AP Chemistry - Slightly Soluble Substances
Equilibria In Saturated Solutions of Slightly Soluble Substances
This unit does not deal with substances that dissolve readily in water or other solvents. Anything that dissolves at least 1 g/100 mL of solvent is considered to be soluble. This unit deals with substances that are slightly soluble to those that are considered to be insoluble (unable to dissolve easily in a solvent). There are also some questions related to precipitates (slightly soluble substances) that are of interest to us.
Some of these questions are:
1) How can the approximate solubility of a precipitate be calculated?
2) How can we predict whether or not a precipitate forms when two solutions are mixed?
3) Which reagents or methods can be used to dissolve precipitates?
The Solubility Product and Precipitate Formation
Consider a saturated solution of silver chloride.
The equilibrium can be represented by:   AgCl(s)  <=====> Ag+(aq) + Cl-(aq)
The equilibrium law expression is:   Ksp = [Ag+][Cl-]
The Ksp equation resembles an equilibrium expression of a heterogeneous equation. The Keq value is changed to a Ksp because we are now dealing with solubilities and precipitates. The 'sp' means solubility product constant and you can see that the [Ag+] and [Cl-] are multiplied together to create a product.


The Ksp equation does not contain a term concerning the solid AgCl(s). Once a solution has reached saturation, the addition of more solid will not change the [] of the ions already in solution.

The Ksp value for AgCl is 1.8 x 10-10. This shows that the [Ag+] and [Cl-] are very low in a saturated solution of AgCl. Equilibrium is established before a significant amount of AgCl dissolves.

Another way of saying this would be that the Ag+ ions and the Cl- ions can exist together at Equilibrium in the same solution only when their concentrations are low enough so that the product of their concentrations does not exceed the Ksp value.

eg. Calculate the [Ag+] and [Cl-] in a saturated solution of the salt at 25oC. What is the approximate solubility of AgCl in mol/L at this temperature?
The ions Ag+ and Cl- react together on a 1:1 basis therefore any AgCl that does dissolve will product 'x' amount of each of the ions.
AgCl(s) <======>  Ag+(aq) + Cl-(aq)
                                   x              x
Ksp = [Ag+][Cl-]

1.8 x 10-10 = (x)(x)

1.8 x 10-10 = x2

x = 1.3 x 10-5 mol/L

Therefore 1.3 x 10-5 moles of AgCl dissolves in 1 L of H2O. The equilibrium []'s of Ag+ and Cl- are each 1.3 x 10-5 mol/L. Since we are dealing with concentrations we can automatically substitute in the units of concentration, moles/Litre.
Follow-up Problems
Determine the solubility of AgI, Ag2CrO4, and Zn2[Fe(CN)6] in
a) moles/L and b) g/L at 25oC. You will have to refer to the Ksp tables in your databook.

Predicting Precipitate Formation
The Ksp value is the product of the upper limit of the product of the concentration of the soluble ions. When the product of the concentration of the ions exceed the value of Ksp they cannot exist in equilibrium anymore. They will form a precipitate in order to reduce the concentrations of the ions in solution back to the equilibrium value.
To determine whether the Ksp is exceeded, just substitute the ion concentrations into an expression similar to the Ksp expression and determine an experimental Trial Ion Product. (T.I.P.)
If the T.I.P. exceeds the Ksp then a precipitate forms. If the T.I.P. is larger than the values of Ksp then the ions are above the saturation line and the ions are in a super-saturated situation. The ions will precipitate until their individual concentrations multiplied together equal the value of Ksp.
If the T.I.P. does not exceed the value of Ksp then the ions are in an unsaturated situation, which means there aren't enough ions dissolved to form a precipitate.
If the T.I.P. equals the value of Ksp then the solution is saturated. The solution can hold no more dissolved ions without a precipitate forming.
Does a ppt. of AgCl form when 1 mL of 0.1 mol/L AgNO3 is added to a beaker containing 1 L of tap water with a Cl- ion concentration of 1.0 x 10-5 mol/L?

