Chemical Equilibrium

Name: ______________________

Chemistry 1 Notes, 2012–13; P. Holloman

Topic 10 – Equilibrium

2.1.1 – Explain the energetic nature of phase changes.

2.1.3 – Interpret the data presented in phase diagrams.

3.1.2 – Explain the conditions of a system at equilibrium.

3.1.3 – Infer the shift in equilibrium when a stress is applied to a chemical system.

I. Physical Equilibrium

· Most chemical reactions do not go to completion. Some (but not all) of the reactants react to form products, and these products can recombine (in the reverse reaction) to reform the original reactants, given the appropriate conditions.

· ex: 2FeO(s) 2Fe(s) + O2(g)

· In a closed system (where matter and energy cannot enter or escape from the system), the concentrations of reactants and products will eventually become constant. At this point, the system is in a state of dynamic equilibrium.

· In an open system, some of the products could escape (especially if one of them were a gas), and equilibrium could never be reached.

· Equilibrium is a dynamic (changing) condition in which two opposing physical or chemical changes occur at equal rates. Take note: This is an equilibrium of rates – not concentrations!

· Look again at the decomposition reaction above. If this reaction is at equilibrium, that does not mean you have equal mass, or volume of reactants and products. It simply means that the rate at which FeO decomposes into Fe and O2 is equal to the rate at which Fe and O2 combine to reform FeO. Essentially, at equilibrium, for every 2 FeO formula units that decompose, 2 Fe atoms and 1 O2 molecule combine to reform those 2FeO.

Equilibrium & Changes of State

· Think about liquid water in a sealed glass beaker:

· Given enough heat energy (enough to break the hydrogen bonds holding the liquid water molecules together), molecules at the surface can begin to evaporate into water vapor.

· Some vapor molecules will begin to cool – they lose KE, through collisions with each other or container walls, etc. Then they condense, and reenter the liquid phase.

· Given constant T & surface area, the air above the liquid will eventually become saturated with vapor molecules (at 100% humidity, in the case of water). At this point, the rate at which H2O molecules evaporate will equal the rate at which they condense. Now these rates are at equilibrium: H2O(l) + heat energy H2O(g)

· Consider how vapor pressure will change over time, as this water / water vapor system reaches equilibrium:

· At first, the vapor pressure will steadily increase. Since the air above the beaker is not yet saturated, water molecules will continue to evaporate, resulting in more and more water vapor molecules zipping around the inverted beaker. This means more collisions with the container walls, and thus greater pressure.

· Once equilibrium is reached (when the rate of evaporation = the rate of condensation), the amounts of water and water vapor will remain relatively constant, so long as the temperature does not change. Because you have a relatively constant amount of vapor at a constant temperature, the vapor pressure will level off and remain constant. This is called equilibrium vapor pressure.

· If the temperature in this system increases (you warm up the beaker), the vapor pressure will increase for two reasons:

· More evaporation more vapor molecules running into container walls (i.e. – the water vapor molecules are more concentrated).

· More heat = more KE available to vapor molecules = particles colliding with container walls more frequently and with more force.

· However, assuming all the liquid water does not evaporate, the system will eventually reestablish equilibrium between evaporation and condensation, as the warmer air will eventually become saturated with water vapor again. Until equilibrium is reach, we say that the forward reaction (evaporation, here) is favored.

· The exact opposite of all these things is true for a water / water vapor system when you cool it down. The reverse reaction (condensation) will be favored until a new lower equilibrium is established, with a lower vapor pressure.

· Also consider this: The stronger the forces of attraction between liquid molecules, the lower the equilibrium vapor pressure (because the liquid molecules have a strong attraction for each other and won’t readily go into the vapor phase).

Phase Diagrams & Changes of State:

(Frost deposited on a car window in winter.)change of statenameexample

solid liquidmeltingice water

solid gassublimationdry ice CO2 gas

liquid solidfreezingwater ice

liquid gasvaporizationwater steam

gas liquidcondensation water vapor water

gas soliddepositionwater vapor ice

· Phase changes are reversible processes: Liquid water can evaporate into water vapor, given enough kinetic energy, and when the water vapor molecules cool off and lose some of this kinetic energy, they may condense and reenter the liquid phase. Eventually an equilibrium will be established, in which the rate of evaporation equals the rate of condensation.

· ex: H2O(l) + heat energy H2O(g)

(Solid iodine subliming at 70°C, well below its melting point of 114°C. Deposition of the iodine vapor occurs on the colder walls of the flask.)

