1301 Dynamic Equilibrium, Keq, and the Mass Action Expression

The Equilibrium Process

Dr. Fred Omega Garces Chemistry 111 Miramar College

1 Equilibrium 05.2015 Concept of Equilibrium & Mass Action Expression

Extent of a Chemical Reaction Many reactions do not convert 100% of reactants to products. There is often a point in a reaction when the products will back react to form reactants. The extent of the reaction i.e., 20% or 80%, can be determined by measuring the concentration of each component in solution. In general the extent of the reaction is a function of:

Temperature, Concentration and degree of organization which is monitored by some constant value called the -

The equilibrium constant (Keq).

2 Equilibrium 05.2015 (Dynamic) Equilibrium

Chemical Equilibrium is a dynamic state in which the rates of the forward and the reverse reaction are equal.

A teeter-totter is an analogy of a system in balance. The left is balanced with the right. In a chemical reaction this means that the Left (reactant) is changing at the same rate as the right (products).

Reactant Product

3 Equilibrium 05.2015 How Equilibrium is achieved Consider the reaction A → B

Forward A → B rate = kf[A]

Reverse B → A rate = kr[B]

Overall A D B rate forward (kf[A]) = rate reverse (kr[B])

Example: N2O4 + E D 2 NO2

Effect of temperature on the

N2O4/NO2 equilibrium. The tubes contain a mixture of

NO2 and N2O4 . As predicted by Le Chatelier’s principle, the equilibrium favors

colorless N2O4 at lower temperatures, but shifts to

the darker brown NO2 at higher temperature.

4 Equilibrium 05.2015 Equilibrium: Mass Action Expression When equilibrium is establish A D B

That is rate forward = rate reverse or kf[A] = kr[B]

Rearranging this equation can be rearrange to kf/kb = [B] /[A]

The mass action expression defines the equilibrium constant (Keq): Keq = [B] / [A]

For any general chemical process at equilb. aA + bB D pP + qQ A mass action expression can be written:

p q P • Q K = [ ] [ ] eq a b [ A ] • [ B ]

This is also referred to as the Law of Mass Action

5 Equilibrium 05.2015 Heterogeneous Systems (1)

Substances are in different phases at equilibrium i.e., solid and aqueous.

Which solid is more concentrated?

100 g (1 cup) 200 g (2 cup)

Concentration of a solid (and pure liquid) is always a constant; The concentration does not change !

6 Equilibrium 05.2015 Heterogeneous Systems Consider the following reaction:

CaCO3(s) D CaO (s) + CO2 (g)

CaO • CO [ ] [ 2] Keq = CaCO [ 3]

but, [CaO] = constant (solid)

and CaCO = constant (solid) [ 3]

Constant • CO [ 1 ] [ 2] Keq = Constant [ 2]

K = K = CO eq p [ 2] 7 Equilibrium 05.2015

€ Haber Process

(D signify that the reactants are not N2(g) + 3H2 (g) D 2NH3 (g) completely (100%) converted to products.

As soon as NH3 is form, some of it decomposes and form N2 and H2 the starting chemical. This takes place until the amount consumed is equal to the amount produced, i.e., equilibrium is reached.

- 1.0 - -.60 - Law of Mass Action: -.20 2 !NH # " 3$ K = = K eq 3 p !N #•!H # " 2$ " 2$

8 Equilibrium 05.2015 Meaning of Keq Which is favored: Reactant or Product?

p q x [P ] • [Q ] [Product ] K eq = a = y [ A ] • [ B ]b [ Reactant ]

For Keq > 1, the ratio [Product]x > [Reactant]y For Keq = 1, the ratio [Product]x = [Reactant]y For Keq < 1, the ratio [Product]x < [Reactant]y

For Keq > 1, at equilibrium Product is favored. For Keq = 1, at equilibrium Product and Reactant are equal. For Keq < 1, at equilibrium Reactant is favored.

9 Equilibrium 05.2015 Calculating Keq from reaction

• The mass action expression is equal to the equilibrium constant.

• Equilibrium constants are generally written without units.

For Keq >> 1, at equilibrium Product is favored.

For Keq < 1, at equilibrium Reactant is favored.

10 Equilibrium 05.2015 Equilibrium Constants: Keq, Kc, Kp, Ka, Kb, Ksp, Kf, Kw…

Kc versus Kp: For reactions that undergoes reversibility, the efficiency of the reaction

(extent to which the reaction forms product) is indicated by its equilibrium constant, Keq.

If the equilibrium constant (Keq) express through its mass action expression is based on molar concentration, then the equilibrium constant is written as Kc. If the equilibrium

constant (Keq) express through its mass action expression is based on partial pressure, then the equilibrium constant is written as Kp.

