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Equilibrium Constant Expressions

>> Parts of this equation/concept include:

A - Homogenous Reactions
B - Heterogenous Reactions
C - KP Expressions
D - Converting Between KC and KP, Equation 15.17
E - Effect of Reaction Variations on K
F - Combining Equilibrium Expressions

 

A. Homogenous Reactions

The equilibrium constant expression uses the balanced reaction as written. Changing the way the reaction is written will change the value of the equilibrium constant, so the reaction should not be changed to look more appealing.

Since an equilibrium can be approached from either direction, it is convenient to define reactants as the substances to the left of the arrow and products as the substances to the right. The concentration of each substance is raised to its stoichiometric coefficient. The product of these values for the products is divided by the product of these values for the reactants. Hint: This process is very hard to explain but very obvious from the examples.

The units on the equilibrium constant will be concentration units raised to some power. The power will vary with the equilibrium constant expression. It is customary to define KC as unitless if the concentration is expressed in units of molarity (M).

Gases and solutions may have concentrations that are expressed in the equilibrium constant expression.

>> Example 1

What is the equilibrium constant expression (as KC) for the following gaseous reactions?

  1. C2H4(g) + H2(g) C2H6(g)
  2. Xe + 3 F2 XeF6
  3. 2 N2 + O2 2 N2O

Solution:

  1. KC  = 
    [C2H6]
    [C2H4][H2]

    The square brackets represent the concentration of the substance within the brackets. Since the stoichiometric coefficient on each product and reactant is an understood 1, it is the exponent on the concentration of each substance.

  2. KC  = 
    [XeF6]
    [Xe][F2]3

    The stoichiometric coefficient becomes the exponent. Remember that the stoichiometric coefficient is not part of the formula, so it should not be included within the brackets.

  3. KC  = 
    [N2O]2
    [N2]2[O2]

    The order does not matter in multiplication; therefore an equally correct answer is

    KC  = 
    [N2O]2
    [O2][N2]2

 

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B. Heterogenous Reactions

In heterogeneous equilibria, the products and reactants are not all in the same physical state. The concentrations of solids and liquids are constant, and are included in the value of K rather than in the concentration part of the expression. Another (and sometimes more mathematically useful) way of looking at it is that [solid] = [liquid] = 1.

A solvent is defined as the substance in greater concentration. Normally, the concentration of solvent is significantly higher than that of solute. Consequently, if the solvent participates in the reaction, very little is used up (or produced) and its concentration is, for all practical purposes, unchanged. Therefore, like solids and liquids, the concentration of the solvent is included in the equilibrium constant expression or defined as 1.

>> Example 2

What is the equilibrium constant expression (as KC) for the following reactions?

  1. CH3NH2(aq) + H2O(l) CH3NH3+(aq) + OH–(aq)
  2. H+(aq) + OH–(aq) H2O(l)
  3. Fe2S3(s) 2 Fe3+(aq) + 3 S2–(aq)
  4. C(s) + 2 F2(g) CF4(g)
  5. Hg(l) + H2S(g) HgS(s) + H2(g)

Solution:

  1. "(aq)" means the solute is dissolved in the solvent water. Since water is the solvent, it is not included in the equilibrium constant expression

    KC  = 
    [CH3NH3+][OH–]
    [CH3NH2]

  2. The reaction occurs in an aqueous solution, so the concentration of water is included in the constant. However, having no value on the top of a fraction is not mathematically appropriate, so a 1 is used.

    KC  = 
    1
    [H+][OH–]

  3. In this expression, the solid on the bottom is included in the value of K instead of the concentration part of the expression. Since it is a more aesthetic expression to not use a 1 on the bottom, the expression is

    KC = [Fe3+]2 [S2–]3

  4. It is equally appropriate to use the concentration of gases as the concentration of aqueous solutions, so the equilibrium constant expression is

    KC  = 
    [CF4]
    [F2]2

  5. Mercury metal is a pure liquid, so the equilibrium constant expression is

    KC  = 
    [H2]
    [H2S]

 

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C. KP Expressions

KP expressions are written in the same way as KC expressions except that partial pressures are used instead of the concentration brackets. If the units used for pressure are atmosphere, KP is usually defined as unitless.

>> Example 3

What is the equilibrium constant expression (as KP) for the following reactions?

  1. C2H4(g) + H2(g) C2H6(g)
  2. Xe(g) + 3 F2(g) XeF6(g)
  3. C(s) + 2 F2(g) CF4(g)
  4. Hg(l) + H2S(g) HgS(s) + H2(g)

Solution:

  1. KP  = 
    PC2H6
    PC2H4PH2

  2. KP  = 
    PXeF6
    PXePF23

  3. KP  = 
    PCF4
    PF22

    Like KC, solids, liquids, and solvents are not included in the K expression.

