Things to know for Exam 2

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Gases are interesting materials to work with because of their properties. Since all gases occupy the same volume at the same temperature and pressure for the same number of moles, one can measure gases in volumes for reactions. For instance, the formation of water from hydrogen gas and oxygen gas is shown below.

2HThe reaction tells us that 2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of water. Now it is quite difficult to weigh out 4 grams (2 moles) of hydrogen gas or even 32 g of oxygen gas. It is quite serendipitous that instead one can measure out 2 volumes of hydrogen and one volume of oxygen to produce a stoichiometric reaction. The term volume means any volume. It could me 2 mL to 1 mL or it could be 44.8 L to 22.4 L so long as the volumes are the same base. FYI, 22.4 L is the volume occupied by 1 mole of any gas at 1 atm pressure and at 20øC._{2}(g) + O_{2}(g) --> 2H_{2}O(l)

So on to working with gases!

A simple equilibrium is:

HThe double-headed arrow indicates that there are forward and reverse reactions. Since this is a commonly occuring equilibrium in nature, most people are aware of the fact that it happens. Water in an open container in a sealed system will evaporate to a certain point where water in the air begins to condense back to the liquid state. If the system is let undistured it appears that the water level in the open container is doesn't change. In actuality, specific water molecules in the open container escape into the vapor phase and at the same rate other water molecules in the vapor phase condense to form liquid water and maintain the water level. The system is said to be in equilibrium._{2}O(l) <--> H_{2}O(g)

Equilibrium is recognized to occur in many other systems. Some are more complex because of the number of reactants and products involved. It has been found that a simple mathematical concept can be used to understand the equilibrium of complex systems. That mathematical relationship is known as "K". A "K" is a constant developed from empirical (laboratory) measured data. The mathematical relationship is simple. The amount (concentration or pressure) of the products are multiplied together and divided by the multiplied reactant amounts. Here is the ever-famous chemistry example.

aA(g) + bB(g) <--> cC(g) + dD(g)The small letters symbolize the number of moles of each substance in a stoichiometric reaction. The capital letters are representative of different substances. The constant develops as follows.

Recognize that "a" means there are "a" moles of "A", etc. That concentration of "A" occurs "a" times in the reaction. So go the others. This means there are "c" moles of "C" and "d" moles of "D". The constant, K, is the product of the pressures of the products divided by the product of the pressures of the reactants.

product of the pressures of the products K = ----------------------------------------- product of the pressures of the reactantsFor the "famous" reaction above, the constant is:

(PThe value of this constant is determined by actually measuring the pressures of the various products and reactants. Pressures are measured at different times and the researcher watches for the pressures to stop changing. When the pressures of each constituent (called partial pressures since each pressure is part of the total pressure) are stable, the reaction is considered to be at equilibrium. One can also see that a stiochiometric, balanced reaction must be identified in order to find a K._{C})^{c}(P_{D})^{d}K = ------------ (P_{A})^{a}(P_{B})^{b}

Similar equilibrium constants can be developed for every equilibrium known. These constants tell the researcher the extent to which a reaction will take place. They also tell us which way a reaction will go when a particular change is made to the products or reactants. We can compare K's to a similar mathematical relationship called a "Q". A "Q" is calculated based upon starting concentrations. The value of "Q" is compared to the "K" to see which way the reaction will go to reestablish equilibrium. "Q" is called the reaction Quotient.

The reaction:

NInitially 1.00 atm of N_{2}O_{4}(g) <--> 2NO_{2}(g)

First the K is developed:

(PNext the variables are replaced with the known pressures at equilibrium._{NO2})^{2}K = ------- P_{N2O4}

(1.56)This constant tells us that no matter what the starting pressures of these two gases, the final pressures, when plugged into the above K equation and the calculations performed, will^{2}K = ----------- = 11.06 0.22

Now suppose we had a system at equilibrium where we had 3.12 atm
of NO_{2} and 0.88 atm N_{2}O_{4}. You
can verify equilbrium by comparing the K calculated with these
values to the one above.

Now, let's add 2 atm of NO_{2} so that the
P_{NO2} is now 5.12 atm. Calculate Q
first:

(5.12)Q > K so the reaction will proceed to the left (products reacting to form reatants). Had Q been less than K the reaction would go to the right. When Q=K then nothing changes.^{2}Q = ----------- = 29.79 0.88

This is the central concept of Le Chatelier's Principle. A system in equilibrium will respond to a change so as to reestablish the equilibrium.

In order to differentiate Ks, the K above is usually called a
K_{p} in contrast to K_{a} for acids or
K_{b} for bases, etc.

So what are the pressures when EQ is reestablished?
We know that the NO_{s} will be less but not how much
less so we let the drop be "2x". Since the amount of
N_{2}O_{4} formed is half the number of moles of
the amount lost, we let x be the new N_{2}O_{4}
pressure.

Substituting:

(5.12 - 2x)This turns in a quadratic equation.^{2}K = 11.06 = -------------- 0.88 + x

11.06(0.88 + x) = 26.21 - 20.48x + 4xSo our roots are:^{2}9.73 + 11.06x = 26.21 - 20.48x + 4x^{2}4x^{2}- 31.54x + 16.48 = 0

-(-31.56) +/- ((-31.56)^{2}-4(4)(16.48))^{1/2}------------------------- 2(4) 31.56 +/- 27.06 = --------------- = 7.33 or 0.56 8 We know it can't be 7.33 since this is more pressure than the total system pressure so the x must be 0.56. Thus the final pressures are:PRecheck K:_{N2O4}= 1.44 atm P_{NO2}= 4.00 atm

(4.00)FYI, the K for the reverse of any reaction is 1/K for the forward reaction.^{2}K = --------------- = 11.11 (very close to 11.06) 1.44 (probable rounding error)

(P_{NO2})^{2}K_{forward}= ------- P_{N2O4}P_{N2O4}K_{reverse}= ------- (P_{NO2})^{2}Coefficient rule:Let's say you needed to double a reaction because you were going to add it to another reaction.

Original reaction:

NDoubled reaction:_{2}O_{4}(g) <--> 2NO_{2}(g) (P_{NO2})^{2}K = = 11.06 P_{N2O4}

2NLets manipulate things a bit:_{2}O_{4}(g) <--> 4NO_{2}(g) (P_{NO2})^{4}K = ---------- = 11.06 (P_{N2O4})^{2}

(PThe student should note that doubling the reaction has the effect of squaring the top AND the bottom of the K. Thus the K for the doubled reaction is the square of the K for the single reaction._{NO2})^{4}((P_{NO2})^{2})^{2}K = ---------- = ----------- (P_{N2O4})^{2}(P_{N2O4})^{2}

K" = KA general rule is that the multiplier of the reaction, n, takes the constant, K, to the nth power.^{2}

K" = K^{n}Combining reactions:Sometimes Ks are available for certain reactions but NOT for the one being studied. It is possible to use Ks from individual reactions to produce a K for a combined reaction. When reactions are added (combined), the Ks are multiplied to produce the new constant, K, for the cmobined reaction.

Here's an example:

2SOThe K is found thusly:_{2}(g) + O_{2}(g) <--> 2SO_{3}(g) K = 4.4 2NO_{2}(g) <--> 2NO(g) + O_{2}(g) K = 4.0 ------------------------------ 2SO_{2}(g) + 2 NO_{2}(g) <--> 2 SO_{3}(g) + 2NO(g)

K_{overall}= (K_{SO2})(K_{NO2}) = (4.4)(4.0) = 17.6