· A phase diagram is a graph of pressure vs. temperature showing when a substance is a gas, liquid, or solid.

· Note, then, that temperature and pressure are determining factors for what phase of matter a substance is in at any given point in time.

(Point A is the triple point of substance. (T & P conditions at which all 3 phases of a substance can coexist at equilibrium).Point C is the critical point of a substance.Critical temperature is the T above which the substance cannot exist as a liquid, no matter how much you increase the pressure.)

· There are three curves in a phase diagram:

· curve AB – T & P conditions where solid and vapor coexist at equilibrium

· curve AC – T & P conditions where liquid and vapor coexist at equilibrium

· curve AD – T & P conditions where liquid and solid coexist at equilibrium

· Phase diagrams also indicate the normal boiling point and freezing point of a substance (“normal” means at 1 atm of pressure):

· melting = transition from solid to liquid at 1 atm

· boiling = transition from liquid to gas at 1 atm

· Be prepared to interpret the phase diagrams for water and carbon dioxide. This might include questions about (1) phase at a given temperature and pressure; (2) boiling point or melting at a given pressure; (3) triple point of a material; and (4) sublimation in carbon dioxide.

(Phase diagram for water)

(Phase diagram for CO2)

Heating & Cooling Curves vs. Phase Diagrams

· Consider the similarities and differences of information communicated by both the heating and cooling curve and phase diagram for water, shown below:

(Phase diagram for water)

· Take a look at this excellent animation of changes of state and heating curves:

http://cwx.prenhall.com/petrucci/medialib/media_portfolio/text_images/031_ChangesState.MOV

II. Chemical EquilibriumThe position of equilibrium – The Equilibrium Constant (Kc)

· Suppose 2 reactants, A & B, react to form products C & D:

· A + B C + D

· At equilibrium, the relationship between the molar concentrations of [A], [B], [C], and [D] is:

Kc = =

· Kc is known as the equilibrium constant.

· Kc has a specific value (at a specific temperature) for every reaction.

· When the T changes, K changes.

· If the value of Kc = 1…

· There are roughly equal concentrations of products and reactants at equilibrium.

· If the value of Kc > 1…

· Products exceed reactants at equilibrium (reactants are mostly converted to products).

· When Kc >> 1, the reaction goes almost to completion.

· If the value of Kc < 1…

· Reactants exceed products at equilibrium (forward reaction occurs only slightly; few reactants converted to products).

· When Kc << 1, the forward reaction barely proceeds at all.

· What about coefficients in the reaction?

aA + bB cC + dD

(ex: H2 + I2 2HI Kc = )

· ex: 3A + B 2C + 3D

Kc =

· Substitute [1] for the concentration of solids or liquids when solving for K – you cannot

change the concentration of solids and liquids; only gases and aqueous solutions.

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

Kc = = = [CO2]

Le Chatelier’s Principle

· Le Chatelier’s Principle – when any factor affecting equilibrium is changed, the system will

shift to minimize that change (and reestablish equilibrium)

· 3 important stresses to consider for Le Chatelier’s Principle:

1) Change in Concentration

· A + B C + D

· Increasing concentration:

· If you increase the concentration of a reactant (increase the amount of A or B), the equilibrium shifts forward to produce more products (C, D), but Kc does not change (the initial Kc value and the final Kc value are the same, because the relative ratios remain the same).

· If you increase the concentration of a product, the reverse reaction is favored, producing more reactants, but, again, Kc does not change.

· Decreasing concentration:

· If you decrease the concentration of a reactant, the reverse reaction is favored, producing more reactants, but Kc does not change.

· If you decrease the concentration of a product, the forward reaction is favored, producing more products, but Kc does not change.

· Short version: There is an equilibrium shift away from an increase in concentration

and toward a decrease in concentration.

· Changes in concentration do not affect Kc.

(i.e. – The position of the equilibrium may change without ultimately affecting Kc.)

· Consider the following diagram, showing changes in concentrations when N2 is added to an equilibrium mixture of N2, H2, and NH3. N2 and H2 are consumed to produce more NH3, until a new equilibrium is established.

2) Change in Pressure

· This applies to gases only.

· Consider the manufacture of ammonia as an example:

N2(g) + 3H2(g) 2NH3(g)

(Reactions such as this one, where all reactants and products are in the same phase

are known as homogeneous.)

