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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, \(K_p\) is used, whereas when dealing with concentrations (molarity), \(K_c\) 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:

\[ aA + bB \rightleftharpoons cC + dD \]

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 \(K_c\) 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.

\(K\) is the universal equilibrium constant for reactions. There is also \(K_c\) and \(K_p\), however \(K_c\) applies to reactions involving solutions, whereas \(K_p\) applies to reactions involving gases.

The equilibrium constant expression is the ratio of the concentrations of the products over the reactants

Because \(K\) has no units, we relate the concentrations for \(K\) as activities. For the following reaction at equilibrium:

\[ A\;(aq)+4B\;(aq) \rightleftharpoons 2C\;(aq)+3D\;(aq) \tag{1}\]

The equilibrium constant expression would be:

 \[ K= \dfrac{(a_C)^2(a_D)^3}{(a_A)(a_B)^4} \tag{2}\]

\[ K_c= \dfrac{[C]^2[D]^3}{[A][B]^4} \tag{3}\]

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 \(K_p\), would follow the same formula except with the pressures of the gases written in. In that case, \(K_p\) is only applied to reactions involving gases. Recall the equation relating \(K_p\) and \(K_c\):

\[ K_p = K_c (RT)^{Δn} \tag{4}\]

where

  • R is the Ideal Gas Constant (0.0821 L atm mol-1 K-1),
  • T is the temperature in Kelvins, and
  • \(\Delta n\) 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:

\[ 2A\;(s)+2B\;(aq) \rightleftharpoons C\;(l)+3D\;(aq) \]

\[ K= \dfrac{(a_C)^1(a_D)^3}{(a_A)^2(a_B)^2}\]

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]

\[ K= \dfrac{(1)^1(a_D)^3}{(1)^2(a_B)^2}\]

\[ K_c= \dfrac{[D]^3}{[B]^2}\]

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:

\[ 2SO_2 \; (g) + O_2 \; (g) \rightleftharpoons 2SO_3 \;(g) \]

A commonly used liquid example is the esterification reaction between an organic acid and an alcohol:

\[ CH_3COOH \; (l) + CH_3CH_2OH \; (l) \rightleftharpoons CH_3COOCH_2CH_3 \; (l) + H_2O \]

 
Example 1: Esterification

A typical equation of the esterification reaction equilibrium might be:

\[ CH_3COOH \; (l) + CH_3CH_2OH \; (l) \rightleftharpoons CH_3COOCH_2CH_3 \; (l) + H_2O \]

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.

As long as you keep the temperature the same, whatever proportions of acid and alcohol you mix together, once equilibrium is reached, \(K_c\) always has the same value. At room temperature, this value is approximately 4 for this reaction.

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:

\[ CH_3COOCH_2CH_3 \; (l) + H_2O  \rightleftharpoons CH_3COOH \; (l) + CH_3CH_2OH \; (l) \]

The \(K_c\) expression is:

\[ K_c = \dfrac{[CH_3COOH] \, [CH_3CH_2OH]}{[CH_3COOHCH_2CH_3][H_2O]}\]

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:

\[ 2SO_2 \; (g) + O_2 \; (g) \rightleftharpoons 2SO_3 \; (g) \]

This time the \(K_c\) expression will include some visible powers:

\[ K_c = \dfrac{[SO_3]^2}{[SO_2]^2[O_2]}\]

Although everything is present as a gas, you still measure concentrations in mol dm-3 (M). There is another equilibrium constant called \(K_p\) which is more frequently used for gases. You will find a link to that at the bottom of the page.

Example 4: Haber Process

The equation for Haber Process equilibrium is

\[ N_2 \; (g) + 3N_2 \; (g) \rightleftharpoons 2NH_3 \; (g) \]

and the \(K_c\) expression is:

\[ K_c = \dfrac{[NH_3]^2}{[N_2][H_2]^3}\]

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 \(K_c\) 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.

\[ H_2O \, (g) + C \, (s) \rightleftharpoons H_2\, (g) + CO \, (g)\]

The equilibrium produced on heating carbon with steam

\[ H_2O \, (g) + C \, (s) \rightleftharpoons H_2\, (g) + CO \, (g)\]

Everything is exactly the same as before in the equilibrium constant expression, except that you leave out the solid carbon.

\[ K_c = \dfrac{H_2][CO]}{[H_2O]}\]

Example 6: Solution-Solid Equilibria

If you shake copper with silver nitrate solution, you get this equilibrium involving solids and aqueous ions:

\[ Cu \, (s) + 2Ag^+ \, (aq) \rightleftharpoons Cu^{2+} \, (aq) + 2Ag \, (s)\]

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.

\[ K_c = \dfrac{[Cu^{2+}]}{[Ag^+]^2}\]

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.

\[ CaCO_3 \, (s) \rightleftharpoons CaO\, (s) + CO_2\,(g)\]

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.

