Chemical Equilibrium

Most reactions do not go to completion – instead they form a mixture of reactants and products.

· Reactants continue to form products, but products are simultaneously forming reactants – the reaction proceeds in both the forward and reverse direction.

· At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction.

· No net change in amount of reactant or product occurs even though molecules are still reacting – a dynamic equilibrium has been reached.

· An equilibrium situation is symbolized with a double arrow in the equation.

aA + bB ¾ cC + dD

· An equilibrium is said to lie “to the right” if more product is present than reactant, or “to the left” if more reactant is present than product.

· As an equilibrium reaction moves from the left to the right, reactants are used up and products are formed until the preferred equilibrium position is established.

Ex. (500 °C) N_{2 (g)} + 3 H_{2 (g)} ¾ 2 NH_{3 (g)}

Init. conc. 1.000 M 1.000 M 0 M

Change

Eq. conc. 0.157 M

· The equilibrium constant (K_c) expresses the ratio of product to reactant molar concentrations.

· For a reaction: aA + bB ¾ cC + dD

K_c = [C]^c[D]^d

[A]^a[B]^b (all conc’s = eq. conc’s)

· The thermodynamic definition of the equilibrium constant is similar, but uses activities rather than equilibrium conc’s.

· For pure liquids or pure solids, activity =1.

· For ideal solutions, activity = eq conc./1 M

· For ideal gases, activity = partial pressure/1 atm

· Activities are unitless and K_c is unitless.

· Calculate the equilibrium constant for the reaction between nitrogen and hydrogen above

K = [0.157]² =

[0.921]¹[0.763]³

· Values of K can vary with temperature because changes in temperature can cause the equilibrium position to shift.

· Values of K represent the overall proportion of an equilibrium mixture which is reactants and products.

K >>1 … mostly products

K <<1 … mostly reactants

Ex. (500 °C) N_{2 (g)} + 3 H_{2 (g)} ¾ 2 NH_{3 (g)}

Init. conc. 0 M 0 M 1.000 M

Change

Eq. conc. 0.203

K =

(At 25 °C, K = 3.6 x 10⁸ for this rxn.)

· If the equilibrium was written in reverse, the value of K would change.

Ex. N_{2 (g)} + 3 H_{2 (g)} ¾ 2 NH_{3 (g)}

K = [NH₃]²

[N₂]¹[H₂]³

2 NH_{3 (g)} ¾ N_{2 (g)} + 3 H_{2 (g)}

K’ = [N₂]¹[H₂]³ = 1

[NH₃]² K

· Values of coefficients also make a difference

½ N_{2 (g)} + 3/2 H_{2 (g)} ¾ NH_{3 (g)}

K’’ = [NH₃]¹ = K^1/2

[N₂]^1/2[H₂]^3/2

· Equilibrium constants can only be calculated after equilibrium has been established; reaction quotients (Q) can be calculated at any time and signify whether the reaction will proceed to the left or right to reach equilibrium

· For a reaction: aA + bB ¾ cC + dD

Q = [C]^c[D]^d

[A]^a[B]^b

· If Q = K, then the reaction is at equilbrium

· If Q> K, then the current situation has a higher proportion of products than at equilibrium … the reaction needs to proceed to the left to reach equilibrium

· If Q< K, then the current situation has a higher proportion of reactants than at equilibrium … the reaction needs to proceed to the right to reach equilibrium

Calculate Q for each of the following situations involving the synthesis of ammonia at 500 °C. Determine whether the reaction proceeds to the left or right.

a) [NH₃] = 1.0 x 10^-3 M, [N₂] = 1.0 x 10^-5 M, [H₂] = 2.0 x 10^-3 M

b) [NH₃] = 2.00 x 10^-4 M, [N₂] = 1.50 x 10^-5 M, [H₂] = 3.54 x 10^-1 M

c) [NH₃] = 1.0 x 10^-4 M, [N₂] = 5.0 M, [H₂] = 1.0 x 10^-2 M

Le Chatelier’s Principle

· if stress is applied to an equilibrium, the system shifts to reduce the stress and move to a new state of equilibrium

· Stresses can include

Changes in concentration

Changes in pressure or volume for gas rxns

Changes in temperature

Changes in concentration

• Adding or removing reactants or products will change the value of Q, but not of K. Equilibrium will eventually be reestablished with the same value for K.

A ¾ B K = [B]/[A]

· Adding a reactant causes the equilibrium to proceed to the right to reestablish K.

