If you like us, please share us on social media, tell your friends, tell your professor or consider building or adopting a Wikitext for your course.
In a balanced chemical equation, the total number of atoms of each element present is the same on both sides of the equation. Stoichiometric coefficients are the coefficients required to balance a chemical equation. These are important because they relate the amounts of reactants used and products formed. The coefficients relate to the equilibrium constants because they are used to calculate them. For this reason, it is important to understand how to balance an equation before using the equation to calculate equilibrium constants.
There are several important rules for balancing an equation:
Example 1:  

\[H_2\; (g) + O_2 \; (g) \rightleftharpoons H_2O \; (l) \]

Example 2  

\[Al \; (s) + MnSO_4 \; (aq) \rightleftharpoons Al_2(SO_4)_3 + Mn ; (s) \]

Example 3  

\[P_4S_3 + KClO_3 \rightleftharpoons P_2O_5 + KCl + SO_2 \]

Balanced chemical equations can now be applied to the concept of chemical equilibrium, the state in which the reactants and products experience no net change over time. This occurs when the forward and reverse reactions occur at equal rates. The equilibrium constant is used to determine the amount of each compound that present at equilibrium. Consider a chemical reaction of the following form:
\[ aA + bB \rightleftharpoons cC + dD\]
For this equation, the equilibrium constant is defined as:
\[ K_c = \dfrac{[C]^c [D]^d}{[A]^a [B]^b} \]
The activities of the products are in the numerator, and those of the reactants are in the denominator. For K_{c}, the activities are defined as the molar concentrations of the reactants and products ([A], [B] etc.). The lower case letters are the stoichiometric coefficients that balance the equation.
An important aspect of this equation is that pure liquids and solids are not included. This is because their activities are defined as one, so plugging them into the equation has no impact. This is due to the fact that pure liquids and solids have no effect on the physical equilibrium; no matter how much is added, the system can only dissolve as much as the solubility allows. For example, if more sugar is added to a solution after the equilibrium has been reached, the extra sugar will not dissolve (assuming the solution is not heated, which would increase the solubility). Because adding more does not change the equilibrium, it is not accounted for in the expression.
The following are concepts that apply when adjusting K in response to changes to the corresponding balanced equation:
A balanced equation is very important in using the constant because the coefficients become the powers of the concentrations of products and reactants. If the equation is not balanced, then the constant is incorrect.
For gasphase equilibria, the equation is a function of the reactants' and products' partial pressures. The equilibrium constant is expressed as follows:
\[ K_p = \dfrac{P_C^c P_D^d}{P_A^a P_B^b} \]
P represents partial pressure, usually in atmospheres. As before, pure solids and liquids are not accounted for in the equation. K_{c} and K_{p} are related by the following equation:
\[ K_p = K_c(RT)^{\Delta n} \]
where
\[ \Delta n = (c+d)  (a+b) \]
This represents the change in gas molecules. a,b,c and d are the stoichiometric coefficients of the gas molecules found in the balanced equation.
Note: Neither K_{c} nor K_{p} have units. This is due to their formal definitions in terms of activities. Their units cancel in the calculation, preventing problems with units in further calculations.
Example: Using K_{c}  

\[ PbI_2 \rightleftharpoons Pb \; (aq) + I \; (aq) \] First, balance the equation.
Next, calculate find Kc. Use these concentrations: Pb^{} 0.3 mol/L, I^{} 0.2 mol/L, PbI_{2} 0.5 mol/L \[ K_c = \dfrac{(0.3) * (0.2)^2}{(0.5)} \] \[K_c= 0.024\] Note: If the equation had not been balanced when the equilibrium constant was calculated, the concentration of I^{ }would not have been squared. This would have given an incorrect answer. 
Example  

\[SO_2 \; (g) + O_2 \; (g) \rightleftharpoons SO_3 \; (g) \] First, make sure the equation is balanced.
Calculate K_{p}. The partial pressures are as follows: SO_{2} 0.25 atm, O_{2} 0.45 atm, SO_{3} 0.3 atm \( K_p = \dfrac{(0.3)^2}{(0.25)^2 \times (0.45)} \) \( K_p= 3.2\) 
Unless otherwise noted, content in the UC Davis ChemWiki is licensed under a Creative Commons AttributionNoncommercialShare Alike 3.0 United States License. Permissions beyond the scope of this license may be available at copyright@ucdavis.edu. Questions and concerns can be directed toward Prof. Delmar Larsen (dlarsen@ucdavis.edu), Founder and Director. Terms of Use