\(K_c = [CO_2]\)

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 \(K\) is very helpful, for when it is compared with the Reaction Quotient \(Q\), 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

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 \(K_c\) including its units (which vary from case to case). Alternatively you might have to calculate equilibrium concentrations from a given value of \(K_c\) and given starting concentrations.

1) What is \(K\) and \(K_p\) for the following reaction at equilibrium: \( 3A\;(g)+B\;(s) \rightleftharpoons 2C\;(g)+4D\;(s) \)

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 \( A\; (s) \rightleftharpoons B\;(s) + 2C\;(aq)  \) with \(K_c = 4.5 \times 10^2\)

3) At equilibrium, the following reaction takes place: \(2A(aq)+ 3B(l) \rightleftharpoons C(aq)\). The concentrations of A and C are 0.2M and 0.4 M respectively.  What is Kc for the reaction?

4) The equilibrium constant, \(K_p\), of the decomposition of \(CaCO_3\) (solid limestone) into solid quicklime (CaO) and \(CO_2\, (g)\)  at 1200 K is 1. What must be the partial pressure of \(CO_2\, (g)\)? Find \(K_c\) 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) \(\rightleftharpoons\)C(aq)+D(aq). The molarity, at equilibrium, of A is 0.0250 M and \(K_c= 3.8 \times10^{-4}\).  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

\[K= \dfrac{(a_C)^2(a_D)^4}{(a_A)^3(a_B)^1}\]

Since B and D are solids, their activities are equal to 1.  So,

\[ K= \dfrac{(a_C)^2}{(a_A)^3}\]

Substitute the partial pressures for the activities. 

\[ K_p= \dfrac{(P_C)^2}{(P_A)^3}\]

 

Solution 2: Since A and B are both solids, their activities are equal to 1.  So, the equilibrium constant expression would be,

\[ K_c= \dfrac{(1)[C]^2}{(1)}\]

\[ 4.5 \times 10^2= [C]^2 \]

\[ \sqrt{4.5 \times 10^2}= [C]  \]

\[ [C] = 21.2 \]

Solution 3: The equilibrium constant expression can first be written in terms of the activities.

\[K= \dfrac{(a_C)}{(a_A)^2(a_B)^3}\]

Since B is a liquid, then it's activity is equal to 1 and it may be left out of the equation.

\[K= \dfrac{(a_C)}{(a_A)^2}\]

Substitute the concentrations to obtain an expression for \(K_c\).

\[K_c = \dfrac{[C]}{[A]^2}\]

Substitute the given concentrations at equilibrium. 

\[K_c= \dfrac{0.4}{0.2^2} = 10\]

 

Solution 4: The reaction can be written \( CaCO_3\;(s) \rightleftharpoons CaO\; (s) + CO_2\; (g) \) with  \(K_p= 1\)

 \[ K= \dfrac{(a_{CaO})(a_{CO_2})}{a_{CaCO_3}}\]

Since the limestone and quicklime are both solids, their activities can be set as 1. Therefore the partial pressure of CO2 should equal \(K_p\). 1 atm= Partial Pressure of CO2

To find Kc of the reaction, we use the formula

Kp= Kc(RT)((Sum of Coefficients of gaseous Products)-(Sum of Coefficients of gaseous Reactants))

with

  • Kp= 1,
  • 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: \( CaCO_3\, (s) \rightleftharpoons CaO\, (s) + CO_2\,(g)\).  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

\[ K_c=   \dfrac{1}{(0.08206)(1200)}\]

   = 0.010155171 or 1.02X10-2

 

Solution 5: Begin by writing out the equation accompanied with an ICE table.

\[ A\; (aq) + B\;(l) \rightleftharpoons C\;(aq)+D\;(aq)\]

I     0.0250 M               --       --                               

C         -x                   +x      +x

E   (0.0250-x)              x         x

Set up the equilibrium constant expression.

\(K_c\) = 3.8x10-4=  \(\dfrac{(x)(x)}{(0.0250-x)}\)

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: 

\(\dfrac{-b+ \sqrt{b^2-4ac}}{2a}\) and  \(\dfrac{-b-\sqrt{b^2-4ac}}{2a}\)

with

  • a= 1
  • b= 3.8X10-4
  • c= -9.5X10-6

\(\dfrac{-3.8x10^{-4} \pm \sqrt{(3.8x10^{-4})^2-4(1)(-9.5x10^{-6})}}{2} \)

\[ \dfrac{-3.8X10^{-4} \pm \sqrt{3.81444x10^{-5}}}{2}\]

\[ \dfrac{-3.8x10^{-4} \pm (0.0061761153)}{2}\]

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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Last Modified
14:22, 18 Nov 2013

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