[Cl-] = 1.0 x 10-5 mol/L

[Ag+] calculation. Start with the given info, 1 mL of 0.1 mol/L

Set up a ratio that states, if there are 0.1 moles of Ag+ in 1000 mL how many moles will be in 1 mL?
0.1 mol  x
1000mL   1 mL

x = 0.0001 moles of Ag+ are present.

(This is how many moles there are, this is not the concentration)

The [Ag+] will be this many moles of Ag+ dissolved in the volume of solution in this question. Therefore

[Ag+] =      0.0001 mole  = 0.0001 mole = 1.0 x 10-4 mol/L
             1000 mL + 1 mL       1.001 L

T.I.P. = [Ag+][Cl-]
          = (1.0 x 10-4)(1.0 x 10-5)
          = 1.0 x 10-9

The Ksp for AgCl is 1.8 x 10-10 therefore T.I.P > Ksp so a ppt. will form.

Follow-up Problem
Does a ppt of AgI form when 10 mL of 0.1 mol/L AgNO3 gets added to 90 mL of a solution containing 1.0 x 10-10 mol/L of KI?

What is the highest concentration of I- ions that can exist in equilibrium with 0.01 mol/L Ag+ ions? How many grams of KI are present in 1 litre of a solution with the I- ion concentration found above?
AgI(s) <====> Ag+(aq) + I-(aq)

Ksp = [Ag+][I-]

8.3 x 10-17 = (1.0 x 10-2)[I-]

There [I-] = 8.3 x 10-17 = 8.3 x 10-15 mol/L
                   1.0 x 10-2

If [I-] exceeds 8.3 x 10-15 mol/L then a ppt. will form.

K+ + I- ------> KI

If we have 8.3 x 10-15 moles of I- we also have 8.3 x 10-15 moles of K+ and hence we also have 8.3 x 10-15 moles of KI.

g = moles x molecular mass
   = 8.3 x 10-15 mole x 166.04 g/mole
   = 1.38 x 10-12 grams

Follow up Problems
1. What is the highest concentration of Cl- ions that can exist in equilibrium with 0.001 mol/L Tl+ ions at 25oC?

2. Calculate the [CrO42-] required to begin precipitation of these metal ions from solutions containing 1 g of the metal ions per L of solution.

a) Ag+ b) Ca2+ c) Pb2+

Go to the Solubility & Ksp Worksheet

The Common Ion Effect and Solubility
Le Châtelier's Principle indicates that it should be possible to vary the solubility of a solid by varying the concentration of the ions at equilibrium.
i.e. To decrease the [Ag+] in a solution of AgBr, we can add excess Br- ions in the form of a soluble salt such as NaBr. This excess puts a stress on the equilibrium system which adjusts itself by shifting left. As it consumes Br- ions and forms AgBr(s).
The Ksp is a constant. If one ion gets larger then the other must get smaller.
Ksp = [Ag+][Br-]
Ksp = [Ag+][Br-]
Each Br- ion that reacts ties up a Ag+ ion which removes it from the solution.

Ksp = [Ag+][Br-] = 4.8 x 10-13

The value of Ksp for KBr is so small that even a little Br- from the NaBr will cause a ppt. The overall effect is that the solubility of AgBr is decreased. AgBr is less soluble in a solution of NaBr then it would be in pure water.
Compare the molar solubility of AgBr in pure water and in a 0.10 M NaBr solution.

AgBr(s) <====> Ag+(aq) + Br-(aq)

In pure water

Let 'x' be the [Ag+] and the [Br-] in a saturated solution of AgBr.

Therefore   Ksp = [Ag+][Br-]
                         = (x) (x)
        4.8 x 10-13 = x2

                         = 6.9 x 10-7 mol/L

Therefore in a solution made with pure water the [Ag+] = [Br-] = 6.9 x 10-7 mol/L.

In a 0.10 M NaBr soln.

