equilibrium notes.notebook april 06, 2016 · equilibrium notes.notebook 11 april 06, 2016 the...

Equilibrium Notes.notebook

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April 06, 2016

Properties of a system at equilibrium:

• rate forward = rate reverse

• concentrations of reactants and products remain constant

• constant macroscopic properties (those we can observe) – system appears static (unchanging)

• dynamic process at the molecular level


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April 06, 2016

Let’s consider a hypothetical reaction and graph how the concentrations of reactants and products change over time until equilibrium is reached. Recall that the coefficients are proportional to the relative rates.

2A + B à C

concentration (M)

Time (min)


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April 06, 2016

To quantify an equilibrium (the Law of Mass Action):

Keq =

equilibrium constant = mass action expression

example: aA + bB = cC + dD

Keq =

The brackets [ ] indicate molar concentrations. Keq is a temperature dependent constant. Changing the temperature of an equilibrium system changes both the concentration of participants and the numeric value of Keq. When the concentration of one of the participants is changed, the concentration of the others varies in such a way as to maintain a constant value for Keq.


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April 06, 2016

Let’s consider the reaction between hydrogen gas and iodine vapor reacting to yield hydrogen iodide. For a fixed volume at a constant temperature of 490oC an equilibrium is established that yields a numeric value = 45.9.

H2(g) + I2(g) = 2HI(g) Keq = 45.9 at 490oC

We will conduct 5 experiments in which we start with differing concentrations of reactants and products. After equilibrium is established we will measure the concentrations of all species and plug the molar concentrations into the mass action expression. Below are the results:


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April 06, 2016

The value of Keq tells us something about the relative concentrations of reactants and products at equilibrium. A large value of Keq tells us we have mostly products at equilibrium while a small value for Keq indicates mostly reactants at equilibrium.

Equilibrium Problems

Problem 1

A certain equilibrium system is represented by the following chemical equation:

A(g) + 2B(g) = 2C(g)

a) Write the equilibrium expression for the system.

b) At a particular temperature, [A] = 0.88 M, [B] = 0.45 M, and [C] = 0.0022 M. Calculate the value of Keq.

c) Are the reactants or products favored at equilibrium? How can you tell?


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April 06, 2016

Problem 2

Given the following Equilibrium system:

2 SO2(g) + O2(g) = 2 SO3(g)

Exactly 5.00 mole of SO3 is placed in a 2.00 liter vessel at 500.oC. It then decomposes and 0.500 mole of O2 is present when equilibrium is established.

a) Calculate the molar concentrations of all three species at equilibrium. (Let’s do this using an ICE approach)

2 SO2(g) + O2(g) = 2 SO3(g)

initial I

change C

equilibrium E


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April 06, 2016

If 0.500 mole of O2 is present at equilibrium, how many moles of SO2 must be present (we started initially with all SO3 and SO3 decomposes in a 2:2 stoichiometric ratio with SO2 and a 2:1 ratio with O2)?

How many moles of SO3 must remain at equilibrium? Keep in mind the stoichiometric ratios!

b) Calculate the value of Keq


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April 06, 2016

Problem 3

Given the following equilibrium system:

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

Exactly 50.0 moles of ammonia is introduced into a 10.0 liter flask at constant temperature. The ammonia decomposes and when equilibrium is established 15.0 mole of nitrogen gas is present.

a) calculate the molar concentrations of all species present at equilibrium

b) calculate the value of Keq.


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April 06, 2016

Le Chatelier’s Principle

For reactions that are reversible, a state a thermodynamic stability will be reached when the reactions are at equilibrium. Since they are stable at equilibrium they will tend to maintain this condition. The chemist Le Chatelier studied equilibria in detail and proposed the following model, now known as Le Chatelier’s Principle:

If a stress is applied to a system at equilibrium, the equilibrium will shift in the direction that relieves the stress. A stress will make the rates of the forward and reverse reactions unequal – eventually equilibrium will be reestablished at a new point.


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April 06, 2016

Stresses

1. Concentration Changes – an increase in concentration (of reactant or product) will make the equilibrium shift in the direction that reduces that increase.

2. Pressure Changes – only effective for gaseous systems. An increase in pressure will shift the equilibrium to the side with the smaller volume (i.e. smaller number of moles of gas) and vice versa. If the equilibrium involves equal volumes (equal moles) a change in pressure will have no effect on the equilibrium.

3. Temperature Changes – an increase in temperature favors the endothermic direction; a decrease in temperature favors the exothermic direction. Keep in mind that changes in temperature will also have the effect of changing the value of the equilibrium constant.

4. Adding a Catalyst – recall a catalyst lowers the activation energy for both the forward and reverse reactions. Thus, adding a catalyst to system at equilibrium will have no effect. If the system is not at equilibrium, adding a catalyst will make the reaction reach equilibrium faster.