· When pressure on the whole system increases, equilibrium shifts in the direction that produces the fewest gas molecules. (The converse would also be true. Decreasing pressure on the whole system shifts equilibrium in the direction producing the greater number of gas molecules.)

· When the partial pressure of an individual gas increases, equilibrium shifts in the opposite direction (away from that gas; the same effect as increasing the

concentration of the gas).

· Pressure changes are often related to volume changes:

· Increasing pressure in a gaseous reaction often happens by decreasing the volume occupied by the gas molecules (such as a piston compressing in your car’s cylinders). Thus, increasing pressure on the whole system will cause an equilibrium shift in the direction with the fewest molecules of gas.

· The volume of liquids and solids is essentially fixed, but gases expand to fit the shape of their container. Reducing the # of moles of gas in the system minimizes this tendency, helping to offset the increase in pressure.

· Either way (whether increasing pressure on the whole system or just one gas), since the relative ratios of reactants and products will remain unchanged, changes in pressure do not affect Kc (i.e. – the position of the equilibrium may change without ultimately affecting Kc).

· The figure below shows the qualitative effect of pressure and volume on the equilibrium: N2(g) + 3H2(g) 2NH3(g).

· Figure (a) shows a mixture of gaseous N2, H2, and NH3 at equilibrium.

· Figure (b) shows that when the pressure is increased by decreasing the volume, the mixture is no longer at equilibrium (Qc < Kc).

· In figure (c), under these conditions, the reverse reaction is favored, decreasing the total number of gas molecules until equilibrium is re-established (Qc = Kc).

3) Change in Temperature

· Reversible reactions are exothermic in one direction and endothermic in the other direction.

· Just treat heat energy just like any other reactant or product:

· Increasing the temperature of the system always favors the endothermic rxn.

· Cooling favors the exothermic rxn.

· Consider the Haber process: N2(g) + 3H2(g) 2NH3(g) + 92kJ

· As the graph below illustrates, heating this reaction up favors the endothermic reaction (reverse rxn. in this case).

· Changes in temperature are the only changes that affect Kc.

· To figure out how they affect Kc, here’s a simple shortcut: Whatever happens to the products as a result of the equilibrium shift also happens to Kc. If [products] goes up, Kc goes up; if [products] goes down, Kc goes down.

(Here is a video illustrating Le Chatelier’s Principle and T, with the cobalt ion complex:http://cwx.prenhall.com/petrucci/medialib/media_portfolio/text_images/066_LeChateliersPr.MOV )

Entropy & Equilibrium

· We also need to consider equilibrium in terms of the entropy of a system.

· Entropy (S) can be described as the disorder in a system. Entropy is a measure of the amount of randomness, or freedom of movement of particles in a system.

· The more disordered (more random) a system = higher entropy. The more orderly and organized particles are, the less entropy they have.

· All systems & reactions in the universe have a natural tendency toward low enthalpy (low energy), but high entropy (more random and disordered).

· In any reversible reaction, the change in enthalpy and change in entropy must have the same sign.

· Practically speaking, this means that (when you are looking at a thermochemical equation for a reaction) the side with the heat on it is opposite from the side with the highest entropy

· So, for any equilibrium, if you increase the temperature, the equilibrium will shift to the side of the equation with more entropy. Likewise, if you decrease the temperature, the equilibrium will shift to the side of the equation with less entropy.

· Consider how this makes sense using the equation for the Haber Process shown above…

Catalysts & Equilibrium

· Catalysts change the rate of a reaction but do not affect Kc.

· Nor do catalysts affect the position of the equilibrium!

· Catalysts affect forward and reverse reactions equally.

· The potential energy diagrams are shown below for a reaction whose activation energy is lowered by the presence of a catalyst. The activation energy for the catalyzed pathway (red curve) is lower than that for the uncatalyzed pathway (blue curve).

· The catalyst lowers the activation energy barrier for the forward and reverse reactions by exactly the same amount. (The catalyst decreases the PE of the activated complex, often by splitting the reaction up into multiple steps, each of which form activated complexes that have lower PE than the original pathway. Both the forward and reverse reactions must form these same activated complexes, so the both reactions are equally affected.) The catalyst therefore accelerates the forward and reverse reactions by the same factor, and the final composition of the equilibrium mixture is unchanged – it just gets there faster now.

1

]

[Reactants

[Products]

[A][B]

[C][D]

]

][I

[H

[HI]

2

2

2

[B]

[A]

[D]

[C]

3

3

2

]

[CaCO

]

[CaO][CO

3

2

[1]

]

[1][CO

2