Ksp: If the rxn is a solid dissolving to ions, i.e., solubility of a salt, then Keq is written as Ksp.

Kf: If the reaction is a formation of a transition complex, then Keq is written as Kf.

Ka: If the reaction involves an acid losing a proton, then Keq is written as Ka.

Ka2: If the rxn involves an diprotic acid losing its second proton, then Keq is written as Ka2.

Ka3: If the reaction involves a polyprotic acid losing third proton, then Keq is written as Ka3 .

Kb: If the reaction involves an base, then Keq is written as Kb.

Kw: If the reaction involves water, then Keq is written as Kw.

11 Equilibrium 05.2015 Keq: Kc versus K p

If the equilibrium constant (Keq) express through its mass action expression is based on molar concentration, then the equilibrium

constant is written as Kc. If the equilibrium constant (Keq) express through its mass action expression is based on partial pressure, then the equilibrium constant is written as Kp.

−Δn +Δn K =K RT K =K RT c p( ) p c( ) [RT is positive exponent (+Δn)]

where Δn = ∑(gas moles) − ∑(gas moles) € € product reactant

12 Equilibrium 05.2015 € Derivation Between Kc and Kp Consider the Reaction:

COCl2 (g) D CO (g) + Cl2 (g)

[CO] [Cl ] CO * Cl K = 2 or (p ) (p 2 ) c KP = [COCl2] COCl (p 2 ) It can be shown that:

€ € −Δn +Δn K =K RT K =K RT c p( ) p c( ) [RT is positive exponent (+Δn)]

where Δn = ∑(gas moles) − ∑(gas moles) € € product reactant

13 Equilibrium 05.2015

€ 1. Calculating Keq: Kc - Kp relationship

Carbon dioxide is placed in a 5.00-L high-pressure reaction vessel at 1400.°C. Given the following

reaction, CO (g) + 1/2 O2 (g) D CO2 (g) , calculate the equilibrium constant Kc and Kp after equilibrium

is achieved with 4.95 mol of CO2 , 0.0500 mol of CO, and 0.0250 mol of O2 found in the container.

T = 1400°C CO + 1 O ! CO (g) 2 2 (g) 2 (g) V = 5.0 L i Δ mol e 0.050 mol 0.025mol 4.95mol Vol = 5.0 L -2 -3 -1 [e] 1.00 ⋅10 M 5.00 ⋅10 M 9.90 ⋅10 M 9.90 ⋅10-1M [CO2 ] [ ] kc = 1/2 = 1/2 CO ∗ O -2 -3 €[ ] [ 2 ] [1.00 ⋅10 M]∗[5.00 ⋅10 M] 3 kc =1.40 •10

€ 14 Equilibrium 05.2015 2. Calculating Keq iCe - method

At 300°C, 0.600 mol of PCl5 is placed in a 100mL reaction vessel.

During the course of the reaction PCl5 converts to PCl3 and Cl2 gas. At

equilibrium the amount of PCl3 formed is 0.200mol. What is the

equilibrium constant for the following reaction. PCl5(g) D PCl3 (g)+ Cl2 (g)

Note: 0.600mol/0.100L = 6.00 M and 0.200mol/0.100L = 2.00M

PCl ↔ PCl + Cl 5 (g) 3 ( g ) 2 ( g ) i 6.00 M 0 0 V = 100 mL Δ -X +X +X = 0.100L "e$ ? 2.00 ? # %

15 Equilibrium 05.2015 Animation: PCl5 D PCl3 + Cl2

PCl D PCl + Cl C 5(g) 3 (g) 2 (g) Cl Cl P C C Cl Cl Cl C Cl P P Cl Cl Cl Cl Cl Cl Cl P C Cl Cl Cl Cl Cl P C C Cl Cl Cl Cl Cl Cl P P C Cl Cl Cl Cl Cl Cl P Cl Cl PCl ↔ PCl + Cl 5 (g) 3 ( g ) 2 ( g ) i 6.00 M 0 0 V = 100 mL Δ -X +X +X = 0.100L "e$ ? 2.00 ? # % 4.00 2.00

16 Equilibrium 05.2015 Summary: Equilibrium constant

• The mass action expression is equal to the equilibrium constant.

• Equilibrium constants are generally written without units.

For Keq >> 1, at equilibrium Product is favored.

For Keq < 1, at equilibrium Reactant is favored.

17 Equilibrium 05.2015 LeChatelier’s Principle

Changing the concentration or pressure of a substance in a reaction at equilibrium is creating a stress. The removal of something creates a “deficit stress”, and adding something creates an “excess stress.” Reactions will always respond to stress by replacing the deficit (making more of what was removed) or by conversion of the excess (consuming some of what was added). This description works regardless of which side of the reaction is changed.