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April 06, 2016

The Chicken Eggs Equilibrium A True Story

As reported in the prestigious Journal of Chemical Education

Chickens cannot perspire, so when they get hot they pant. This seemingly trivial fact leads to a serious economic loss for egg producers. In hot weather, chickens lay eggs with thin shells which are easily (and frequently) broken. The chicken egg shell is composed of calcium carbonate (CaCO3).

A little reflection shows that this is an inevitable consequence of Le Chatelier’s Principle and the wellknown carbonate equilibrium system:

Ca2+CO2(g) = CO2(aq) = H2CO3(aq) = H+(aq) + HCO3

(aq) = H+(aq) + CO32(aq) = CaCO3(s)

(chicken breath) (eggshell) When the chicken pants, the equilibrium is perturbed by the rapid loss of carbon dioxide. Because this effect cascades through all of these equilibria, the effect is a loss of solid calcium carbonate which ultimately produces weaker egg shells.

So what do we do with chickens in hot weather?

• set up fans in the chicken coops?• play them Mozart?• throw the chickens into a big refrigerator?

Ted Odum, while a graduate student at the University of Illinois, found the deceptively simple “solution” to this problem – give the chickens carbonated water. Now the equilibrium has been perturbed in the opposite direction. The addition of aqueous carbon dioxide shifts all the equilibria to the right and results in stronger egg shells. Moreover, the chickens seem to like the carbonated water, and there are rumors that they spend their spare time singing familiar jingles about “spirit” and “the real thing.” Philosophical questions about which came first are left to your philosophy class, but in this case, at least, Le Chatelier’s Principle comes before the egg (shell).


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April 06, 2016

The Haber Process

The Haber process involves the synthesis of ammonia from hydrogen and nitrogen. This is without doubt one of the most important chemical reactions that chemists have ever carried out. The reason for this is the fact that plants require nitrogen for growth in a “fixed” form – that is, in a compound form that is usable by plants. The only widely available source of nitrogen is the N2 present in our atmosphere. Recall nitrogen gas is extremely unreactive (triple bond between N atoms). In nature, fixation is carried out by nitrogenfixing bacteria that grow on the roots of certain plants.

In 1912, the German chemist Fritz Haber developed a process of synthesizing ammonia directly from its elements. The process requires high temperatures, high pressures, and a catalyst and is illustrated on the following page. This process is interesting because it is an equilibrium system that is net exothermic:

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

Based on Le Chatelier’s principle and the illustration, predict four optimum conditions for the production of ammonia:

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

1. ________________________________ 2. _________________________

3. ________________________________ 4. _________________________


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April 06, 2016

Consider the following equilibrium:

2SO3(g) + heat = 2SO2(g) + O2(g)

stress shift after new equilibrium point is established

add SO3

what will be the effect of an increase in SO3 on the equilibrium constant, Keq?

remove O2

add SO2

increase temp

what will be effect on the equilibrium constant after an increase in temp?

decrease temp

what will be effect on the equilibrium constant after a decrease in temp?

increase pressure

decrease pressure

add catalyst


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April 06, 2016

Solubility Product Constant

In a saturated solution of an ionic solid, an equilibaqueous ions and the solid. For example, consider sil

AgCl(s) = Ag+(aq) + Cl(aq)

At constant temperature,

Ksp = [Ag+][Cl]

The equilibrium constant is designated with an “sp” suand recall that the solid does not enter into the massamount of a solid changes but not its concentration).

What will happen if you add more AgCl(s) to the above

Write the Ksp expression for a saturated solution of s


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April 06, 2016

The usefulness of a Ksp value is to distinguish relatively insoluble compounds. Consider two sulfates, for example:

BaSO4 Ksp = 1.1 X 1010

CaSO4 Ksp = 9.1 X 106

Although both compounds are fairly insoluble but calcium sulfate is four orders of magnitude more soluble compared to barium sulfate.

Solubility Product Constant Problems

We will consider three types of problems in dealing with the solubility product constant:

(a) given the Ksp, determine the solubility

(b) given the solubility, determine the Ksp


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April 06, 2016

Problem 1

Determine the solubility of barium sulfate in moles per liter and grams per liter given the above Ksp value. What are the concentrations of barium ion and sulfate ion in a saturated solution of barium sulfate?


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April 06, 2016

Problem 2

Determine the molar solubility of magnesium hydroxide. It has a Ksp = 8.8 X 1012.


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April 06, 2016

Problem 3

The solubility of cadmium sulfide is 1.0 X 1014 mole/liter. Determine the Ksp for cadmium sulfide.


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April 06, 2016

The Common Ion Effect

Consider AgCl(s) = Ag+(aq) + Cl(aq)

Using Le Chatelier’s Principle, predict what will occur if you add aqueous KCl to this equilibrium solution.


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April 06, 2016

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