In a 0.10 mol/L solution, the [Br-] = 0.10 mol/Lalready

The small quantity of Br- furnished from the dissolving of the AgBr is insignificant since in pure water it was only 6.9 x 10-7 M. Since dissolution is repressed by the presence of the common Br- ion, it will be even less. Therefore we can forget about it.

Since [Br-] = x + 0.10 mol/L
                 = 6.9 x 10-7 mol/L + 0.10 mol/L
                 = 0.10000069 mol/L

Therefore Ksp = [Ag+][Br-]
     4.8 x 10-13 = (x)(0.10 mol/L)
                    x = 4.9 x 10-12 mol/L

Since 0.10 + 4.8 x 10-12 0.10 our assumption is valid. AgBr is about 100000 times more soluble in pure water than it would be in a 0.10 M NaBr solution.

Follow-Up Problem
Compare the molar solubility of PbI2 in pure water and in 0.10 NaI.
Two or More Precipitate Ions in a Single Solution
When a precipitating agent is slowly added to a solution that contains more ions than precipitates, then the compound whose solubility product is exceeded first will precipitate first. If I-1 ion solution is slowly added to a solution containing 0.010 mol/L Ag+1  and 0.010 mol/L Pb2+, AgI precipitates as soon as the [I-1] exceeds:
            Ksp = [Ag+][I-]
8.3 x 10-17 = (1.0 x 10-2)[I-]
             [I-] = 8.3 x 10-15 mol/L
The lead ions do not precipitate until [I-] exceeds

Ksp = [Pb2+][I-]2 since PbI2 <=====> Pb2+ + 2 I-

[I-]2    =   Ksp   =   7.9 x 10-9
              [Pb2+]     1.0 x 10-2

[I-]2 = 7.9 x 10-7


Follow-Up Problem
A solution contains 0.01 mol/L Ag+1 ions and 0.01 mol/L Sr2+ ions. Which ion precipitates first, when K2CrO4 is slowly added to the mixture? 
Dissolving Precipitates  OR 
How to Get Rid of Those Really Nasty Stains
Applying Le Châtelier's Principle to a solubility equilibrium indicated that a slightly soluble solid can be dissolved by decreasing the [ ] of one or more of the ions in equilibrium with the solid.
Consider a saturated solution of CaCO3 solution.
CaCO3(s) <====>  Ca2+(aq) + CO32-(aq)
The dissolution of CaCO3 can be enhanced by adding any reagent that ties up and decreases the concentration of either Ca2+ ions or CO32- ions below the equilibrium value in the saturated solution.
i.e. take some of the dissolved ions away and make room for more of the solid to dissolve.
Diluting with water does decrease the [] of the dissolved ions but not enough to make any appreciable difference on solids with very low Ksp values.
Besides the dilution method, there are 5 ways which reduce the [] of an ion in equilibrium with a slightly soluble solid.
1. Formation of a weakly dissociated species.
2. Formation of a volatile substance.
3. Formation of a precipitate which is even less soluble than that of the original substance.
4. Formation of a complex ion.
5. An oxidation or reduction by a redox agent.
There is some overlap between these five methods and we will look at only the first 4. We'll save the fifth method for the unit on electrochemistry.
Formation of a Weakly Dissociated Species and/or Gas
We can cause reactions to occur if we create a product that is either a gas and or a product that we consider to be weakly dissociated. The acids that we consider strong like HCl and, HNO3 are considered to be strong because they release tremendous amounts of H+ ions and their anions are so very weak. Most reactions that you have worked with in Grade 11 where reactions that had water or a gas as a product. In a few cases in the double displacement reactions or single displacement reactions had products that were weaker than the reactants. This means that once they where formed they did not have a strong tendency to breakdown and reform the reactants.
CaCO3 dissolves in HCl and forms a gas and a soluble product.
i.e. CaCO3 + 2 HCl -----> CaCl2 + CO2 + H2O
Notice the single line. This reaction is not reversible since we allow the CO2 to escape. (an open system)
If we did this under pressure in a closed system then it would set up it's own unique equilibrium.
The H3O+ ions reduce the [CO32-] below its equilibrium value by forming CO2 (a gas) and H2O (a weakly dissociated species).
Many slightly soluble salts of weak acids may be dissolved by strong acids.
It is often necessary to eliminate all the spectator reactants from a reaction and show only the major pieces that actually react. We can write what we call a net ionic equation that shows only the bare bones parts of the working reaction. For example:
CaCO3 + 2 HCl -----> CaCl2 + CO2 + H2O (full equation)
CO32- + 2 H+ -----> CO2 + H2O (net ionic equation)
You'll note that the Ca2+ ion and 2 Cl- ions are what we call spectator ions. They appear as themselves on both sides of the equation and they play no real part and therefore they can be eliminated. The net ionic equation only shows the ions and molecules that are really important in a particular equation.
To write a net ionic equation for the reaction(or any reaction), of HCl with CaCO3 follow the rules listed below
1. Write the formula of any strong acid or strong base in dissociated (ionic) form.

