HOME
Course Chapters
Calculator Fundamentals
Mathematics Review
Basic Concepts
Advanced Concepts
Section Tests
Pretest
Posttest
Useful Materials
Glossary
Online Calculators
Redox Calculator
Kinetics Arrhenius Calculator
Thermodynamics Calculator
Nuclear Decay Calculator
Linear Least Squares Regression
Newton's Method Equation Solver
Compressibility Calculator
Units Conversion Calculator
Nomenclature Calculator
Related Information Links
Texas Instruments Calculators
Casio Calculators
Sharp Calculators
Hewlett Packard Calculators
Credits
Contact Webmaster

Equilibrium
In stoichiometry calculations, we assume that reactions run to completion. However, when a chemical reaction is carried out in a closed vessel, the system achieves equilibrium.
Equilibrium occurs when there is a constant ratio between the concentration of the reactants and the products. Different reactions have different equilibria. Some may appear to be completely products, however, all reactions have some reactants present. A reaction may look "finished" when equilibrium is reached, but actually the forward and reverse reactions continue to happen at the same rate. A reverse reaction is when the written reaction goes from right to left instead of the forward reaction which proceeds from left to right. This is why equilibrium is also referred to as "steady state".
It is possible to write an equilibrium expression for a reaction. This can be expressed by concentrations of the products divided by the concentration of the reactants with the coefficients of each equation acting as exponents. It is important to remember that only species in either the gas or aqueous phases are included in this expression because the concentrations for liquids and solids cannot change. For the reaction:
jA + kB > lC + mD
the equilibrium expression is:
Where:
K is the equilibrium constant
[A], [B], etc. are the molar concentrations of A, B, etc.
l, m, etc. are the coefficients of the balanced reaction
For every reaction at a specific temperature, there is only one value for K. A large value of K implies that there are more products than reactants and that the equilibrium lies to the right. A small K value implies there are more reactants than products and the reaction lies to the left. It is critical to remember that the only thing that changes K is changing temperature.
For reactions in the gas phase, equilibrium positions can also be expressed in terms of pressure. K_{p}, the equilibrium constant in terms of pressure, is related to K by the equation:
K_{p}=K(RT)^{Δn}
Where:
Δn is the sum of the coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants.
R is the gas law constant (see the gas laws page)
T is the temperature in kelvins
The reaction quotient, Q, is an expression which deals with initial values instead of the equilibrium value that K deals with. We compare Q and K to determine which direction the reaction will proceed to obtain equilibrium. If Q is greater than K, the system will shift to the left. If Q is less than K, the system will shift to the right. If Q is equal to K than the system is already at equilibrium so it will not shift in either direction.
To find the concentrations which characterize an equilibrium, it is best to proceed through the "start, change, equilibrium" process. This implies starting with the initial concentrations, determining the change, and using K to find the equilibrium concentrations.
Example "Start, change, equilibrium" problem:
C_{2}H_{6}(g) + Cl_{2}(g) <> C_{2}H_{5}Cl(s) + HCl(g)
If we had six (6) mols of C_{2}H_{6}(g) and six (6) mols of Cl_{2}(g) originally in a three (3) liter container at 10° C, determine the equilibrium concentration. K for this reaction at 10° C is 0.10.
Answer: 
we must first convert our values to molarity: 

6 mols
3 L 
= 2 M 
fill in the "start, change, equilibrium" chart: 
 
C_{2}H_{6} (g) 
+ 
Cl_{2} (g) 
→ 
C_{2}H_{2}Cl (s) 
+ 
HCl (g) 
Start: 
2 M 

2 M 

 

0 M 
Change: 
 x 

 x 

 

+ x 
Equilibrium: 
2  x 

2  x 

 

x 
now substitute what we know into the equilibrium constant expression and solve for x: 
K = 
[HCL]
[C_{2}H_{2}][Cl_{2}] 
= 0.10 = 
(x)
(2x)(2x) 
= 
x
4  4x + x^{2} 

.40  .40x + .1x^{2} = x 
.10x^{2}  1.4x + .4 = 0 
x = 
1.25 atm 
= 
.02573 mol
.07719 mol 

