| Syllabus / Schedule | 
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| Chem 109 |
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Summer 2002 |
| Lecture Notes::17 July |
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| PREVIOUS |
As a general statement can say that all chemical reactions are equilibrium reactions and go toward a state of equilibrium or approach equilibrium. Some reactions reach equilibrium rapidly, some slowly, and some favor products to such an extent that the reactions "go to completion." Regardless of initial concentrations, systems will reach equilibrium given time.
What is an equilibrium - very dynamic situation in chemistry (Great Crabapple War overheads)
Let's look at some classic examples of reactions approaching equilibrium.
Vast numbers of chemical reactions operate at equilibrium in the natural world, and it is frequently essential to be able to understand and to predict their behavior. Qualitatively we can get an idea of how an equilibrium system will behave by using Le Châtelier's Principle.
Le Châtelier's Principle: If stress is applied to a system at equilibrium, the equilibrium will shift in such a way as to relieve the stress. e.g. if the pressure of carbon dioxide is increased over a solution of carbon dioxide in water, more carbon dioxide will dissolve, reducing the pressure increase. As an example let's look at the response of the ammonia synthesis reaction (above) to the introduction of nitrogen (overhead: Russell 14.3)
Quantitatively we can look at the relationship in a reaction at equilibrium represented by
The Equilibrium Expression is then:
Generalizing for the equation (Note that the simple derivation above does not generalize, since rates and stoichiometry are not generally correlated - the relationships between rates at equilibrium and stoichiometry are beyond our study.):
aA + bB + ... Æ cC + dD + ... Keq = [C]c[D]d/ [A]a[B]b.
A similar expression is the Mass Action Expression:
The mass action expression is algebraically identical to the equilibrium expression, but it applies to a more general case. That is, the equilibrium expression requires that the values in the expression give the equilibrium constant, whereas the mass action expression allows any set of values. Thus the mass action expression is used to describe a system which has not yet reached equilibrium, while the equilibrium expression is the special case of the mass action expression for a system at equilibrium.
Let's look at an equilibrium system, and try to predict its response using Le Châtelier's Principle and/or the equilibrium expression.
Example: Consider the reaction
CO + NO2 ´ NO + CO2 + heat (226 kJ) Note that heat appears on the product side - the system is giving up heat, \ DH is negative, DH = - 226 kJ
K = [NO][CO2]/[CO][NO2] So, what will happen to [CO2] if:
- CO is added?
- [CO2] will increase. The increased [CO] will react more rapidly with NO2, driving the reaction to the right and reducing the amount of CO added. (Looking at K, [CO2] must increase to keep K constant.)
- NO2 is added?
- [CO2] will increase. The increased [NO2] will react more rapidly with CO, driving the reaction to the right and reducing the amount of NO2 added. (Looking at K, [CO2] must increase to keep K constant.)
- NO is added?
- [CO2] will decrease. The increased [NO] will react with CO2, driving the reaction to the left. (Looking at K, [CO2] must decrease to keep K constant.)
- T is increased?
- [CO2] will decrease. If T increases heat must have been added, therefore the reaction will use up some of the extra heat, driving the reaction to the left and lowering the temperature back toward its original value.
- V is increased?
- [CO2] will decrease in proportion. Although a larger volume will result in a lower pressure (fewer collisions), we have equal numbers of reactants on both sides, so neither side will be favored. (Looking at K, all concentrations reduced by the same factor, so K remains constant without changing proportions of players.)
- Ar is added?
- No change. Ar is unreactive and will have no effect.
I want to look at solving equilibrium problems, so we'll look at a number of examples and see how to use the equilibrium expression to follow the equilibrium process.
Example: Consider the gas phase reaction:
PCl5 ´ PCl3 + Cl2 K = 5.0 x 10-2 @ 150 °C (What will happen to this reaction if the volume is increased? Shift to right to bring the total concentration of particles back up.) Find [Cl2] if [PCl5] = 0.40 M and [PCl3] = 0.20 M.
Know that K = [PCl3] [Cl2] / [PCl5] Substituting, K = (0.20)[Cl2] / (.040) = 5.0 x 10-2 Rearranging, [Cl2] = (5.0 x 10-2)(0.40)/(0.20) = 1.0 x 10-1 M
A very common type of reaction is one in which a dissociation takes place.
Example: Consider the gas phase dissociation of carbonyl chloride to carbon monoxide and chlorine @ 100 °C.
If 0.20 moles of carbonyl chloride (COCl2) is placed in a 2.5 L container at 100 °C calculate the concentrations of all species at equilibrium. K = 2.6 x 10-10 @ 100 °C.
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Notice that this will give a quadratic equation - yuk! This is to be avoided if possible, if for no other reason than more steps means more opportunities to make a mistake! So we will take advantage of the fact that we are working with experimental work - there is always an error, and if x is smaller than our error, then we can simplify.
Thus the trick is to assume x<< 0.080 and if we look at K, its value (2.6 x 10-10) is small enough that it is likely that our assumption is correct. So:
K = x2/ 0.080 = 2.6 x 10-10 x2 = (0.080)(2.6 x 10-10) = 2.08x 10-11 and x = 4.6 x 10-6 Note that x = 4.6 x 10-6<< 0.080, so our assumption is correct! \ [CO] = [Cl2] = 4.6 x 10-6 M [COCl2] = 0.080 M
Pressure vs. Concentration in Equilibria: In gas phase equilibria we can use either concentrations or pressures, since by the Ideal Gas Law P = (n/V)RT = [ ] RT
So how are the K's for concentration related? It turns out that this depends on the stoichiometry. If the sums of the coefficients on both sides are equal, then K = KP. More generally (see derivation in Zumdahl 5th, p 620-621), KP = K(RT)Dn, where n = the number of moles appearing on each side of the stoichiometric equation, and Dn = nproducts - nreactants.
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© R A Paselk
Last modified 17 July 2002