2. Write the formula of any weak acids, weak bases, water, gases, and other weakly dissociated species in combined (molecular) form.

3. Write the formulas of soluble ionic salts in dissociated (ionic) form. Refer to the solubility chart to help identify soluble and slightly soluble substances.

4. Write the formulas of slightly soluble ionic salts in undissociated (molecular) form and identify then as slightly soluble by writing '(s)' after the formula.
5. Identify and write the formulas of the products.
6. Delete from the equation any ions that appear on both sides.
7. Use the inspection method to balance. The atoms on the reactant side must be equal to the atoms on the product side. The net charge on the reactant side must be equal to the net charge on the product side as well.
eg. CaCO3 + HCl
HCl + H2O -----> written as H3O+(aq) + Cl-(aq)
products CaCl2 -----> written as Ca2+(aq) + 2 Cl-(aq)
H2CO3 -----> a weak unstable acid written as H2O(l) + CO2(g)
CaCO3(s) + 2 H3O+(aq) 2 Cl-(aq) ---->  Ca2+(aq) + CO2(aq) + 3 H2O(l) + 2 Cl-(aq)
The net ionic equation is:
CaCO3(s) + 2 H3O+(aq) ------> Ca2+(aq) + CO2(g) + 3 H2O(l) 
Formation of Weak Acids
The reaction of a strong acid with an ionic salt containing the negative ion of a very weak acid produces the weak acid and goes essentially to completion. The weaker the acid formed the more nearly complete is the reaction. You have a table of  acid strengths in your databook.
Dissolution of a Precipitate by Complex Ion Formation
A number of slightly soluble salts do not dissolve readily in acids.  eg. AgCl will not dissolve in nitric acid.
In order to dissolve AgCl, the equilibrium of  AgCl(s)  <=====>  Ag+(aq) + Cl-(aq)
must be shifted to the right.
Two things can do this:
NH3 or CN- form co-ordinate bonds with the Ag+ ions and remove them from the solution. They create complex ions that are even more insoluble then the original AgCl.
The ammonia solution has Ag(NH3)2+ in it, and the cyanide solution has the Ag(CN)2- ion in it.     These two polyatomic ions are weakly dissociated.
The part of the ion which attaches by a co-ordinate bond is called a ligand.
eg. AgCl(s) + 2 NH3(aq) <======> Ag(NH3)2+(aq) + Cl-(aq)
AgCl(s) + 2 CN-(aq) <=======>  Ag(CN)2- + Cl-(aq)
A special nomenclature has been developed for complex ions.
The name of the Ag(NH3)2+ ion is the diamminesilver(I) ion.
The name of the Ag(CN)2- ion is the dicyanoargenate(I) ion.
The rules for naming complex ions are as follows:
1. Use the mono, di, tri system for naming the ligand attachments.
