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Equilibrium
I. Dynamic equilibrium
¨ Many chemical reactions are reversible and never go to completion: The reactants react to form products, which can recombine (in the reverse reaction) to reform the original reactants, given the appropriate conditions.
Ø In a closed system, the concentrations of reactants and products will eventually become constant (in a state of dynamic equilibrium).
Ø A closed system is one in which neither matter nor energy can be gained or lost from the system. In other words, its macroscopic properties remain constant. If the system were open, some of the products could escape, and equilibrium would never be reached.
¨ equilibrium – dynamic (changing) condition in which two opposing physical or chemical
changes occur at equal rates
Ø It’s like customers entering and leaving a department store on the day of a big sale
Ø At first, everybody à in (forward reaction favored)
About lunchtime = equilibrium between people coming from lunch and going to lunch.
At closing time, everyone à out (reverse reaction favored)
¨ Equilibrium & changes of state:
Ø Think about liquid in a sealed glass beaker.
Ø Given enough heat energy, molecules at the surface can evaporate.
Ø Some vapor molecules will begin to cool, condense, and reenter the liquid phase.
Ø Given constant T & surface area, the rate at which molecules evaporate will eventually equal the rate at which they condense.
Ø Then, these rates are at equilibrium:
liquid + heat energy vapor
¨ Also: consider equilibrium as it would relate to ice and water in a sealed container!
¨ In a chemical reaction, equilibrium is reached when the rate of the forward reaction equals the rate of the reverse reaction.
Ø At equilibrium, the concentrations of all reactants and products remain constant, but both forward and reverse reactions are still proceeding.
Ø Equilibrium can be approached from both directions (when a reaction is begun or when the equilibrium is disturbed).
¨ Le Chatelier’s Principle – when any factor affecting equilibrium is changed, the system will
shift to minimize that change (and reestablish equilibrium)
Ø ex: When you add heat to a system
(put the beaker from the previous example in a sunny window):
liquid + heat energy vapor
· The forward reaction is favored, producing more vapor.
· The reaction proceeds forward (liquid water à vapor) more quickly until a new higher equilibrium is established.
· The reverse is true for cooling, here – if you take away heat energy (ex: put the beaker in a refrigerator), the reverse reaction is favored (to the left).
· The reaction proceeds backward (vapor à liquid water) until a new lower equilibrium is reached.
¨ equilibrium vapor pressure – pressure exerted by a vapor (gas) in equilibrium with its liquid
(at a given temperature)
Ø e.v.p. increases with increasing temperature
(higher T à more evaporation à more gas molecules running into container walls à
more pressure, at equilibrium)
Ø The stronger the forces of attraction between liquid molecules à lower e.v.p.
(because the liquid molecules hold on to each other and don’t want to go into the vapor phase)
· In the diagram below, vapor pressure is shown as a function of temperature for several liquids. For all, the vapor pressure increases with temperature. Shown are:
(a) carbon disulfide, CS2
(b) methanol, CH3OH
(c) ethanol, CH3CH2OH
(d) water, H2O
(e) aniline, C6H5NH2
· The temperature at PE = 760 mmHg on each vapor pressure curve is the normal boiling point of that liquid.
II. The Position of Equilibrium
¨ Suppose 2 reactants, A & B, react to form products C & D:
Ø A + B C + D
¨ At equilibrium, the relationship between the molar concentrations of [A], [B], [C], and [D] is:
[C] x [D] [products]
[A] x [B] [reactants]
¨ Kc is known as the equilibrium constant
¨ Kc has a specific value (at a specific temperature) for every reaction.
Ø When the T changes, K changes.
¨ If the value of Kc = 1…
Ø There are roughly equal concentrations of products and reactants at equilibrium.
¨ If the value of Kc > 1…
Ø Products exceed reactants at equilibrium (reactants are mostly converted to products).
Ø When Kc > 1, the reaction goes almost to completion.
¨ If the value of Kc < 1…
Ø Reactants exceed products at equilibrium (forward reaction occurs only slightly; few reactants converted to products).
Ø When Kc < 1, the forward reaction barely proceeds at all.
¨ What about coefficients in the reaction?
aA + bB cC + dD
[C]c x [D]d
[A]a x [B]b
¨ ex: 3A + B 2C + 3D
[C]2 x [D]3
[A]3 x [B]
¨ ex: H2 + I2 2HI
[HI]2
[H2] x [I2]
¨ Substitute [1] for the concentration of solids or liquids when solving for K – solids and
liquids will not change concentration; only gases.
CaCO3(s) CaO(s) + CO2(g)
[ CaO(s)][CO2(g)] [ 1 ][CO2(g)]
[CaCO3(s)] [1] = [CO2]
¨ Le Chatelier’s Principle – If a system at equilibrium is subjected to a stress, the equilibrium
shifts to relieve that stress.
¨ 3 important stresses to consider for Le Chatelier’s Principle:
1) Change in Concentration
Ø A + B C + D
Ø 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 K value and the final K 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 Kc does not change.
Ø 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.
Ø Changes in concentration do not affect K (i.e. – the position of the equilibrium may change without ultimately affecting Kc).
Ø Consider the following diagram, showing changes in concentrations when N2 is added to an equilibrium mixture of N2, H2, and NH3. Net conversion of N2 and H2 to NH3 occurs until a new equilibrium is established. That is, N2 and H2 concentrations decrease, while the NH3 concentration increases.
