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Chapter-4- chemical equilibrium
يئايميكلا نزاوتلا -4-لصفلا
In this chapter, we focus:* The concept of chemical potential (يئايميكلا دهجلا) and shows how it is used to account for the
equilibrium composition of chemical reactions ( يئايميكلا ت لاعافتلل يبيكرتلا نزاوتلا).
* Establishing the relation between the equilibrium constant (نزاوتلا تباث) and the standard
Gibbs energy of reaction (يئايميك لعافتل يسايقلا سبج قاط).
* Establishing the quantitative effects of changes in the conditions ( وا طورشلا رييغتب يمكلا ت اريثأتلا
فورظلا).
* Description of the thermodynamic properties of reactions that take place in electrochemical
cells ( يئايميكرهكلا ايلاخلا), in which the reaction drives electrons through an external circuit ( ةرادلا يجراخلا يئابرهكلا).
* Using thermodynamic arguments to derive the electric potential expression of electrochemical cells and its relation to the composition of electrochemical cell.
The definition and tabulation of standard potentials ( يسايقلا دوهجلا)
The use of the standard potentials to determine the equilibrium constants and other thermodynamic properties of chemical reactions.
* Using thermodynamics to predict the equilibrium composition under any reaction conditions and understand the underlying molecular processes.
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Part-1-Spontaneous chemical reactions يئاقلتلا يئايميكلا ت لاعافتلا
The direction of spontaneous change (يئاقلتلا ريغتلا هاجتا) at constant temperature and pressure is
towards lower values of the Gibbs energy, G (سبج قاط).
4.1 The Gibbs energy minimum
The equilibrium composition of a reaction mixture and its corresponding composition are
located by calculating the minimum Gibbs energy of reaction mixture.
4.1. The reaction Gibbs energy
Consider the equilibrium 𝐴 ⇌B.
Suppose an infinitesimal amount dξ (رغصلا يهانتم يمك) of A turns into B; then the change in the amount of A present is dnA = −dξ and the change in the amount of B present is dnB = +dξ.
The quantity ξ (xi) in moles, is called the extent of reaction or degree of reaction or degree of
advancement of reaction.
When the extent of reaction, ξ changes by a finite amount Δξ, then: The amount of A present changes from nA,0 to nA,0 − Δξ
The amount of B changes from nB,0 to nB,0 + Δξ.
Brief illustration
If initially 2.0 mol A is present and we wait until Δξ = +1.5 mol, then the amount of A remaining will be 0.5 mol. The amount of B formed will be 1.5 mol.
The reaction Gibbs energy, ΔrG, is defined as the slope of the graph of the Gibbs energy plotted
against the extent of reaction:
∆rG = (∂G
∂ζ)p,T 𝟒. 𝟏 Here, Δ signifies a derivative, the slope of G with respect to ξ.
Suppose the reaction advances by dξ. The corresponding change in Gibbs energy is: 𝐝𝐆 = 𝛍𝐀𝐝𝐧𝐀+ 𝛍𝐁𝐝𝐧𝐁
3 dnA and dnB change in the amounts of A and B.
we have dnA = -dζ and dnB = dζ, thus
dG = μAdnA+ μBdnB = −μAdζ + μBdζ = (μB−μA)dζ we reorganize this equation, we obtain:
(∂G
∂ζ)p,T= μB−μA That is,
∆𝐫𝐆 = 𝛍𝐁−𝛍𝐀 𝟒. 𝟐
Because chemical potentials vary with composition, the slope of the plot of Gibbs energy against extent of reaction, and therefore the reaction Gibbs energy, changes as the reaction proceeds.
The spontaneous direction of reaction lies in the direction of decreasing G (that is, down the
slope of G plotted against ξ ). Thus we see from eqn 4.2 that:
The forward reaction A→B is spontaneous when μA > μB
The reverse reaction B→A is spontaneous when μB > μA
The reaction is at equilibrium when μB = μA, thus the slope is zero, hence
∆𝐫𝐆 = 𝟎 𝟒. 𝟑
Fig. 6.1 The equilibrium corresponds to zero slope
It follows that, if we can find the composition of the reaction mixture that ensures μB = μA, then
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4.1.2 Exergonic and endergonic reactions
We can express the spontaneity of a reaction at constant temperature and pressure in terms of the reaction Gibbs energy:
If ΔrG < 0, the forward reaction is spontaneous.
If ΔrG > 0, the reverse reaction is spontaneous.
If ΔrG = 0, the reaction is at equilibrium.
