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C. Reactions Main
Introduction Limiting Reagents and Theoretical Yield |
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K, the Equilibrium Constant
While we have so far discussed reactions in terms of "product-" or "reactant-favored," most reactions do not fit entirely into either category: few convert all reactants into products or refuse to proceed at all. Instead, most reactions proceed to an equilibrium point; before equilibrium, more products are being converted to reactants than vice versa. At equilibrium, the reaction is still proceeding on a microscopic scale, but the conversion from products to reactants exactly equals the reverse reaction. Scientists have proven that a given atom or molecule may shift between the product or reactant category using radioactive tracers; the reaction does not stop, but simply reaches a balance point. The double-arrow,
, represents the fact that the reaction continues, with both products and reactants still present.
To provide a convenient way to measure how far a reaction proceeds before reaching equilibrium, chemists use the equilibrium constant. It is one of the rare concepts that permeates all of chemistry. It is especially prominent in acid/base reactions, oxidation-reduction reactions, and thermodynamics, but influences most of the science in some way or another. So what is this magic letter, you ask? Put simply, it is just a ratio of the products of a reaction to the reactants.
For the reaction W A + X B
K = ([C]Y[D]Z)/([A]W[B]X)
where A and B are the reactants, W and X are their coefficients (respectively), C and D are the products, and Y and Z are their coefficients. The rule to remember for equilibrium constants is "products over reactants," not vice versa. Therefore, if K is much greater than one, the reaction is product-favored; if K is much less than one, the reaction is reactant-favored. If K is about one, the reaction tends to have similar concentrations of products and reactants.
Note that the value of K is dependent only on temperature. No matter how much the pressure, volume, or concentration of species change or time elapses, K will remain the same. However, you need to be conscious of the temperature that K values are given at; don't automatically assume that the K value given is for the appropriate temperature.
K is sometimes referred to as Keq, the subscript "eq" meaning "equilibrium." In most equilibrium equations, the values in brackets represent the concentration in moles per liter of that species. If all species are measured in a concentration of moles per liter (see note below), the subscript "c" is often used (Kc). Other flavors of K include: Kp, for use with gas partial pressures (discussed next chapter); Ksp, for solubility products; Ka, for acid dissociation constants; Kb, for base dissociation constants; Kw, for the autoionization of water; and Kf, for complex ion formation constants. (Ka, Kb, and Kw discussed in the Acid/Base Reactions chapter.)
Note that solids are never included in an equilibrium constant (since concentrations vary in solids), nor are pure liquids for the same reason (this second rule usually applies only to water). Liquids are also not included when used as solvents. If any of these species is present and has to be eliminated from the equilibrium expression, the subscript "c" is dropped.
Another symbol using the same expression as K is widely used as well: Q, the reaction quotient. To calculate Q, simply plug the concentrations of all species into the equilibrium expression and find the value that results--it need not be the same as the equilibrium constant. If Q is less than K, the concentration of reactants is too high, and the reaction will proceed to equilibrium. If Q is bigger than K, then there are too many products, and the reaction will run backwards until equilibrium is established. If Q is equal to K, then the system is at equilibrium and the net reaction ceases. Q is touched on again in the "Redox Reactions" chapter, so keep it in mind.
Sometimes, it is convenient to manipulate the equilibrium constant mathematically to make a problem easier to solve. If you reverse the reaction, you must take the reciprocal of the equilibrium constant (if K for one reaction were 2/7, K for the reverse would be 7/2). If you multiply the reaction's coefficients by some number, you must raise K to that power as well (multiplying the reaction by 3 means you must cube K). Finally, dividing the reaction by a number forces you to take that root of K (dividing the reaction by 3 means taking the cube root of K). Other mathematical properties of K will be discussed later.
As an example of using K, we'll first write the equilibrium equation for the following reaction:
2 SO2 (g) + O2 (g)
The equilibrium equation for this reaction, following the format above, is:
Kc = [SO3]2/([SO2]2[O2])
In this case, the value of the equilibrium constant is 1.28 x 104 1/mol * L at 852 K. Let's say we start with 1 mole per liter of SO2 and 2 moles per liter of O2. How much of each species will exist when the reaction is complete? First, create a table similar to that below:
SO2 O2 SO3 Initial 1 M 2 M 0 M Change - 2x - x + x Equil. 1 M - 2x 2 M - x x Answer ? ? ? Now, substitute in the "Equilibrium" values for the concentrations in the equilibrium equation:
Kc = 1.28 x 104 = (x)2/((1-2x)2(2-x))
From here, you can either multiply out the variables and solve algebraically, or use a calculator to find the value for x. We recommend the second method, simply because this problem involves more variables than you are likely to see on an AP or other exam. In any case, the value for x can be 1.999965, .501812, or .498203. Which is it? Well, since the 1 M - 2x term cannot be negative, x must be .498203. Substituting this for x, we get final values of:
SO2 O2 SO3 Answer 0.003594 M 1.501797 M 0.498203 M The next page in this chapter deals with various ways we can perturb reactions and how they react to our interference.
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