Stress/Deficit will be in the form of: (1) Concentration, (2) Pressure and (3) Temperature variation

18 Equilibrium 05.2015 LeChatelier Principle

Methods of disturbing Equilibrium and LeChatelier response.

Stress on Reactant Stress on Product Relief on Reactant Relief on Product

Reactant D Product

Equilibrium Revisited: Equilibrium - Situation in which changes occur at equal rates so no net change is apparent.

LeChatelier Principle: A stress on the a system at equilibrium causes changes to the system to reduce the stress until a new equilibrium is established.

19 Equilibrium 05.2015 Reactant Change: Stress added Consider the affect on Equilibrium as a result of increasing the concentration of the reactant.

Stress on the reactant

Which direction does the reaction shift towards?

i) A system at ii) A stress is equilibrium. added to the reactant.

The reaction shifts to the right or left?

20 Equilibrium 05.2015 Reactant Change: Stress added Consider the affect on Equilibrium as a result of increasing the concentration of the reactant.

Stress on the reactant

Which direction does the reaction shift towards?

i) A system at ii) A stress is iii) The system equilibrium. iv) Until a new added to the self-adjust to equilibrium is re- reactant. reduce the established stress.

The reaction shifts to the right. The reaction favors the product. 21 Equilibrium 05.2015 Product Changes: Stress added

Consider the affect on Equilibrium as a result of increasing the concentration of the product.

Stress on the product

Which direction does the reaction shift towards?

i) A system at ii) A stress is equilibrium. added to the product.

The reaction shifts to the right or left?

22 Equilibrium 05.2015 Product Changes: Stress added

Consider the affect on Equilibrium as a result of increasing the concentration of the product.

Stress on the product

Which direction does the reaction shift towards?

i) A system at ii) A stress is iii) The system equilibrium. iv) Until a new added to the self-adjust equilibrium is re- product. itself to reduce established. the stress.

The reaction shifts to the left. The reaction favor the reactant. 23 Equilibrium 05.2015 Reactant Change: Stress Relief Consider the affect on Equilibrium as a result of decreasing the concentration of the reactant. What shift in direction occurs on the reaction ?

Deficit on the reactant

i) A system at ii) A de-stress equilibrium. is place to the reactant.

The reaction shifts to the right or left?

24 Equilibrium 05.2015 Reactant Change: Stress Relief Consider the affect on Equilibrium as a result of decreasing the concentration of the reactant. What shift in direction occurs on the reaction ?

Deficit on the reactant

i) A system at ii) A de-stress iii) The system equilibrium. iv) Until a new is place to the self-adjust itself equilibrium is re- reactant. to replace the established. missing reactants.

The reaction shifts to the left. The reaction favors the reactant. 25 Equilibrium 05.2015 Product Change: Stress Relief

Consider the affect on Equilibrium as a result of decreasing the concentration of the Product. What shift in direction occurs on the reaction ?

Deficit on the reactant

i) A system at ii) A de-stress equilibrium. is place to the product.

The reaction shifts to the right or left?

26 Equilibrium 05.2015 Product Change: Stress Relief

Consider the affect on Equilibrium as a result of decreasing the concentration of the Product. What shift in direction occurs on the reaction ?

Deficit on the product

i) A system at ii) A de-stress iii) The system self- equilibrium. iv) Until a new is place to the adjust itself to equilibrium is re- product. replace the displaced established. products.

The reaction shifts to the right. The reaction favors to the product.

27 Equilibrium 05.2015 LeChatelier’s Principle (1): Concentration Effect If a chemical system is at equilibrium and then a substance is added (either a reactant or product), the reaction will shift so as to re- establish equilibrium by subtracting part of the added substance. Conversely, removal of a substance will result in the reaction moving in the direction that forms more of the substance.

Consider the Haber reaction at

equilibrium: N2 + 3H2 → 2NH3

If some H2 is added to the reaction which was at equilibrium, the system self-adjust to remove

the excess H2 by converting it to

NH3 until equilibrium is re-

establish; in the process some N2 is also consumed.

28 Equilibrium 05.2015 LeChatelier’s Principle (2): Volume / Pressure Effect

Increasing the pressure (by reducing the volume) of a gaseous mixture causes the system to shift in the direction that reduces the number of moles of gas.*

Consider the reaction :

N2O4 (g) D 2NO 2 (g) + P Suppose the volume of the container is reduced?

The equilibrium shifts to the side that reduces the total number of moles of

gas involved in the reaction. In this example, 2 moles of product (NO2) will

change to 1 mol of reactant (N2O4). The total moles in the reaction mixture is reduced to compensate for the increase in pressure.

Note: Adding an inert gas will not shift the equilibrium because the inert gas is not involved in the overall reaction.

29 Equilibrium 05.2015 LeChatelier’s Principle (2): Volume / Pressure Effect