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Writing Equilibrium Constant Expressions Involving Solids and Liquids - Chemwiki
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ChemWiki: The Dynamic Chemistry E-textbook > Physical Chemistry > Equilibria > Chemical Equilibria > The Equilibrium Constant > Calculating
An Equilibrium Concentrations > Writing Equilibrium Constant Expressions Involving Solids and Liquids
Writing Equilibrium Constant Expressions Involving Solids
and Liquids
The equilibrium constant expression is the ratio of the concentrations of a reaction at equilibrium. Each equilibrium constant expression has a
constant value known as K, the equilibrium constant. When dealing with partial pressures, is used, whereas when dealing with concentrations
(molarity),
is employed as the equilibrium constant. Reactions containing pure solids and liquids results in heterogeneous reactions in which the
concentrations of the solids and liquids are not considered when writing out the equilibrium constant expressions.
Introduction
We are going to look at a general case with the equation:
No state symbols have been given, but they will be all (g), or all (l), or all (aq) if the reaction was between substances in solution in water. If you
allow this reaction to reach equilibrium and then measure the equilibrium concentrations of everything, you can combine these concentrations into an
expression known as an equilibrium constant.
The equilibrium constant always has the same value (provided you do not change the temperature), irrespective of the amounts of A, B, C and D you
started with. It is also unaffected by a change in pressure or whether or not you are using a catalyst.
Compare this with the chemical equation for the equilibrium. The convention is that the substances on the right-hand side of the equation are written
at the top of the expression, and those on the left-hand side at the bottom. The indices (the powers that you have to raise the concentrations to for example, squared or cubed or whatever) are just the numbers that appear in the equation.
There are two different types of equilibria (homogeneous and heterogeneous) separately, because the equilibrium constants are defined differently.
A homogeneous equilibrium has everything present in the same phase. The usual examples include reactions where everything is a gas, or
everything is present in the same solution.
A heterogeneous equilibrium has things present in more than one phase. The usual examples include reactions involving solids and gases, or
solids and liquids.
is the universal equilibrium constant for reactions. There is also and , however applies to reactions involving solutions, whereas applies to reactions involving gases.
The equilibrium constant expression is the ratio of the concentrations of the products over the
reactants
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Because
Writing Equilibrium Constant Expressions Involving Solids and Liquids - Chemwiki
has no units, we relate the concentrations for
as activities. For the following reaction at equilibrium:
The equilibrium constant expression would be:
Notice how each concentration of product or reactant is raised to the power of its coefficient. For example, the concentration of D is raised to the
power of 3 since it is 3D in the balanced reaction (eq. 1).
If the reaction involved partial pressures instead of concentrations, then the equilibrium constant, now , would follow the same formula except
with the pressures of the gases written in. In that case, is only applied to reactions involving gases. Recall the equation relating
and
:
where
R is the Ideal Gas Constant (0.0821 L atm mol-1 K-1),
T is the temperature in Kelvins, and
is the Sum of Coefficients of gaseous Products - Sum of Coefficients of gaseous Reactants.
The activities of pure solids and liquids are equal to 1
Pure solids and liquids are not included in the equilibrium constant expression. This is because they do not affect the reactant amount at equilibrium
in the reaction, so they are disregarded and kept at 1.
For the following reaction reaction at equilibrium:
Remember that the activity, a, of any solid or liquid in a reaction is equal to 1. So, the activities of A and C will be set equal to 1. aA(s)=1, aB(aq)= [B], aC(l)=1, aD(aq)=[D]
Homogeneous Equilibria
This is the more straightforward case and applies where everything in the equilibrium mixture is present as a gas, or everything is present in the same
solution. A good example of a gaseous homogeneous equilibrium is the conversion of sulfur dioxide to sulfur trioxide at the heart of the Contact
Process:
A commonly used liquid example is the esterification reaction between an organic acid and an alcohol:
Example 1: Esterification
A typical equation of the esterification reaction equilibrium might be:
There is only one molecule of everything shown in the equation. That means that all the powers in the equilibrium constant expression are "1". You
don't need to write those into the Kc expression.
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As long as you keep the temperature the same, whatever proportions of acid and alcohol you mix together, once equilibrium is reached,
has the same value. At room temperature, this value is approximately 4 for this reaction.
always
It is really important to write down the equilibrium reaction whenever you talk about an equilibrium constant. That is the only way that you can be
sure that you have got the expression the right way up - with the right-hand substances on the top and the left-hand ones at the bottom.
Example 2: Hydrolysis
The equilibrium in the hydrolysis of esters is the reverse of the last reaction:
The
expression is:
If you compare this with Example 1, you will see that all that has happened is that the expression has been reversed. Its value at room temperature
will be approximately 1/4 (0.25).
Example 3: The Contact Process
The equation for the Contact Process equilibrium is:
This time the expression will include some visible powers:
Although everything is present as a gas, you still measure concentrations in mol dm-3 (M). There is another equilibrium constant called is more frequently used for gases. You will find a link to that at the bottom of the page.
which
Example 4: Haber Process
The equation for Haber Process equilibrium is
and the
expression is:
Heterogeneous Equilibria
A heterogeneous equilibrium is more complicated than homogeneous examples discussed above and has species present in more than one phase.
Writing an expression for
for a heterogeneous equilibrium is similar to homogeneous reactions, with the important difference that you don't
include any term for a solid in the equilibrium expression. Taking another look at the two examples above, and adding a third one:
Example 5: Gas-Solid Equilibria
The equilibrium established if steam is in contact with red hot carbon. Here we have gases in contact with a solid.
