Chemical Equilibrium Reactions Go Both Ways Equilibrium Plural

Chemical Equilibrium Reactions Go Both Ways

Equilibrium • Plural is equilibria (for you Latin fans) • Reactions are reversible – they go forward and backward • There is always some product and reactant • How much of each is there? – Depends on the reaction

Equilibria is when…. • The rate of the forward reaction and reverse reaction are the same • Sealed jar example

Add liquid to empty, dry jar When added, the air hasnot nochange, liquid vapor. The ratefirst of evaporation does Thethevapor pressure of the liquid starts to increases populate the vapor but rate of condensation continuously Withthe molecules Until rate of evaporation = the rate of condensation

Reactions Go Both Ways 2 NO 2 (g) Brown N 2 O 4 (g) Colorless 2 NO 2 (g) Colorless Brown

In an Equilibrium The forward reaction rate is the same as the reverse reaction

Chemical equilibrium Different types of arrows are used in chemical equations associated with equilibria. Single arrow • Assumes that the reaction proceeds to completion as written. Two single-headed arrows • Used to indicate a system in equilibrium. Two single-headed arrows of different sizes. • May be used to indicate when one side of an equilibrium system is favored.

Chemical equilibrium • Homogeneous equilibria - Equilibria that involve only a single phase. Examples. • All species in the gas phase H 2 (g) + I 2 (g) 2 HI (g) • All species are in solution. HC 2 H 2 O 2 (aq) H+ (aq) + C 2 H 3 O 2 - (aq)

Chemical equilibrium • Basic steps for reaching equilibrium: • For the general reaction: A+B C We can view the reaction as occurring in three steps. • Initial mixing • Kinetic region • Equilibrium region

Chemical equilibrium Initial mixing. • When A and B are first brought together, there is no C present. • The reaction proceeds as A + B C • This is just at the very start of the reaction. Things change as soon as some C is produced.

Chemical equilibrium Kinetic region. • As soon as some C has been produced, the reverse reaction is possible. A + B C • Overall, we still see an increase in the net concentration of C. • As we approach equilibrium, the rate of the forward reaction becomes slower.

Chemical equilibrium Equilibrium region. • A point is finally reached where the forward and reverse reactions occur at the same rate. A + B C • There is no net change in the concentration of any of the species.

• Concentration Chemical equilibrium • Kinetic • Region • Equilibrium • Region • C • B • A • Time

• Rate of Formation Equilibrium • A point is ultimately • reached where the • rates of the forward • and reverse changes • are the same. • Reactants • Products • Time • • • At this point, equilibrium is reached.

Equilibrium • Concentration • An equilibrium exists when no further change in concentration occurs. Note that the concentrations of products and reactants do not have to be equal! • Products • Kinetic • Region Equilibrium Region • Reactants • Time

• The Equilibrium Constant • No products • No reactants • Also note that the equilibrium concentrations of the reactants and products are the same regardless of the whether or not you start with only the reactants or only the products.

Equilibrium Constants • The equilibrium constant is related to the rate laws of the forward and reverse reactions. • Because the forward and reverse reactions are equal in rate, the rate laws must be equal to each other. • kf [N 2 O 4] = kr[NO 2]2 • Isolating the concentrations and rate constants together… • [NO 2]2/[N 2 O 4] = kf/kr = a constant (Keq), or the equilibrium constant.

Equilibrium Constant • The position of the equilibrium can be described mathematically • We use the balanced chemical equation to develop a mathematical expression • Use molarity to describe concentration – Remember, that’s moles per liter of solution

Law of Mass Action p. A + q. B r. C + t. D K = [C]r • [D]t [A]p • [B]q [ ] = molarity K = equilibrium constant Products “over” reactants

Law of Mass Action 4 NH 3(g) + 7 O 2 (g) 4 NO 2 (g) + 6 H 2 O (g) K = [C]r • [D]t [A]p • [B]q = [NO 2]4 • [H 2 O]6 [NH 3]4 • [O 2]7

Equilibrium Constant Y Reactant Keq = X Product (g) [products]x [reactants]y • More products means a bigger K – Reaction lies to the right • More reactants means smaller K – Reaction lies to the left

N 2(g) + 3 H 2(g) 2 NH 3(g) K= [NH 3]2 [N 2] • [H 2]3 K = 6. 02 x 10 -2

N 2(g) + 3 H 2(g) 2 NH 3(g) K= [NH 3]2 [N 2] • [H 2]3 K = 6. 02 x 10 -2

N 2(g) + 3 H 2(g) 2 NH 3(g) K= [NH 3]2 [N 2] • [H 2]3 K = 6. 02 x 10 -2

Equilibria • It doesn’t matter where you start • The “ratio” of products to reactants is the same – Because the equilibria is established by the rate of the forward reaction vs the rate of the reverse reaction

Equilibrium Constant, Keq • The equilibrium constant, Keq, is the ratio of the concentrations of the products compared to the concentrations of the reactants. • Keq > 1 indicate that forming products is favored. Keq < 1 indicate that reactants don’t react easily.