A reaction for which ΔrG < 0 is called exergonic (i.e., workproducing).
A reaction for which ΔrG > 0 is called endergonic (i.e., work-consuming).
4.2 The description of equilibrium
4.2.1 Perfect gas equilibria
The chemical potential expression of a perfect gas is:
𝜇 = 𝜇°+ 𝑅𝑇 ln 𝑝 , 𝑤ℎ𝑒𝑟𝑒 𝑝 𝑖𝑛𝑡𝑒𝑟𝑝𝑟𝑒𝑡𝑒𝑑 𝑎𝑠 𝑝/𝑝°
When A and B are perfect gases we can use the potential chemical expression to write: ∆rG = μB−μA = (𝜇𝐵° + 𝑅𝑇 ln p B) − (𝜇𝐴° + 𝑅𝑇 ln pA) ∆rG = ∆rG°+ 𝑅𝑇 ln pB pA 𝟒. 𝟒 where, ∆rG°= 𝜇 𝐵 ° − 𝜇 𝐴 °, and 𝑅𝑇 ln p B− 𝑅𝑇 ln pA= 𝑅𝑇 lnppB A
Denote the ratio of partial pressures by Q = pB/pA, we obtain
∆𝐫𝐆 = ∆𝐫𝐆°+ 𝑹𝑻 𝐥𝐧 𝐐 𝒘𝒉𝒆𝒓𝒆 𝑸 =𝐩𝐁
𝐩𝐀 𝟒. 𝟓
The ratio Q is a reaction quotient (لعافتلا لصاح).
The standard reaction Gibbs energy, ΔrG°, is defined (like the standard reaction enthalpy) as the
difference in the standard molar Gibbs energies of the reactants and products. For our reaction
∆𝐫𝐆° = 𝐆 𝐦 ° (𝐁) − 𝐆 𝐦 ° (𝐀) = 𝛍 𝐁 ° − 𝛍 𝐀 ° 𝟒. 𝟔
5 Note that in the definition of ΔrG°(chapter 3)
∆𝐫𝐆° = 𝚫𝐆 𝐟
°(𝐁) − 𝚫𝐆 𝐟
°(𝐀) 𝟒. 𝟕
At equilibrium ΔrG = 0. The ratio of partial pressures at equilibrium is denoted K, and eqn 4.5
becomes: 0 = ∆rG°+ 𝑅𝑇 ln K we rearrange to 𝑹𝑻 𝐥𝐧 𝐊 = −∆𝐫𝐆° 𝑲 = ( 𝐩𝐁 𝐩𝐀)𝒆𝒒 𝟒. 𝟖
This relation is a special case of one of the most important equations in chemical thermodynamics: it is the link between tables of thermodynamic data, and the chemically important equilibrium constant, K.
We see from eqn 4.8 that:
When ΔrG° > 0, K < 1. Therefore, at equilibrium the partial pressure of A exceeds that of B,
which means that the reactant A is favoured in the equilibrium.
When ΔrG° < 0, K > 1, so at equilibrium the partial pressure of B exceeds that of A. Now the
product B is favoured in the equilibrium.
4.2.2 The general case of a reaction
The eqn 4.4 can be extend to a general reaction. A chemical reaction may be expressed symbolically in terms of stoichiometric numbers as:
∑ 𝝂𝑱
𝑱
𝑱 = 𝟎 𝟒. 𝟗
where J denotes the substances and the νJ are the corresponding stoichiometric numbers in the
chemical equation (A stoichiometric number is positive for products and negative for reactants).