2) Change in Pressure
Ø This applies to gases only.
Ø Consider the manufacture of ammonia as an example:
N2(g) + 3H2(g) 2NH3(g)
Ø 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.)
Ø 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).
Ø Pressure changes are often related to volume changes:
· If there is an overall volume change in a gaseous reaction, then increasing the pressure will move the equilibrium towards the side with less volume (i.e. – the side with fewer moles of gas).
· This shift reduces the total number of molecules in the equilibrium system and so
tends to minimize the pressure.
Ø Either way (whether increasing pressure on the whole system or just one gas), since the relative ratios of reactants and products remain unchanged, changes in pressure do not affect K (i.e. – the position of the equilibrium may change without ultimately affecting Kc).
Ø 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 forward reaction is favored, decreasing the total number of gas molecules until equilibrium is re-established (Qc = Kc).
3) Change in Temperature
Ø Reversible reactions are exothermic in one direction and endothermic in the other direction.
Ø Increasing temperature of the system favors the endothermic reaction.
Ø Cooling favors the exothermic reaction:
N2(g) + 3H2(g) 2NH3(g) + 92kJ
Ø (Treat heat energy just like any other reactant or product.)
Ø The graph below shows temperature dependence of the equilibrium constant for the reaction: N2(g) + 3H2(g) 2NH3(g). Note that Kc (the equilibrium constant) is plotted on a logarithmic scale and decreases by a factor of 1011 on raising the temperature from 300 K to 1000 K.
· Reaction rates are temperature dependent. Since equilibrium is based on rates of forward and reverse reactions, equilibrium is also temperature dependent.
¨ Catalysts change the rate of a reaction but do not affect Kc.
Ø Nor do catalysts affect the position of the equilibrium!
Ø They 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 therefore accelerates the forward and reverse reactions by the same factor, and the composition of the equilibrium mixture is unchanged.
The Haber Process
¨ Nitrogen-containing compounds are needed to make everything from fertilizers to explosives.
¨ During WWI, the major source of nitrates (NaNO3, used as a starting point for many of these other compounds) were deposits in Chile.
Ø Germany was dependent on these deposits as well, but when the Allies blockaded South America, Germany was forced to look for another source.
Ø A German chemist named Fritz Haber investigated directly combining nitrogen and hydrogen gases to form ammonia:
N2(g) + 3H2(g) 2NH3(g) + 92kJ
¨ The reaction worked, BUT – Problem:
Ø This reaction reached a very low equilibrium, long before any significant amount of
ammonia had been produced.
¨ The Solution: Use Le Chatelier’s Principle to keep producing more ammonia!
N2(g) + 3H2(g) 2NH3(g) + 92kJ
¨ The Haber process forces the continued production of ammonia by:
1. Removing ammonia as it is produced.
· This has the effect of continually decreasing [NH3], driving the forward reaction.
2. Keeping the pressure on the system high.
· Increasing the pressure favors the direction producing the fewest moles of gas (forward reaction, in this case).
· Increasing pressure also increases the # of particles per unit volume. This
increases the rate at which equilibrium is reached.
3. Keeping the temperature of the system low.
· Decreasing temperature always favors the exothermic reaction.
· However, at low T, this reaction is so slow that the process becomes uneconomical.
· Adding heat will speed the reaction up, but lower the percent yield, so a happy medium must be found.
Ø All of these factors cause a forward reaction shift, producing more ammonia.
Ø Consider the following relationship between temperature and percent yield of ammonia:
¨ Here is another diagram that plots % yield of ammonia vs. pressure. It conveys the same information as the previous diagram, just in a slightly different form…
¨ Haber discovered that using an iron oxide catalyst eliminates the need for excessively high temperatures.
Ø The catalyst increases the rate at which equilibrium is reached (but not the % yield).
Ø The catalyst is used in a finely divided form (small pieces) so that the surface area is maximized, to increase its efficiency.
¨ A modern industrial plant using a modified Haber process might operate with a T of about 450°C and a pressure of about 250 atm.
Ø Under these conditions, the percent yield of ammonia is about 40%.
¨ The hydrogen gas required for the Haber Process is obtained from natural gas (methane), and the nitrogen gas from the fractional distillation of air.
The Contact Process
¨ Sulfuric acid is the most industrially produced chemical in the world. Over 150 million tones of H2SO4 are produced globally every year.
Ø It is used for fertilizers, paints, detergents, and fibres, and as a feedstock for other chemicals.
¨ Most of the sulfuric acid manufactured is produced using the Contact Process, which involves the catalytic oxidation of sulfur dioxide, SO2, to sulfur trioxide, SO3.
Ø Solid sulfur, S(s), is burned in air to form sulfur dioxide gas, SO2:
S(s) + O2(g) SO2(g)
Ø These gases are mixed with more air, then cleaned by electrostatic precipitation to remove any particulate matter.
Ø The mixture of sulfur dioxide and air is heated to 450oC and subjected to a pressure of
1 – 2 atmospheres in the presence of a vanadium (or platinum) catalyst to produce sulfur
trioxide, SO3(g), with a yield of 98%.
2SO2(g) + O2(g) 2SO3(g) + heat
Ø Sulfur trioxide, SO3(g) is dissolved in 98% (18M) sulfuric acid, H2SO4, to produce disulfuric acid, or pyrosulfuric acid, also known as fuming sulfuric acid or oleum, H2S2O7.
SO3(g) + H2SO4(l) H2S2O7(l)