• Equilibrium Constant, Keq • 3 H 2(g)+N 2(g) 2 NH 3(g) • The equilibrium constant, Keq, for the reaction above is determined by: • Keq = [ NH 3 ] 2 [ H 2 ] 3 [ N 2 ] • At 472 C, Keq = 0. 105

• Equilibrium Constant, Keq 2 NH 3(g) 3 H 2(g)+N 2(g) • Reversing a reaction will result in the new equilibrium constant, Keq new, that is equal to 1 / Keq old • Keq new = [ H 2 ] 3 [ N 2 ] = 1 / 0. 105 [ NH 3 ] 2 or 9. 52

• Equilibrium Constant, Keq NH 3(g) 1. 5 H 2(g)+0. 5 N 2(g) • Changing the number of moles of reactants and products will exponentially change the n equilibrium constant: K = (K )^ eq new eq old Keq new = [ H 2 ]1. 5 [ N 2 ]0. 5 = (9. 52)^0. 5 [ NH 3 ] or 3. 09

Partial pressure equilibrium constants At constant temperature, the pressure of a gas is proportional to its molarity. Remember, for an ideal gas: PV = n. RT and molarity is: M = mole / liter or n/V so: P = MR T where R is the gas law constant T is the temperature, K.

Partial pressure equilibrium constants For equilibria that involves gases, partial pressures can be used instead of concentrations. a. A (g) + b. B (g) Kp = e. E (g) + f. F (g) • p E e p F f • p A a p B b Kp is used when the partial pressures are expressed in units of atmospheres.

Partial pressure equilibrium constants In general, Kp ≠ Kc, instead Kp = Kc D n (RT) g Dng is the number of moles of gaseous products minus the number of moles of gaseous reactants. Dng = (e + f) - (a + b)

Partial pressure equilibrium constants For the following equilibrium, Kc = 1. 10 x 107 at 700. o. C. What is the Kp? • 2 H 2 (g) + S 2 (g) Kp = Kc (RT) ng T R = 700 + 273 = 973 K = 0. 08206 • atm L • mol K ng = ( 2 ) - ( 2 + 1) = -1 2 H 2 S (g)

Partial pressure equilibrium constants Kp = Kc (RT)Dng [ = 1. 10 x 107 (0. 08206 = 1. 378 x 105 • atm L • mol K) ] (973 K) -1

Equilibrium constants and expressions Heterogeneous equilibria • Equilibria that involve more than one phase. Ca. CO 3 (s) Ca. O (s) + CO 2 (g) • Equilibrium expressions for these types of systems do not include the concentrations of the pure solids (or liquids). Kc = [CO 2]

Equilibrium constants and expressions Heterogeneous equilibria - We don’t include the pure solids and liquids because their concentrations do not vary. These values end up being included in the K value. • CO 2 • Ca. O & • Ca. CO 3 As long as the temperature is constant and some solid is present, the amount of solid present has no effect on the equilibrium.

Writing an equilibrium expression • Write a balanced equation for the equilibrium. • Put the products in the numerator and the reactants in the denominator. • Omit pure solids and liquids from the expression • Omit solvents if your solutes are dilute (<0. 1 M). • The exponent of each concentration should be the same as the coefficient for the species in the equation.

Writing an equilibrium expression Example. • What would be the equilibrium expression for the following? (NH 4)2 CO 3(s) 2 NH 3(g) + CO 2(g) + H 2 O(g) • Kc = [NH 3]2 [CO 2] [H 2 O] • (NH 4)2 CO 3 is a pure solid so is not included Kc • We use [NH 3]2 because the coefficient for NH 3(g) in the equation is 2.

Equilibrium constant alphabet soup • The equilibrium constant expressed in c concentration units • K • The equilibrium constant expressed p in pressure units • K

Equilibrium constant alphabet soup • For the limited dissociation of sp insoluble solids. • K Ksp is called the • solubility product constant.