Brief illustration
For instance, in the reaction
2 A + B→3 C + D We have νA=−2, νB=−1, νC=+3, and νD=+1.
6 We define the extent of reaction ξ so that, if it changes by Δξ, then the change in the amount of any species J is νJΔξ. With these points in mind and with the reaction Gibbs energy, ΔrG, defined
in the same way as before (eqn 4.1) that the Gibbs energy of reaction can always be written: ∆𝐫𝐆 = ∆𝐫𝐆°+ 𝑹𝑻 𝐥𝐧 𝐐 𝟒. 𝟏𝟎 where ∆𝐫𝐆°= ∑ 𝛎𝐉 𝐏𝐫𝐨𝐝𝐮𝐜𝐭𝐬 ∆𝐟𝐆°− ∑ 𝛎𝐉 𝐑𝐞𝐚𝐜𝐭𝐚𝐧𝐭𝐬 ∆𝐟𝐆° = ∑ 𝛎𝐉 𝐉 ∆𝐟𝐆° 𝟒. 𝟏𝟏𝐚 Or ∆𝐫𝐆° = ∑ 𝛎𝐉 𝐉 ∆𝐟𝐆°(𝐉) 𝟒. 𝟏𝟏𝐛
Q is the reaction quotient, it has the form:
𝑸 =𝐚𝐜𝐭𝐢𝐯𝐢𝐭𝐢𝐞𝐬 𝐨𝐟 𝐩𝐫𝐨𝐝𝐮𝐜𝐭𝐬 𝐚𝐜𝐭𝐢𝐯𝐢𝐭𝐢𝐞𝐬 𝐨𝐟 𝐩𝐫𝐨𝐝𝐮𝐜𝐭𝐬 𝟒. 𝟏𝟐𝒂 More formally, 𝑸 = ∏ 𝒂𝑱𝝂𝑱 𝑱 𝟒. 𝟏𝟐𝐚
Note: For pure solids and liquids, the activity is 1, so such substances make no contribution to Q even though they may appear in the chemical equation.
Brief illustration
Consider the reaction 2 A + 3 B →C + 2 D, in which case νA = −2, νB = −3, νC = +1, and νD = +2.
The reaction quotient is then
𝑄 = 𝑎𝐴−2𝑎 𝐵−3𝑎𝐶1𝑎𝐷2 = 𝑎𝐶1𝑎 𝐷 2 𝑎𝐴2𝑎 𝐵 3
At equilibrium, the slope of G is zero: ΔrG= 0.
7 𝐊 = (∏ 𝐚𝐉𝛎𝐉 𝐉 ) 𝐞𝐪𝐮𝐢𝐥𝐢𝐛𝐫𝐢𝐮𝐦 𝟒. 𝟏𝟑
This expression has the same form as Q but is evaluated using equilibrium activities.
Note: For K we use equilibrium values and for Q we use the values at the specified stage of the reaction.
An equilibrium constant K expressed in terms of activities (or fugacities) is called a thermodynamic equilibrium constant. Note that, because activities are dimensionless numbers, the thermodynamic equilibrium constant is also dimensionless.
In elementary applications, the activities that occur in eqn 4.13 are often replaced by: Molalities, by replacing aJ by bJ/b° where b° = 1 mol kg−1
Molar concentrations, by replacing aJ by [J]/c°, where c° = 1 mol dm−3
Partial pressures, by replacing aJ by pJ/p°, where p° = 1 bar
Brief illustration
Consider the heterogeneous equilibrium
CaCO3(s) ⇌ CaO(s) + CO2(g)
Its equilibrium constant K is
K = aCaCO−1 3(s)aCaCO(s)1 a1CO2(𝑔) =aCaCO(s)
1 ⏞ 1 a CO2(𝑔) 1 a1CaCO3(s) ⏟ 1 = a1CO2(𝑔)
Provided that the carbon dioxide can be treated as a perfect gas, we can go on to write 𝐾 =𝑝CO2(𝑔)
𝑝°
and conclude that in this case the equilibrium constant is the numerical value of the decomposition vapour pressure of calcium carbonate.
8 At this point we set ΔrG = 0 in eqn 4.10 and replace Q by K. We immediately obtain
𝑹𝑻 𝐥𝐧 𝐊 = −∆𝐫𝐆° 𝟒. 𝟏𝟒
This is an exact and highly important thermodynamic relation, for it enables us to calculate the equilibrium constant of any reaction from tables of thermodynamic data, and hence to predict the equilibrium composition of the reaction mixture.
Example 4.1 Calculating an equilibrium constant Consider the ammonia synthesis reaction,
N2(g) + 3H2(g) ⇌ 2NH3(g)
Calculate the equilibrium constant for the ammonia synthesis reaction, and show how K is related to the partial pressures of the species at equilibrium when the overall pressure is low enough for the gases to be treated as perfect.
Data: ∆fG°(NH3, g) = −16.5 kJ mol−1
Solution
The standard Gibbs energy of the reaction is ∆rG° = ∑ νJ Products ∆fG°− ∑ νJ Reactants ∆fG° ∆rG°= 2∆ fG°(NH3, g) − 2∆fG°(N2, g) − ∆fG°(H2, g) = 2∆fG°(NH3, g) = 2 × (−16.5 kJ mol−1) = −33 kJ mol−1 𝑅𝑇 ln K = −∆rG° ln K = −∆rG° 𝑅𝑇 = − ∆rG° 𝑅𝑇 = − −33 × 103J mol−1 (8.3145 J K−1 mol−1× 298 K−1) = −33 × 103 (8.3145 × 298 ) Hence, K = 6.1 × 105
This result is thermodynamically exact.
