing the temperature affect the system? aA + bB  cC + dD + Heat

The Effect of Temperature Changes on Equilibrium for Endothermic Reactions  for an endothermic reaction, heat is a reactant  increasing the temperature is like adding heat  according to Le Châtelier’s Principle, the equilibrium will shift away from the added heat  adding heat to an endothermic reaction will decrease the concentrations of reactants and increase the concentrations of products  adding heat to an endothermic reaction will increase the value of K  how will decreasing the temperature affect the system?  for an endothermic reaction, heat is a reactant  increasing the temperature is like adding heat  according to Le Châtelier’s Principle, the equilibrium will shift away from the added heat  adding heat to an endothermic reaction will decrease the concentrations of reactants and increase the concentrations of products  adding heat to an endothermic reaction will increase the value of K  how will decreasing the temperature affect the system? Heat + aA + bB  cC + dD

The Effect of Temperature Changes on Equilibrium The Effect of Temperature Changes on Equilibrium

Not Changing the Position of Equilibrium - Catalysts  catalysts provide an alternative, more efficient mechanism  works for both forward and reverse reactions  affects the rate of the forward and reverse reactions by the same factor  therefore catalysts do not affect the position of equilibrium  catalysts provide an alternative, more efficient mechanism  works for both forward and reverse reactions  affects the rate of the forward and reverse reactions by the same factor  therefore catalysts do not affect the position of equilibrium

Practice - Le Ch âtelier’s Principle  The reaction 2 SO 2 (g) + O 2 (g) Û 2 SO 3 (g) with  H° = -198 kJ is at equilibrium. How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is re- established?  adding more O 2 to the container  condensing and removing SO 3  compressing the gases  cooling the container  doubling the volume of the container  warming the mixture  adding the inert gas helium to the container  adding a catalyst to the mixture  The reaction 2 SO 2 (g) + O 2 (g) Û 2 SO 3 (g) with  H° = -198 kJ is at equilibrium. How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is re- established?  adding more O 2 to the container  condensing and removing SO 3  compressing the gases  cooling the container  doubling the volume of the container  warming the mixture  adding the inert gas helium to the container  adding a catalyst to the mixture

Practice - Le Ch âtelier’s Principle  The reaction 2 SO 2 (g) + O 2 (g) Û 2 SO 3 (g) with  H° = -198 kJ is at equilibrium. How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is re- established?  adding more O 2 to the containershift to SO 3  condensing and removing SO 3 shift to SO 3  compressing the gasesshift to SO 3  cooling the containershift to SO 3  doubling the volume of the containershift to SO 2  warming the mixtureshift to SO 2  adding helium to the containerno effect  adding a catalyst to the mixtureno effect  The reaction 2 SO 2 (g) + O 2 (g) Û 2 SO 3 (g) with  H° = -198 kJ is at equilibrium. How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is re- established?  adding more O 2 to the containershift to SO 3  condensing and removing SO 3 shift to SO 3  compressing the gasesshift to SO 3  cooling the containershift to SO 3  doubling the volume of the containershift to SO 2  warming the mixtureshift to SO 2  adding helium to the containerno effect  adding a catalyst to the mixtureno effect

exam First Exam on Thursday, Topics covered kinetics and chemical equilibrium, exam format multiple choice

Key Concepts Equilibria  The Equilibrium Constant. The relative concentrations of reactants and products are expressed using the equilibrium constant K. K measures how far a reaction proceeds towards products before concentrations attain dynamic equilibrium.  Large K means high concentration of products  Small K means low concentration of products  The Equilibrium Constant. The relative concentrations of reactants and products are expressed using the equilibrium constant K. K measures how far a reaction proceeds towards products before concentrations attain dynamic equilibrium.  Large K means high concentration of products  Small K means low concentration of products

Key Concepts Equilibria  Dynamic Equilibrium. Reactions are generally reversible (can go forwards or backwards). At equilibrium rate of product formation = rate of reactant formation. Net concentrations no longer change. (Concentrations of reactants and products not generally equal.

Key Concepts Equilibria  Equilibrium Constant Expression. Given by the law of mass action it can be expressed in terms of the concentrations or the partial pressures

Key Concepts Equilibria  Equilibrium Constant Expression and States of Matter. The concentrations and partial pressures used are only those reactants and products that exist as gases or solutes. Pure solids and liquids are not included in the expressions for the equilibrium constant

Key Concepts Equilibria  Reaction Quotient. For a reaction with an equilibrium constant Kc the reaction quotient Qc is defined as  Where [A], [B], [C] and [D] are non equilibrium concentrations. When Q c K c reaction will proceed toward reactants. At equilibrium Q c =K c  Reaction Quotient. For a reaction with an equilibrium constant Kc the reaction quotient Qc is defined as  Where [A], [B], [C] and [D] are non equilibrium concentrations. When Q c K c reaction will proceed toward reactants. At equilibrium Q c =K c

Key Concepts Equilibria  Le Châtalier’s Principle. When a system at equilibrium is disturbed - by a change in the amount of a reactant and product, a change in volume, or a change in temperature - the system shifts in the direction that minimizes the disturbance.

Key Concepts Equilibria  Finding Equilibrium Concentrations. Two types of problem  K, initial concentrations and one equilibrium concentration known. This can be re-arranged using the law of mass action and substituting values  K and only the initial concentrations are known. This is solved by defining a variable x that represents the change in concentration. One then solves for x.  Finding Equilibrium Concentrations. Two types of problem  K, initial concentrations and one equilibrium concentration known. This can be re-arranged using the law of mass action and substituting values  K and only the initial concentrations are known. This is solved by defining a variable x that represents the change in concentration. One then solves for x.