Equilibrium constant alphabet soup For the dissociation of a weak acids For the dissociation b of weak bases For the dissociation of w water into H+ and OH- • K • K

Equilibrium and rate of reaction Chemical reactions tend to go to equilibrium providing that reaction takes place at a significant rate. There is no relationship between the magnitude of the equilibrium constant and the rate of a reaction. Example: 2 H 2 (g) + O 2 (g) 2 H 2 O (g) Kc = 2. 9 x 1031 = [H 2 O]2 [H 2]2 [O 2] However, the reaction will take years to reach equilibrium at room temperature.

Determining equilibrium constants Equilibrium constants can be found by experiment. If you know the initial concentrations of all of the reactants, you only need to measure the concentration of a single species at equilibrium to determine the Kc value. Lets consider the following equilibrium: H 2 (g) + I 2 (g) 2 HI (g)

Determining equilibrium constants H 2 (g) + I 2 (g) 2 HI (g) Assume that we started with the following initial concentrations at 425. 4 o. C. H 2 (g) I 2 (g) HI (g) 0. 00500 M 0. 01250 M 0. 00000 M At equilibrium, we determine that the concentration of iodine is 0. 00772 M

Determining equilibrium constants The equilibrium expression for our system is: Kc = [HI]2 [H 2] [I 2] Based on the chemical equation, we know the equilibrium concentrations of each species. I 2 (g) = the measured amount = 0. 00772 M That means that 0. 00478 mol I 2 reacts (0. 01250 M - 0. 00772 M) to produce HI in 1. 00 L of solution.

Determining equilibrium constants I 2 (g) = 0. 00772 M HI (g) = 0. 00478 M 2 mol HI 1 mol I 2 0. 00956 M = H 2 (g) = = Kc 0. 00500 M - 0. 00478 M 0. 00022 M • [HI]2 (0. 00956)2 = = = 54 • [H 2] [I 2] (0. 00022)(0. 00772)

Equilibrium calculations We can predict the direction of a reaction by calculating the reaction quotient. Reaction quotient, Q For the reaction: a. A + b. B Q = e. E + f. F [E]e [F]f [A]a [B]b Q has the same form as Kc with one important difference. Q can be for any set of concentrations, not just at equilibrium.

Reaction quotient Any set of concentrations can be given and a Q calculated. By comparing Q to the Kc value, we can predict the direction for the reaction. Q < Kc Net forward reaction will occur. Q = Kc No change, at equilibrium. Q > Kc Net reverse reaction will occur.

Reaction quotient example For an earlier example H 2 (g) + I 2 (g) 2 HI (g) we determined the Kc to be 54 at 425. 4 o. C. If we had a mixture that contained the following, predict the direction of the reaction. [H 2] = 4. 25 x 10 -3 M [I 2] = 3. 97 x 10 -1 M [HI] = 9. 83 x 10 -2 M

Reaction quotient example Q = = = [ HI ]2 [ H 2 ] [ I 2 ] (9. 83 x 10 -2)2 (4. 25 x 10 -3)(3. 97 x 10 -1) 5. 73 Since Q is < Kc, the system is not in equilibrium and will proceed in the forward direction.

Calculating equilibrium concentrations If the stoichiometry and Kc for a reaction is known, calculating the equilibrium concentrations of all species is possible. Commonly, the initial concentrations are known. One of the concentrations is expressed as the variable x. All others are then expressed in terms of x.

Equilibrium calculation example A sample of COCl 2 is allowed to decompose. The value of Kc for the equilibrium COCl 2 (g) CO (g) + Cl 2 (g) is 2. 2 x 10 -10 at 100 o. C. If the initial concentration of COCl 2 is 0. 095 M, what will be the equilibrium concentrations for each of the species involved?

Equilibrium calculation example COCl 2 (g) CO (g) Cl 2 (g) Initial conc. , M 0. 095 0. 000 Change in conc. due to reaction -X +X +X Equilibrium Concentration, M (0. 095 -X) [CO ] [Cl 2 ] Kc = [ COCl 2 ] X X 2 =(0. 095 - X) X

Equilibrium calculation example Kc = 2. 2 x 10 -10 2 X = (0. 095 - X) • Rearrangement gives X 2 + 2. 2 x 10 -10 X - 2. 09 x 10 -11 = 0 This is a quadratic equation. Fortunately, there • is a straightforward equation for their solution

Quadratic equations An equation of the form a X 2 + b X + c = 0 Can be solved by using the following -b + b 2 - 4 ac x= 2 a Only the positive root is meaningful in equilibrium problems.