9 K = 𝑎𝑁𝐻3 2 𝑎𝑁12𝑎 𝐻2 3 = a2NH3 aN2aH32
This ratio has the value we have just calculated.
At low overall pressures, the activities can be replaced by the ratios pJ/p° and an approximate
form of the equilibrium constant is
K = (pNH3/𝑝°) 2 (pN2/𝑝°)(p H2/𝑝°) 3 = 𝑝NH2 3/𝑝°2 pN2𝑝H32 Homework 4.1
Evaluate the equilibrium constant for below reaction at 298 K. N2O4(g) ⇌ 2NO2(g)
Data: ∆fG°(NO
2, g) = 51.31 kJ mol−1 and ∆fG°(N2O4, g) = 97.89 kJ mol−1
Example 4.2 Estimating the degree of dissociation at equilibrium
The degree of dissociation (or extent of dissociation, α) is defined as the fraction of reactant that has decomposed; if the initial amount of reactant is n and the amount at equilibrium is neq, then
α = (n − neq)/n.
The decomposition reaction of water is
H2O(g) → H2(g) +1
2O2(g)
The standard reaction Gibbs energy for H2O decomposition at 2300 K is +118.08 kJ mol−1.
What is the degree of dissociation of H2O at 2300 K and 1.00 bar?
Solution
The equilibrium constant is obtained from eqn 4.14 in the form ln K = −∆rG° 𝑅𝑇 = − ∆rG° 𝑅𝑇 = − 118.08 × 103J mol−1 (8.3145 J K−1 mol−1× 2300 K−1) = 118.08 × 103 (8.3145 × 2300 ) Hence, K = 2.08 × 10−3
10 H2O (g) H2 (g) O2 (g)
Initial amount n 0 0
Change to reach equilibrium -αn +αn +αn
Amount at equilibrium n-αn= (1-α)n +αn +12αn Total: (1 +12α)n
Mole fraction, xJ 1 − α 1 +12 α α 1 +12 α 1 2 α 1 +12 α Partial pressure (1 − α)𝑝 1 +12 α αp 1 +12 α 1 2 αp 1 +12 α
where, for the entries in the last row, we have used pJ = xJp. The equilibrium constant is therefore
K =pH2𝑝O2
1/2
pH2O
= 𝛼3/2𝑝1/2 (1 − α)(2 + 𝛼)1/2
In this expression, we have written p in place of p/p°, to simplify its appearance. Now make the approximation that α << 1, and hence obtain
K =𝛼3/2𝑝1/2 21/2
Under the stated condition, p = 1.00 bar (that is, p/p° = 1.00), so α ≈ (21/2
×K)2/3 = 0.0205. That is, about 2 per cent of the water has decomposed.
Homework 4.2
Given that the standard Gibbs energy of reaction at 2000 K is +135.2 kJ mol−1 for the same reaction (Example 4.2), suppose that steam at 200 kPa is passed through a furnace tube at that temperature. Calculate the mole fraction of O2 present in the output gas stream.
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4.2.3. The relation between equilibrium constants
Equilibrium constants in terms of activities are exact, but it is often necessary to relate them to concentrations. Formally, we need to know the activity coefficients, and then to use
𝑎𝐽 = 𝛾𝐽𝑥𝐽; 𝑎𝐽 = 𝛾𝐽𝑏𝐽⁄ ; 𝑎𝑏°
𝐽 = 𝛾𝐽[𝐽] 𝑐⁄ ; °
where xJ is a mole fraction, bJ is a molality, and [J] is a molar concentration.
Example:
If we were interested in the composition in terms of molality for an equilibrium of the form: A + B ⇌ C + D
where all four species are solutes, we would write 𝐊 = 𝐚𝐂𝐚𝐃 𝐚𝐀𝐚𝐁= 𝛄𝐂𝛄𝐃 𝛄𝐀𝛄𝐁∗ 𝐛𝐂𝐛𝐃 𝐛𝐀𝐛𝐁 = 𝐊𝛄𝐊𝐛 𝟒. 𝟏𝟓
The activity coefficients must be evaluated at the equilibrium composition of the mixture (for instance, by using one of the Debye–Hückel expressions)
In elementary applications, the assumption is often made that the activity coefficients are all so close to unity that Kγ = 1. Then we obtain the result widely used in elementary chemistry that K ≈ Kb, and equilibria are discussed in terms of molalities (or molar concentrations) themselves.