The other common ligands are:
Ag(NH3)2+ has two ammine ligands therefore its diammine
Ag(CN)2- has two cyano ligands therefore its dicyano
2. The charge on the overall molecule is the sum of the metal's charge and the ligands charge.
Ag(NH3)2+ The silver ion has a +1 charge and the ammine has no charge therefore +1 +0 +0 = +1 which is the overall charge on the complex.
Ag(CN)2- The silver ion still has a +1 charge but now two CN- ions also come into play. +1 -1 -1 = -1 as an overall charge on the complex.
3. The overall charge of the metallic atom help determine the name of the complex. Complexes with a positive or neutral charge get the normal metallic name and the charge in brackets.
Ag(NH3)2+ The overall charge is +1 therefore the metallic portion is named silver(I) because the silver has a charge of +1.
A negative complex has slightly different rules. If the metal is named after an old element use the older name, such as argentum for silver, hydroargentum for mercury, ferrum for iron, etc. Drop the suffix and add the suffix 'ate' which is the signifier of a negative charge. The value of the charge is given in brackets. If the metals name does not stem from an older classical name then use its modern name along with its 'ate' suffix.
Ag(CN)2- The overall charge is -1 and the metal is silver whose symbol comes from the old name of argentum, therefore argentate(I) is its name. The 'ate' signifies that the complex is -ve in charge but the (I) is the value of the positive charge on the silver atom.
Tl(NO3)63- This would be hexanitrothallate(III)
Six nitro groups would be hexanitro
Thallium has no old classical name therefore it must remain thallium. Drop the suffix and adding 'ate' gives thallate.
What is the charge? 6 nitro's donate -6 to the complex. The overall charge is -3, therefore the thallium's charge is +3. This is indicated by the thallate(III) portion of the name.
Examples of complex ions:
  Ligand Formula Name
Ag+ NH3 Ag(NH3)2+ diamminesilver(I)
Ag+ CN-   Ag(CN)2  dicyanoargenate(I)
Ag+  Cl-   Ag(Cl)2 dichloroargenate(I)
Cu2+   NH3 Cu(NH3)42+  tetramminecopper(II)
Cu+2 CN-  Cu(CN)42- tetramminecuprate(II)
Cd2+  NH3 Cd(NH3)42+  tetramminecadmium(II)
Cd2+   CN- Cd(CN)42-  tetracyanocadimate(II)
Co3+ NH3   Co(NH3)63+ hexamminecobalt(III)
Al3+ OH- Al(OH)4-  tetrahydroxoaluminate(III)
Zn2+ NH3 Zn(NH3)42+ tetramminezinc(II)
Zn2+  OH-  Zn(OH)42- tetrahydroxozincate(II)
Follow-Up Problems
Write net ionic equation using complex ions, for the reactions below. In each case name the complex ion formed.