Equilibrium calculation example X 2 + 2. 2 x 10 -10 X - 2. 09 x 10 -11 = 0 a b c X= -b + b 2 - 4 ac 2 a - 2. 2 x 10 -10 + [(2. 2 x 10 -10)2 - (4)(1)(- 2. 09 x 10 -11)]1/2 2 • X = 9. 1 x 10 -6 M

Equilibrium calculation example Now that we know X, we can solve for the concentration of all of the species. COCl 2 = 0. 095 - X = 0. 095 M CO = X = 9. 1 x 10 -6 M Cl 2 = X = 9. 1 x 10 -6 M In this case, the change in the concentration of is COCl 2 negligible.

Summary of method of calculating equilibrium concentrations Write an equation for the equilibrium. Write an equilibrium constant expression. Express all unknown concentrations in terms of a single variable such as x. Substitute the equilibrium concentrations in terms of the single variable in the equilibrium constant expression. Solve for x. Use the value of x to calculate equilibrium concentrations.

You Try! • A 40. 0 g sample of solid (NH 4)2 CO 3 is placed in a closed evacuated 3 L flask and heated to 400 o. C. It decomposes to produce NH 3, H 2 O, and CO 2 : • (NH 4)2 CO 3 ↔ 2 NH 3 + H 2 O + CO 2 • The equilibrium constant, Kp , is 0. 295 • Write the Kp expression. • Calculate Kc • Calculate the partial pressure of NH 3 • Calculate the total pressure in the flask at equilibrium. • Calculate the number of grams of solid at equilibrium. • What is the minimum amount of solid needed to establish equilibrium?

Answers • • • Kp = PNH 32 PH 2 O PCO 2 3. 17 x 10 -8 1. 04 atm 2. 08 atm 37. 3 g More than 2. 72 g

Predicting Shifts: Le. Chatelier Any stress placed on an equilibrium system will cause the system to shift to minimize the effect of the stress. t • You can put s ress on a system by adding or removing something from one side of a reaction. • Co(H 2 O)62+ + 4 Cl 1 - Co. Cl 42 - + 6 H 2 O • How can you cause the color to change from pink to blue?

Predicting shifts in equilibria • Equilibrium concentrations are based on: – The specific equilibrium – The starting concentrations – Other factors such as: • Temperature • Pressure • Reaction specific conditions • Altering conditions will stress a system, resulting in an equilibrium shift.

Le Chatelier’s principle Co(H 2 O)62+ + 4 Cl 1 - • Add HCl • (adds Add heat Cl 1 -) Co. Cl 42 - + 6 H 2 O Note: ∆H = + value • Add acetone • Add • (removes Cl 1 - ) Ag 1+ • (removes • Add ice. H 2 O) • What other stresses could be placed on this system to change the color back and forth between pink and blue?

Le Chatelier’s principle Any stress placed on an equilibrium system will cause the system to shift to minimize the effect of the stress. • You can put stress on a system by adding or removing something from one side of a reaction. N 2(g) + 3 H 2 (g) 2 NH 3 (g) • What effect will there be if you added more • ammonia? How about more nitrogen?

Changes in concentration do not change the value of the equilibrium constant at constant temperature. When a material is added to a system in equilibrium, the equilibrium will shift away from that side of the equation. When a material is removed from a system in equilibrium, the equilibrium will shift towards that sid of the equation.

• Le Châtelier’s Principle • Change in Reactant or Product Concentrations • Consider the Haber process • If H 2 is added while the system is at equilibrium, the system must respond to counteract the added H 2 (by Le Châtelier). • That is, the system must consume the H 2 and produce products until a new equilibrium is established. • Therefore, [H 2] and [N 2] will decrease and [NH 3] increases.

• Le Châtelier’s Principle • Change in Reactant or Product Concentrations

• Le Châtelier’s Principle • Change in Reactant or Product Concentrations • Adding a reactant or product shifts the equilibrium away from the increase. • Removing a reactant or product shifts the equilibrium towards the decrease. • To optimize the amount of product at equilibrium, we need to flood the reaction vessel with reactant and continuously remove product (Le Châtelier). • We illustrate the concept with the industrial preparation of ammonia

• Le Châtelier’s Principle • Change in Reactant or Product Concentrations • The catalyst bed • N 2 and H 2 are is kept at 460 - 550 C pumped into a under high pressure. • The pre-heated chamber. gases are passed through a heating • In the coil to the catalyst refrigeration unit, bed. • The product ammonia liquefies gas stream but not N 2 or H 2. (containing N 2, H 2 and NH 3) is passed over a cooler to a refrigeration unit.