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The equilibrium produced on heating carbon with steam
Everything is exactly the same as before in the equilibrium constant expression, except that you leave out the solid carbon.
Example 6: Solution-Solid Equilibria
If you shake copper with silver nitrate solution, you get this equilibrium involving solids and aqueous ions:
Both the copper on the left-hand side and the silver on the right are solids. Both are left out of the equilibrium constant expression.
Example 7: Gas-Solid Equilibria
The equilibrium produced on heating calcium carbonate is only established if the calcium carbonate is heated in a closed system, preventing the
carbon dioxide from escaping.
The only thing in this equilibrium which is not a solid is the carbon dioxide, which is all that is left in the equilibrium constant expression.
Notes to Remember
Pure solids and pure liquids have activities that are equal to 1.
H2O is one of the most common liquids dealt with in reactions. Remember to set its activity equal to 1 when it is a liquid in a reaction. However, if H2O is written as a gas, then its concentration must be considered.
Knowing is very helpful, for when it is compared with the Reaction Quotient , which can be used to determine the direction the reaction
will proceed, therefore knowing whether or not more products or reactants are being made. This is essential when solving problems involving
ICE tables.
References
1. Petrucci, Ralph H., et al. General Chemistry: Principles and Modern Applications Ninth Edition. Upper Saddle River, NJ: Prentice Hall, 2007. 2. Zumdahl, Steven; Zumdahl, Susan; DeCoste, Don, World of Chemistry, McDougal Littell, Boston, 2002
Outside Links
http://en.wikipedia.org/wiki/Chemical_equilibrium
http://www.wwnorton.com/college/Chemistry/gilbert/concepts/chapter15/ch15_1.htm
http://www.avogadro.co.uk/chemeqm/chemeqm.htm
Problems
There are all sorts of calculations you might be expected to do which are centered around equilibrium constants. You might be expected to calculate a
value for including its units (which vary from case to case). Alternatively you might have to calculate equilibrium concentrations from a given
value of and given starting concentrations.
1) What is and
for the following reaction at equilibrium:
2) In a chemical reaction, one part chemical A(s) forms one part chemical B(s) and 2 parts of chemical C(aq). Find the concentration of C present at
equilibrium. The reaction can be written as
with
3) At equilibrium, the following reaction takes place:
. The concentrations of A and C are 0.2M and 0.4 M
respectively. What is Kc for the reaction?
4) The equilibrium constant, , of the decomposition of
(solid limestone) into solid quicklime (CaO) and
at 1200 K is 1. What
must be the partial pressure of
? Find for the reaction.
5) Aqueous solution A is added to liquid B to form aqueous solution C and aqueous solution D. At equilibrium the solution is written as A(aq) + B(l) http://chemwiki.ucdavis.edu/Physical_Chemistry/Equilibria/Chemical_Equilibria/The_Equilibrium_Constant/Calculating_An_Equilibrium_Concentration_From_An…
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C(aq)+D(aq). The molarity, at equilibrium, of A is 0.0250 M and
. Find the concentrations of C and D.
Solutions
Solution 1: Remember to raise the activities to the power of the respective coefficients. The equilibrium constant expression should be written in
terms of activities
Since B and D are solids, their activities are equal to 1. So,
Substitute the partial pressures for the activities. Solution 2: Since A and B are both solids, their activities are equal to 1. So, the equilibrium constant expression would be,
Solution 3: The equilibrium constant expression can first be written in terms of the activities.
Since B is a liquid, then it's activity is equal to 1 and it may be left out of the equation.
Substitute the concentrations to obtain an expression for .
Substitute the given concentrations at equilibrium. Solution 4: The reaction can be written
with Since the limestone and quicklime are both solids, their activities can be set as 1. Therefore the partial pressure of CO2 should equal
atm= Partial Pressure of CO2
To find Kc of the reaction, we use the formula
. 1
Kp= Kc(RT)((Sum of Coefficients of gaseous Products)-(Sum of Coefficients of gaseous Reactants))
with
Kp= 1,
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T= 1200 K,
R= 0.08206 L atm mol-1 K-1
(RT) is raised to the power of the ((Sum of Coefficients of gaseous Products)-(Sum of Coefficients of gaseous Reactants)). There is 1 of each
molecule in the reaction:
. So, (RT) will be raised to the power of (1-0)= 1
Kp= Kc(RT)((Sum of Coefficients of gaseous Products)-(Sum of Coefficients of gaseous Reactants))
1= Kc((0.08206 L atm mol-1 K-1)(1200 K))1
= 0.010155171 or 1.02X10-2
Solution 5: Begin by writing out the equation accompanied with an ICE table.
I 0.0250 M -- -- C -x +x +x
E (0.0250-x) x x
Set up the equilibrium constant expression.
= 3.8x10-4= 3.8x10-4(0.0250-x)= x2
9.5x10-6-3.8X10-4x= x2
0= x2+3.8X10-4x-9.5X10-6
Use the quadratic formula: and with
a= 1
b= 3.8X10-4
c= -9.5X10-6
The appropriate solution comes out to be x=0.0028980577 or 2.9x10-3 = concentrations of C and D
Contributors
Kyle Catabay (UCD), Emily Pong (UCD)
Jim Clark (ChemGuide)
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