P_{Cl2} = .4167 atm 

Le Chatelier's principle allows us to predict the effects of changes in temperature, pressure, and concentration on a system at equilibrium. It states that if a system at equilibrium experiences a change, the system will shift its equilibrium to try to compensate for the change.
 changing the concentration (only with gases or aqueous solutions):
If you lower the concentration or remove some of a species, the system will shift to produce more of that species. On the other hand, if you increase the concentration or add some of a species, the system will shift to produce less of that species. For example, in the equation:
H_{2}+ I_{2} <>2HI
If we remove some of the H_{2}, the system will shift towards the left (the reverse reaction will happen the most) to produce more H_{2}.
 changing the volume/pressure (only gases):
Increasing the volume has the same effect as decreasing the pressure and vice versa so we are only going to talk about changing the pressure. When you increase the pressure, the system will shift so the least number of gas molecules are formed because the more gas molecules there are, the more collisions there are. These collisions and the presence of gas molecules are what cause the pressure to increase. Likewise, when you decrease the pressure, the system will shift so the highest number of gas molecules are produced. For example, in the equation:
N_{2} (g) + 3H_{2} (g) <> 2NH_{3} (g)
if the pressure is increased, the system will shift to the right because fewer gas molecules are produced in the forward reaction than in the reverse reaction.
 changing temperature:
For every reaction which can go forwards and backwards, one direction is
endothermic and the other is exothermic. A reaction is endothermic if it takes heat from its surroundings. On the other hand, a reaction is exothermic if it gives heat to the surroundings. If you increase the temperature, then the endothermic reaction will be favored because that will take in some of the excess heat. If you decrease the temperature, the exothermic reaction will be favored because it will produce the heat that was lost. For example, in the equation:
PCl_{3}(g) + Cl_{2}(g) <> PCl_{5}(g) + energy
if the temperature was increased, the system would shift to the left and the reverse reaction would happen more because that would use some of the extra energy.
 using a catalyst:
A catalyst increases the speed in which a reaction takes place, however it
never has any effect on the equilibrium.
Solubility is how much material dissolves in a solution, usually given in g/L. K_{sp} is the equilibrium expression related to ion concentration in a solution. For example, in the equation:
AgBr (s) > Ag^{+} (aq) + Br^{} (aq)
K_{sp}=[Ag^{+}][Br^{}]
Adding common ions, acids, or bases affects the solubility of a substance. Taking the above reaction as an example, if you add AgNO_{3}, the [Ag^{+}] would increase causing the system to shift to the left and therefore causing the solubility to decrease. If you add Pb(NO_{3})_{2}, the Pb^{2+} and the Br^{} would form a precipitate which would decrease the [Br^{}] and cause the system to shift to the right and therefore increase the solubility. Look at the following reaction:
BaCO_{3}(s) > Ba^{2+} (aq) + CO_{3}^{2}(aq)
If you add HNO_{3}, the weak acid HF would form which will lower
the F^{} concentration and cause the system to shift to the right and therefore increase the solubility. You should watch out for the formation of a precipitate or the formation of a weak acid when dealing with solubility reactions.
Example K_{sp} Problem:
0.100 L of 0.003 M Pb(NO_{3})_{2} is added to 0.400 L of 0.005 M Na_{2}SO_{4}. Will PbSO_{4} precipitate? K_{sp}=1.6x10^{8}
Practice Equilibrium Problem:
At a certain temperature, a 1.00L flask initially contained 0.298 mol PCl_{3}(g) and 0.00870 mol PCl_{5}(g). After reaching equilibrium, 0.00200 mol Cl_{2}(g) was found in the flask. PCl_{5} decomposes according to the equation:
PCl_{5}(g) <> PCl_{3}(g) + Cl_{2}(g)
Calculate the equilibrium concentrations of the three molecules and the value of K.
equilibrium solution.
Practice K_{sp} Problem:
K_{sp} for SrSO_{4}=7.6x10^{7}. K_{sp} for SrF_{2} =7.9x10^{10}.
a) What is the molar solubility of SrSO_{4} in pure water?
b) What is the molar solubility of SrF_{2} in pure water?
c) Sr(NO_{3})_{2}(aq) is added to 1.0 L of solution containing 0.020 mol F^{} and 0.10 mol SO_{4}^{2} with constant volume. 1: which salt precipitates first? 2: What is [Sr^{2+}] in solution when precipitate forms?
K_{sp} solution.
[Advanced Index]
[Gas Laws]
[Thermodynamics]
[Kinetics]
[Equilibria]
[Redox Reactions]
[Nuclear Chemistry]