• Le Châtelier’s Principle • Change in Reactant or Product Concentrations • The unreacted nitrogen and hydrogen are recycled with the new N 2 and H 2 feed gas. • The equilibrium amount of ammonia is optimized because the product (NH 3) is continually removed and the reactants (N 2 and H 2) are continually being added. • Effects of Volume and Pressure • As volume is decreased pressure increases. • Le Châtelier’s Principle: if pressure is increased the system will shift to counteract the increase by producing fewer moles of gas.

Changes in pressure • Changing the pressure does not change the value of the equilibrium constant at constant temperature. • Solids and liquids are not effected by pressure changes. • Changing pressure by introducing an inert gas will not shift an equilibrium. • Pressure changes only effect gases that are a portion of an equilibrium.

Changes in pressure In general, increasing the pressure by decreasing volume shifts equilibria towards the side that has the smaller number of moles of gas. H 2 (g) + I 2 (g) 2 HI (g) • Unaffected by pressure N 2 O 2 (g) 2 NO 2 (g) • Increased pressure, shift to left

• Le Châtelier’s Principle • Effects of Volume and Pressure • That is, the system shifts to remove gases and decrease pressure. • An increase in pressure favors the direction that has fewer moles of gas. • In a reaction with the same number of product and reactant moles of gas, pressure has no effect. • Consider

• Le Châtelier’s Principle • Effects of Volume and Pressure • An increase in pressure (by decreasing the volume) favors the formation of colorless N 2 O 4. • The instant the pressure increases, the system is not at equilibrium and the concentration of both gases has increased. • The system moves to reduce the number moles of gas (i. e. the forward reaction is favored). • A new equilibrium is established in which the mixture is lighter because colorless N 2 O 4 is favored.

• Le Châtelier’s Principle • Effect of Temperature Changes • The equilibrium constant is temperature dependent. • For an endothermic reaction, H > 0 and heat can be considered as a reactant. • For an exothermic reaction, H < 0 and heat can be considered as a product. • Adding heat (i. e. heating the vessel) favors away from the increase: – if H > 0, adding heat favors the forward reaction, – if H < 0, adding heat favors the reverse reaction.

• Le Châtelier’s Principle • Effect of Temperature Changes • Removing heat (i. e. cooling the vessel), favors towards the decrease: if H > 0, cooling favors the reverse reaction, if H < 0, cooling favors the forward reaction. for which H > 0. – Co(H 2 O)62+ is pale pink and Co. Cl 42 - is blue.

• Le Châtelier’s Principle • Effect of Temperature Changes

• Le Châtelier’s Principle • Effect of Temperature Changes • Co If a light purple room temperature equilibrium mixture is placed in a beaker of warm water, the mixture turns deep blue. Since H > 0 (endothermic), adding heat favors the forward reaction, i. e. the formation of blue Co. Cl 42 -. If the room temperature equilibrium mixture is placed in a beaker of ice water, the mixture turns bright pink. Since H > 0, removing heat favors the reverse reaction which is the formation of pink Co(H 2 O)62+.

• Le Châtelier’s Principle • The Effect of Catalysts • A catalyst lowers the activation energy barrier for the reaction. • Therefore, a catalyst will decrease the time taken to reach equilibrium. • A catalyst does not effect the composition of the equilibrium mixture.

You try! • H 2(g) + S(s) → H 2 S(g) ∆Hrxn = -20. 17 k. J/mol • An amount of solid S and an amount of gaseous H 2 are placed in an evacuated container at 25 o. C. At equilibrium, some solid S remains in the container. Predict and explain the following: • Effect on partial pressure of H 2 S when S is added. • Effect on partial pressure of H 2 when H 2 S is added • Effect on mass of S when volume is increased. • Effect on S when the temperature increased. • Effect of adding a catalyst to initial concentrations.

Answers • Solids have no effect on equilibrium • P H 2 will increase. • S remains the same, equal numbers of moles of gases on both sides means no change in equilibrium. • S decreases. Equilibrium is not effected by solids, but solid amounts can be changed by equilibrium. If more gases are produced, solid must go down. • Catalysts make them reach equilibrium faster, but do not change equilibrium positions.