a) copper(II) hydroxide with excess aqueous ammonia.

b) nickel(II) hydroxide with excess sodium cyanide.

c) silver bromide with excess sodium thiosulphate. The complex formed is Ag(S2O3)23-

d) zinc hydroxide with excess sodium hydroxide

Complex Ion Equilibria
In an aqueous solution, a complex ion exists in equilibrium with it's components, the central metallic ion and the ligands.
Complex ions are formed when Lewis bases (ligands) attach themselves to metal ions in solution. An example is the blue Cu(NH3)42+ ion, which is formed by attaching four ammonia molecules to a Cu2+ ion. Complexes of this type can be quite complicated because as each ligand attaches it forms it's own unique equilibrium. When the equilibrium concentration is large we can simplify the problem by realizing that the equilibrium is just:
Cu2+(aq) + 4 NH3(aq) <======>  Cu(NH3)42+(aq)
The K value is given a new name, the Kform or the K of Formation for the complex.
The Kform is often called a stability constant. The larger it's magnitude the more stable is the complex. Most of the metal ions that form complexes are those of the transition elements. In fact, the tendency to form complex ions is often listed as one of the general properties of the transition metals.
Instability Constants
In many chemical references the stabilities of complex ions are indicated as inverses. The inverses of formation constants are called instability constants, Kinst.
Kinst =                                Kinst is called the instability constant because
            Kform                                the larger it's value the more unstable the complex is.
Effect of Complex Ion Formation on Solubility
The Ksp of AgBr is 5.0 x 10-13. Suppose we have a saturated solution of this salt, with some AgBr(s) left undissolved at the bottom of the beaker. We next add some aqueous ammonia to the system. Its molecules are strong ligands for silver ions, so they begin to form Ag(NH3)2+ ions from the trace amount of Ag+(aq) initially present in solution.
Ag+(aq) + 2 NH3(aq) <======> Ag(NH3)2+(aq)
The ammonia represents upsets the equilibrium present in the saturated solution of
AgBr(s) <======> Ag+(aq) + Br-(aq)
By pulling Ag+ ions out of this equilibrium, the equilibrium must shift to replace them as best it can. i.e. more AgBr(s) must go into solution. The solubility of a slightly soluble salt increases when one of its ions can be changed to a soluble complex ion.
Adding the ammonia introduces another equilibrium and we can sum the two equilibrium present.
Original Equilibrium
AgBr(s) <=====>  Ag+(aq) + Br-(aq)      Ksp = [Ag+][Br-]
New Equilibrium
Ag+(aq) + 2 NH3(aq) <===>  Ag(NH3)2+(aq) 

Sum of the equilibriums

AgBr(s) + 2 NH3(aq) <======> Ag(NH3)2+(aq) + Br-(aq)
The equilibrium constant for this new overall equilibrium is:
The value of Kc can be found by multiplying Kform * Ksp
i.e. Kc = Kform * Ksp
The [Ag+] cancel out. If we know the values of Kform and Ksp we can calculate the new value of Kc.
Calculating the Solubility of a Slightly Soluble Salt in the Presence of a Ligand
How many moles of AgBr can dissolve in 1.0 L of 1.0 M NH3?
AgBr(s) + 2 NH3(aq) <=====>  Ag(NH3)2+(aq) + Br-(aq)
Ksp = 5.0 x 10-13 Kform = 1.6 x 107 Kc = 5.0 x 10-13 * 1.6 x 107

      = 8.0 x 10-6

                    AgBr(s) + 2 NH3(aq) <======> Ag(NH3)2+(aq) + Br-(aq)
initial [ ]'s                         1.0                                  0.0                    0.0

Changes in
[ ] caused
by NH3                         -2x                                   +x                     +x

Becomes                    1.0-2x                                 +x                     +x 1

8.0 x 10-6 = (x)(x)
0.0028 = __x__

x = 0.0028 - 0.0056x

1.0056x = 0.0028

x = 2.78 x 10-3

So 2.8 x 10-3 mol of AgBr dissolves in 1.0 L of 1.0 M NH3. this is not very much but it is approximately 4000 times more than when AgBr dissolves in pure water.
Go to the Solubility Complex Ion Worksheet

Determination of Ksp Values Using the Nernst Equation 
If Keq can be determined then it is reasonable to assume that other equilibrium constants can be determined by using half-cell reaction.  By measuring the voltage of an appropriate cell we can calculate the Eo value.
The electrochemical cell for determining the Ksp for AgCl is below.
In this cell, the anode consists of a piece of metallic silver immersed in a 1 M Cl-1 ion solution with excess AgCl(s).  The cathode is a standard Ag(s) |  Ag+1 half cell. The schematic is:

                            Ag |  AgCl(s) | Cl-1 (1 M) || Ag+1 (1 M) |  Ag(s)

The anode half cell potential at SATP is -0.22 V for the reaction: 
Ag(s)  + Cl-1  --->  AgCl(s) +  e-1

The standard state cathode potential for silver is +0.80 V. 
Ag+1  + e-1 ---> Ag(s)   Eo = +0.80 V

Therefore -0.22 V + 0.80 V= 0.58 V is the overall reaction potential for 
Ag+1  + Cl-1 ---> AgCl(s)