· Adding a product causes the equilibrium to proceed to the left to reestablish K.

· Removing a reactant causes the equilibrium to proceed to the left to reestablish K.

· Removing a product causes the equilibrium to proceed to the right to reestablish K.

Changes in volume and pressure

· Changes in pressure and volume can significantly change the concentration (mol/L) of a gas

· This will change the value of Q, but not of K_c

· PV = nRT or P =(n/V)RT where n/V = M

· (P = MRT)

· M is directly proportional to P, and P is inversely proportional to V

· If T is constant and the volume of a gas increases, M and P will decrease; if the volume of a gas decreases, M and P will increase

Consider the following equilibria: A _(g) ¾ 2 B _(g)

Q = [B]²/[A]

If volume decreases,

· Partial pressure of A and B increase, Molarity of A and B increase

· Q increases (Q > K)

· Rxn must proceed to the left to reestablish equilibrium

If volume increases,

· Partial pressure of A and B decrease, Molarity of A and B decrease

· Q decreases (Q < K)

· Rxn must proceed to the right to reestablish equilibrium

What effect would an increase or decrease in pressure have on the value of Q for the following reaction?

2 A _(g) ¾ B _(g)

Increase in pressure:

Decrease in pressure:

What effect would an increase or decrease in pressure have on the value of Q for the following reaction?

A _(g) ¾ B _(g)

Increase in pressure:

Decrease in pressure:

Summary:

· For rxns where the total # of moles of gas reactant = total # of moles of gas product, changes in V & P have no effect on Q

· For rxns where the total # of moles of gas reactant ¹ total # of moles of gas product,

· A decrease in V/increase in P shifts the rxn toward the side with the smaller # of moles of gas

· An increase in V/decrease in P shifts the rxn toward the side with the larger # of moles of gas

· Changes in total pressure which do not accompany a change in volume will have no effect (ex. pumping in an inert gas which is not part of the reaction) – partial pressure and molarity of each reactant and product are unchanged.

Changes in temperature

· Exothermic reactions produce heat (product); endothermic reactions require heat (reactant).

· For endothermic reactions, an increase in temperature causes the equilibrium to shift right.

· For exothermic reactions, an increase in temperature causes the equilibrium to shift left.

· Since these shifts do cause changes in concentrations of reactants and products, the value of K will change.

· For exothermic reactions an increase in T causes K to decrease; for endothermic reactions an increase in T causes K to increase

Addition of a catalyst

· Catalysts increase the rates of both the forward and reverse directions of an equilibrium reaction. Consequently K remains unchanged.

· Catalysts also do not change Q.

· In the presence of a catalyst equilibrium is established more quickly.

Consider the following exothermic equilibrium reaction. For each change of conditions described, predict what (if anything) happens to Q, K, and in which direction rxn will proceed.

A _(g) + 3 B _(g) ^ß_à 2 C _(g) + 3 D _(g)

a) adding more A

b) adding more C

c) removing some D

d) increase pressure & decrease volume

e) increase temperature

Finding equilibrium concentrations.

H_{2 (g)} + F_{2 (g)} ¾ 2 HF _(g) K = 115

If 3.00 mol of each component are initially present in a 1.50 L flask, find the equilibrium concentrations of each component.

H_{2 (g)} F_{2 (g)} 2 HF _(g)

Init conc.

D conc.

Eq. conc.

K = [HF]²

[H₂] [F₂]

H_{2 (g)} + F_{2 (g)} ¾ 2 HF _(g) K = 115

If 3.00 mol of H₂ is mixed with 6.00 mol F₂ in a 3.00 L flask, find the equilibrium concentrations of each component.

H_{2 (g)} F_{2 (g)} 2 HF _(g)

Init conc.

D conc.

Eq. conc.

K = [HF]²

[H₂] [F₂]

Quadratic formula for ax² + bx + c = 0

Calculations involving changes in equilibrium concs.

H_{2 (g)} + F_{2 (g)} ¾ 2 HF _(g) K = 115

Consider the previous scenario. After reaching equilibrium, an additional 0.900 mol of H₂ is added into the 3.00 L flask. Calculate the new equilibrium concentrations.

H_{2 (g)} F_{2 (g)} 2 HF _(g)

Init conc.

D conc.

Eq. conc.

K = [HF]²

[H₂] [F₂]

N₂O_{4 (g)} ¾ 2 NO_{2 (g)} K = 4.66 x 10^-3

Calculate the equilibrium concentrations which will be obtained by injecting 2.40 mol of N₂O₄ into an empty, closed 3.00 L flask.