A special case arises when we need to express the equilibrium constant of a gas-phase reaction in terms of molar concentrations instead of the partial pressures that appear in the thermodynamic equilibrium constant.
Provided we can treat the gases as perfect, the pJ that appear in K can be replaced by [J]RT, and
K = ∏(aJ)νJ J = ∏[𝐽]νJ J (𝑅𝑇 𝑝°) νJ = ∏[𝐽]νJ J × ∏ (𝑅𝑇 𝑝°) νJ J
The (dimensionless) equilibrium constant Kc is defined as
𝐊𝐂= ∏ ([𝐉] 𝐜°) 𝛎𝐉 𝐉 𝟒. 𝟏𝟔 It follows that
12 𝐊 = 𝐊𝐂× ∏ ( 𝐜°𝑹𝑻 𝒑° ) 𝛎𝐉 𝐉 𝟒. 𝟏𝟕𝐚 If now we write 𝛥𝜈 = ∑ 𝜈𝐽 𝐽
which is easier to think of as ν(products) − ν(reactants), then the relation between K and Kc for a
gas-phase reaction is
𝐊 = 𝐊𝐂× (𝐜°𝑹𝑻 𝒑° )
𝛥𝜈
𝟒. 𝟏𝟕𝐛 The term in parentheses works out as T/(12.03 K).
Brief illustration For the reaction
N2(g) + 3H2(g) → 2NH3(g), ∆𝜈 = 2 − 4 = −2 K = KC× ( 𝑇 12.03𝐾) −2 = KC× ( 12.03𝐾 𝑇 ) 2 At 298.15 K the relation is K = KC× (12.03𝐾 298.15) 2 = KC 614.2
so Kc = 614.2 K. Note that both K and Kc are dimensionless.
4.2.3. Equilibria in biological systems
For biological systems, it is appropriate to adopt the biological standard state, in which aH+ = 10−7
and pH =−log aH+ = 7.
The relation between the thermodynamic and biological standard Gibbs energies of reaction for a reaction of the form
𝐑 + 𝛎𝐇+(𝐚𝐪) → 𝐏 𝟒. 𝟏𝟗𝐚
can be found by using the relation between the thermodynamic and biological standard values of the chemical potential of hydrogen ions.
13 𝜇′= 𝜇°′− 7𝑅 ln 10
First, the general expression for the reaction Gibbs energy of this reaction is Δ𝑟𝐺 = Δ𝑟𝐺°+ 𝑅𝑇 ln 𝑎𝑃
𝑎𝑅𝑎𝐻𝜈+ = Δ𝑟𝐺
°+ 𝑅𝑇 ln𝑎𝑃
𝑎𝑅 − 𝑅𝑇 ln 𝑎𝐻𝜈+
In the biological standard state, both P and R are at unit activity (i.e., 𝑅𝑇 ln𝑎𝑃
𝑎𝑅= 0). Therefore, by using ln x = ln 10 log x, this expression becomes
ΔrG = ΔrG°− νRT ln 10 log aH+ = ΔrG°− νRT ln 10 pH
For the full specification of the biological state, we set pH = 7, and hence obtain
𝚫𝐫𝐆′ = 𝚫𝐫𝐆°− 𝟕𝛎𝐑𝐓 𝐥𝐧 𝟏𝟎 𝟒. 𝟏𝟗𝐛
Note: If hydrogen ions are not involved in the reaction (ν = 0). Thus, there is no difference between the two standard values (ΔrG and ΔrG’).
Brief illustration
Consider the reaction NADH(aq) + H+(aq) → NAD+(aq) + H2(g) at 37°C, for which ΔrG° =
−21.8 kJ mol−1. It follows that, because ν = 1 and 7 ln 10 = 16.1, ΔrG’ = −21.8 kJ mol−1 + 16.1 × (8.3145 × 10−3 kJ K−1 mol−1) × (310 K)
= +19.7 kJ mol−1
Note that the biological standard value is opposite in sign (in this example) to the thermodynamic standard value: the much lower concentration of hydronium ions (by seven orders of magnitude) at pH = 7 in place of pH = 0, has resulted in the reverse reaction becoming spontaneous under the new standard conditions.
Homework 4.3
For a particular reaction of the form A → B + 2 H+
in aqueous solution, it was found that ΔrG° =
