From UCDavis Chemwiki
The equilibrium constant, K, is an expression expressing the relationship between products and reactants of a reaction at equilibrium with respect to a specific unit. There are two different types of equilibria reactions: (1) heterogeneous and (2) homogeneous. Below we illustrate the difference between the two, explain how to write equilibrium constants for both, and introduce calculations involved with both the concentration and the pressure equilibrium constant.
As mentioned above, there are two types of equilibrium reactions. It is important to understand the difference between the two, because we express their equilibrium constants differently.
The simpler one, a homogeneous reaction, is one where the states of matter of the products and reactions are all the same . In most cases, the solvent determines the state of matter for the overall reaction. For example, the synthesis of methanol from a carbon monoxide-hydrogen mixture is a gaseous homogeneous mixture, which contains two or more substances:
`CO(g) + 2H_2(g) -> CH_3OH(g)`
At equilibrium, the rate of the forward and reverse reaction are equal, which is demonstrated by the arrows. The K however tells you the ratio of the units (pressure or concentration) when the reaction is at equilibrium.
The synthesis of ammonia is another example of a gaseous homogeneous mixture:
`N_2(g) + 3H_2(g) -> 2NH_3(g)`
A heterogeneous reaction is one where one or more states within the reaction differ (heteros=different). For example, the formation of an aqueous solution of lead(II) iodide is a heterogeneous mixture dealing with molecules in both thesolid and aqueous states:
`Pbl_2(s) -> Pb^(2+) (aq) + 2l^(-) (aq)`
The decomposition of sodium hydrogen carbonate (baking soda) at high elevations is another example of aheterogeneous mixture, this reaction deals with molecules in both the solid and gaseous states:
`2NaHCO_3 (s) -> Na_2CO_3(s) + H_2O(g) + CO_2(g)`
`C(s) + O_2(g) -> CO_2(g)`
Once again, this difference is emphasized so that students remember that equilibrium constant calculations are different from heterogeneous mixtures.
Before we write an equilibrium constant, we should become familiar with the concept of an equilibrium constant. An equilibrium constant is obtained by letting a single reaction proceed to equilibrium and then measuring the concentrations of each molecule involved in that reaction and creating a ratio of products to reactants. Because we must allow the reaction to proceed to equilibrium, the equilibrium constant will remain the same for each reaction independent of initial concentrations which determine the speed of a reaction (for an ideal reaction). This knowledge allowed scientists to derive a model expression that can serve as a "template" for any reaction. We will now look at the basic "template" form of a homogeneous equilibrium constant.
If we have a hypothetical reaction:
`aA(g) + bB(g) -> gG(g) + hH(g)`
*The lower case letters represent the number of moles of each molecule, the upper case letters represent the molecule itself, and the letters in the parenthesis always represents the state of matter of the molecule.
The Equilibrium Constant of Concentration gives the ratio of concentrations of products over reactants for a reaction that is at equilibrium. This is usually used when the state of matter for the reaction is (aq). The equilibrium constant expression is written as Kc, which would be as follows:
Here, the letters inside the brackets represent the concentration of each molecule. Notice the sum of the products is the numerator of the ratio and the sum of the reactants is the denominator. This will be the case for every equilibrium constant. Also note that each concentration is raised to a power of its coefficient; this will also be seen in every equilibrium constant problem. Keep in mind that this expression was obtained by a homogeneous equilibrium reaction. K represent equilibrium constants and c represents concentration (Kc). This means that every molecule shows up in the expression, as long as it is a solution or a gas.
Gaseous reactions are not expressed by concentration, but instead, by partial pressures. The Equilibrium Constant of Pressure gives the ratio of pressure of products over reactants for a reaction that is at equilibrium. The equilibrium constant is written as Kp, which would be as follows:
The procedure for this is the same as the procedure for the concentration constant above.
If you have Kc and want it converted into Kp use the following Equation:
`K_p = K_c(RT)^(Deltan_(gas))
`R= 0.8206 L*atm*K^(-1)*"mol"^(-1)
`T =` Temperature in Kelvin
`Deltan_gas = "Mole of gas (product)" - "Mole of Gas (reactant)"`
So far, we have dealt with reaction is which all the states of matter were the same. So how do we write heterogeneous reaction constants? Most importantly, solids and pure liquids will always be excluded from the equilibrium constant. In a mathematical perspective, concentration of solids and liquids equal one, which do not affect the overall K value. Solvents also equal one. The molarity of solids, liquids, and solvents remain constant throughout the reaction, which means their value can be denoted as 1. This rule is extremely important to remember, especially in dealing with heterogeneous solutions.
| Example 1 |
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If we have a hypothetical reaction: `aA(s) + bB(l) -> gG(aq) + hH(aq)` What would be the equilibrium constant expression? |
So, what do we do if a reaction is reversed, or a reaction is split into its elementary steps. In this section, you will be taught these two manipulations.
1. When the reaction is reversed, the equilibrium constant expression is inverted. The expression would be written as:
`K^'= 1/(([G]^g [H^h])/([A]^a [B]^b)) = ([A]^a [B]^b)/([G]^g [H^h])`
2. In an equation for a reaction is multiplied by any factor, n, then the original value of Kc is raised to the nth power.
3. If we add together two reactions to obtain a new reaction, the equilibrium constant for the new reaction is the product of the equilibrium constants for the original reactions.
`A + C ⇌ AC` `K_3 = ([AC])/([A]{C])`
`AC + C ⇌ AC_2` `K_4 = ([AC_2])/([AC][C])`
`A + 2C ⇌ AC_2` `K_5 = K-3 * K_4 = ([AC])/([A][C]) * ([AC_2])/([AC][C]) * ([AC_2])/([A][C]^2)`
Subpages (1): Example 1