N₂O_{4 (g)} 2 NO_{2 (g)}

Init conc.

D conc.

Eq. conc.

N₂O_{4 (g)} ¾ 2 NO_{2 (g)} K = 4.66 x 10^-3

If the system was compressed to obtain a volume of 1.50 L, calculate the new equilibrium concentrations.

N₂O_{4 (g)} 2 NO_{2 (g)}

Init conc.

D conc.

Eq. conc.

1.00 mol of SO₂ and 1.00 mol of O₂ are placed in 1.00 L flask at 1000 K. When equilibrium has been achieved, 0.925 mol of SO₃ has been formed. Calculate K_c at 1000 K for the reaction.

2 SO_{2 (g)} + O_{2 (g)} ¾ 2 SO_{3 (g)}

K_P – partial pressures

• The equilibrium constant can be defined based on partial pressures (atm) of gases rather than their molar concentrations.
• PV = nRT or P =(n/V)RT where n/V = M
• (P = MRT)
• Pressure is directly proportional to molar conc.
• For a reaction: aA + bB ¾ cC + dD
• K_P = (P_C)^c (P_D)^d
• (P_A)^a (P_B)^b (all P’s = eq. P’s in atm)
• K_P is related to K_C … P = MRT
• K_P = K_C(RT)^Dn or K_C = K_P(RT)^-Dn
• where Dn = moles of gas products – moles of gas reactants.
• If Dn = 0, K_P = K_C
• Note: since K_C involves molar concentrations (units of moles/L) and K_P uses involves pressures in the units of atmospheres, we must use a value of R which involves these units. R= 0.08206 L·atm/ mol·K

The reaction for the formation of nitrosyl chloride was studied at 25°C. The pressures at equilibrium were found to be as follows. Calculate the value of K_P and convert to K_C.

2 NO (g) + Cl₂ (g) ¾ 2 NOCl (g)

0.050 atm 0.30 atm 1.2 atm

Under equilibrium conditions at 250°C, the following concentrations of the species listed are present.

PCl₅ (g) ¾ PCl₃ (g) + Cl₂ (g)

0.038 M 0.262 M 0.262 M

Calculate K_C and convert to K_P

Assume that gaseous hydrogen iodide is synthesized from hydrogen gas and iodine vapor at a temperature where the equilibrium constant is 1.00 x 10². Suppose HI at 5.000 x 10^-1 atm, H₂ at 1.000 x 10^-2 atm, and I₂ at 5.000 x 10^-3 atm are mixed in a 5.000 L flask. Calculate the equilibrium pressure of all species.
Heterogeneous equilibria

• Involve species in more than one phase
• Pure liquids and pure solids are assigned activities of 1, so they do not appear in the calculation of K.

At 90°C, K_C = 6.8 x 10^-2 for the following rxn.

If 0.15 mol hydrogen and 1.0 mol sulfur are heated to 90.0 °C in a 1.0 L container, what will be the partial pressure of H₂S at equilibrium?

H₂ (g) + S (s) ¾ H₂S (g)

Relationship between DG°_rxn and K

When reactants are mixed together, a number of changes occur which influence the thermodynamics of the system

• Mixing relates to an increase in entropy
• If products are formed, the process can be exothermic or endothermic
• The state of matter of the reactants and products can further influence entropy
• Concentrations of reactants and products are changing

DG°_rxn relates to the free energy change which occurs when reactants are completely converted to products under standard conditions.

DG _rxn relates to the free energy change which occurs at any concentrations of reactants and products and under any conditions.

• At equilibrium DG _rxn = 0.

The following relationship summarized the relationship between DG°_rxn and DG _rxn

• DG _rxn = DG°_rxn + RT lnQ
• Since at equilibrium DG_rxn = 0 and Q = K DG°_rxn = -RT lnK
• R must have units which correspond to those of K (K_C or K_P). Use 8.314 J/mol K for K_P.
• For spontaneous reactions (product favored) DG°_rxn < 0, so lnK > 0 and K > 1.
• For non-spontaneous reactions (reactant favored) DG°_rxn > 0, so lnK < 0 and K < 1.

For equilibrium mixtures, the mixture represents the thermodynamically most stable system.

• The more negative the value of DG°_rxn, the more products are favored (equilibrium proceeds further to the right).
• Reactions which are said to “go to completion” have very negative values for DG°_rxn.