At equilibrium Ecell = 0
Therefore     0 = 0.58 V - 0.059/1 log (1 / [Ag+1][Cl-1])
                    -0.58 V = 0.059 log [Ag+1][Cl-1]
                    -0.58 V = 0.059 log Ksp
                    log Ksp = -9.8
                    Ksp = 10-9.8
                           = 1.6 X 10-10

Calculating Equilibrium Constants From Thermodynamic Data
The position of an equilibrium in a chemical reaction is determined by the value of ΔGo  and the large the value of Kc or Kp for a reaction, the farther the reaction will proceed to completion by the time equilibrium is reached.  There is obviously as relationship between ΔGo and the equilibrium constant.  This relationship is simple, but it does involve logarithms. 

The equations are:

  ΔG = -RT ln K
and ΔGo = -2.303RT log K

These equations are valuable because they allow us to relate ΔGo to K (sometimes called the thermodynamic equilibrium constant) quantitatively.  There is one complicating factor.  For reaction involving gases, the K that is calculated is Kp with partial pressures expressing in atmospheres.  For reactions involving liquid solution, the calculated K is Kc.

Calculating ΔGo From K
e.x. 1   The brownish haze associated with air pollution is caused by nitrogen dioxide, NO2, a red-brown gas.  Nitric oxide, NO, is oxidized to NO2 by oxygen.

2 NO(g)  + O2(g)  <====>  2 NO2(g)

The value of Kp for this reaction is 1.7 X 1012 at 25.0oC.  What is ΔGo for the reaction expressed in joules and kilojoules?

Either version of the equation can be used.   R is 8.314 J/mol K (the Ideal Gas Law Constant using Joules)

      ΔGo = -RT ln Kp
               = -8.314 J/mol K  X  298.2 K ln (1.7 X 1012)
               = -2479.25 J  X  28.16 
               = -6.982 X 104 J
               = -69.82 kJ

Calculating K from ΔGo
e.x. 2  Sulfur dioxide reacts with oxygen when it passes over the catalyst in automobile catalytic converters.  The product is SO3.

2 SO2(g)  +  O2(g)  <===>  2 SO3(g)

For this reaction, ΔGo = -139.75 kJ at 25oC What is the value of Kp?

Solution:  The ΔGo = -RT ln Kp equation is needed here.

R = 8.314 J/mol K;  T = 298.2 o

                           -139.75 kJ = -(8.314 J/mol K)(298.2) ln Kp

ln Kp =           -139750 J               
             -(8.314 J/mol K)(298.2 K)

ln Kp = 56.4 

Kp = e56.4 
      = 3 X 1024 

Calculating Equilibrium Constants at Temperatures Other Than 25oC
Temperature is the only thing that will change the value of the K constant for a reaction.   The relative proportions of reactants and products at equilibrium are determined by ΔGo and the above equation show that we can calculate the value of the equilibrium constant at 25oC from ΔGo.    As the temperature moves away from 25oC, the position of the equilibrium also changes because the value of ΔG changes.  To indicate ΔG at other than 25oC we use ΔG'.   As it happens we use exactly the same equation. 

ΔG' = -RT ln K

e.x.   The decomposition of nitrous oxide, N2O, has Kp = 1.8 X 1036 at 25oC.  The equation is:

2 N2O(g) <===>  2 N2(g) + O2(g)

For this reaction, ΔHo = -163 kJ and ΔSo = +148 J/k. What is the approximate value for Kp for this reaction at 40oC?

First Calculate the value of ΔG'  
     ΔG' = ΔHo - T ΔSo
              = -163,000 J - (313 K)(+148 J/K)
              = -163,000 J - 46,300 J
              = -209,000 J

Secondly, use this value of ΔG' to find the value of Kp at 40oC

ΔG' = -RT ln K
-209 000  = -(8.314)(313) ln Kp
ln Kp = 80.3
Kp = e80.3
      = 7 X 1034

Notice that at this temperature, N2O is actually slightly more stable than at 25oC, as reflected in the slightly smaller value for the equilibrium constant for its decomposition.

Go to the Solubility Unit Review Worksheet