  4. KP  = 
    PH2
    PH2S

 

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D. Converting Between KC and KP, Equation 15.17

The relationship between KC and KP is based on the ideal gas law. Rearranging the ideal gas law gives P = MRT. Substituting that into the KC and KP expressions gives Equation 15.17:

KP = KC(RT)n

In this equation, n = moles of product – moles reactant. The moles of products are calculated by adding the stoichiometric coefficients of the gaseous products, including the "understood 1s." Similarly, the moles of reactant are the sum of the stoichiometric coefficients of the gaseous reactants. Since solids and liquids are not included in the equilibrium constant expression, they are not included in n.

Because this equation comes from the ideal gas law, the value for R is 0.08206 L•atm/mol•K, as it is in the ideal gas law. This is consistent with using units of atmosphere in KP and units of molarity (mol/L) in KC. This value also requires that temperature be in units of Kelvin.

>> Example 4

What is the value of KP for the following equations?

  1. C2H4(g) + H2(g) C2H6(g)       KC = 0.99 at 500 K
  2. 2 NOCl(g) 2 NO(g) + Cl2(g)       KC = 4.4 x 10–6 at 250 °C

Solution:

  1. n = (1) – (1 + 1) = –1, so

    KP = KC(RT)–1 = (0.99)[(0.08206)(500)]–1 = (0.99)[41.03]–1 = 0.99/41.03 = 0.024

  2. n = (2 + 1) – (2) = 1, so

    KP = KC(RT)1 = 4.4 x 10–6[(0.08206)(523)] = 1.9 x 10–4

>> Example 5

What is the value of KC for the following equations?

  1. Si(s) + 2 F2(g) SiF4(g)       KP = 1.4 x 1082 at 1000 K
  2. CO2(g) + 2 Cl2(g) CCl4(g) + O2(g)       KP = 6.4 x 10–18 at 550 °C

Solution:

  1. n = 1 – 2 = –1. The moles of silicon does not count because it is a solid.

    1.4 x 1082 = KC[(0.08206)(1000)]

    1.4 x 1082/82.06 = KC

    1.7 x 1080 = KC

  2. n = 2 – 3 = –1.

    6.4 x 10–18 = KC[(0.08206)(823)]–1

    (6.4 x 10–18)(0.08206)(823) = KC

    4.3 x 10–16 = KC

 

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E. Effect of Reaction Variations on K

Since reversing the reaction changes the position of the products and reactants, the equilibrium constant is the reciprocal of the original equation. Multiplying by a factor raises the equilibrium constant by the same factor.

>> Example 6

If the equilibrium constant of the reaction, 2 NOCl(g) 2 NO(g) + Cl2(g), is KC = 4.4 x 10–6 at 250 °C, what is the equilibrium constant of

  1. 2 NO + Cl2 2 NOCl
  2. NOCl NO + 1/2 Cl2
  3. 6 NO + 3 Cl2 6 NOCl

Solution:

  1. This reaction is the reverse of the original reaction. Therefore

    KCa  = 
    1
    KC
    		 = 
    1
    4.4 x 10–6
    		 =  2.3 x 105

  2. This reaction is the original reaction multiplied by 1/2. Therefore

    KCb = (KC)1/2 = 2.1 x 10–3

  3. This reaction is both reversed (compared with the original) and multiplied by 3. Therefore

    KCc  = 
    1
    KC3
    		 = 
    1
    (4.4 x 10–6)3
    		 =  1.2 x 1016

 

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F. Combining Equilibrium Expressions

The equilibrium constant for a reaction that is the sum of two reactions is the product of the equilibrium constants of those two reactions.

>> Example 7

What is the equilibrium constant for the aqueous reaction:

H2CO3 2 H+ + CO32–

if

K1 = 4.3 X 10–7 for H2CO3 H+ + HCO3–

and

K2 = 5.6 x 10–11 for HCO3– H+ + CO32–

Solution:

If you add the two reactions, bicarbonate, HCO3–, cancels, and the desired reaction results. The equilibrium constant is the product of the two reaction that were added together, so

KC = K1K2 = (4.3 x 10–7)(5.6 x 10–11) = 2.4 x 10–17

>> Example 8

What is the equilibrium constant for the reaction (the ions are aqueous):

PbCl2 + 2 Ag+ 2 AgCl + Pb2+

if

AgCl(s) Ag+ + Cl–       K1 = 1.8 x 10–10

and

PbCl2(s) Pb2+ + 2 Cl–       K2 = 1.2 x 10–4

Solution:

To get the desired reaction, the silver chloride reaction must be doubled and reversed. Therefore

2 Ag+ + 2 Cl– AgCl       K3 = (1/1.8 x 10–10)2 = 3.1 x 1019

and this reaction is added to the lead chloride reaction. So

KC = K3K2 = (3.1 x 1019)(1.2 x 10–4) = 3.7 x 1014

 

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