Atkins & De Paula: Elements of Physical Chemistry 6E

Atkins & de Paula

Atkins & de Paula: Elements of Physical Chemistry 6e

Chapter 7: Chemical equilibria: the principles

• Reaction equation for A reacting with B forming C and D A + B C + D (1) - general expression of a reaction v A + v B = v C + v D (2) - quantitative representation of a reaction A B → C D vAA + vBB vCC + vDD (3) - representing an equilibrium controlled reactio • Reaction stoichiometry ⇄ – vA, vB, vC and vD are numbers, called stoichiometry coefficients – The concept of mole (the number of molecules) in a chemical reaction – Determination of stoichiometry coefficients - balancing equation (equal number o each atom on both sides of the equation)

e.g. 2NO + O2 = 2NO2 we have: 2 N, 4 O on both side Q. For the same reaction can we write the equation as

4 NO + 2 O2 = 4NO2 or

NO + ½O2=NO2 ?

Atkins & de Paula: Elements of Physical Chemistry 6e Direction of a Reaction

Q. A reaction: A + B C + D. Will it proceed in the direction indicated? Is there a general way to know reaction direction? → A. We can tell from the change of Gibbs free energy in a reaction, G ∆ ° < reaction can proceed (we don’t know how fast it will be!) for a reaction at  GT 0  Δ constant T, P, ∆GT ° = 0 reaction at equilibrium (no change in system - ‘dead’ state) we have  ∆GT ° > 0 reaction will NOT proceed (or can proceed backward!) The 2nd Law of Thermodynamics - Processes occur in a direction of decreasing quality of energy. We need to study reaction system in terms of energies and their changes • The energies associated with a reaction system – Many energy terms, a, U, H, G, - all depend on 4 basic parameters: P, T, V, S – Usually P, T, V are specified by given reaction conditions. S is related to the substances in the reaction (reactants/products) and reaction conditions (P, T, V) – Knowing P, T, V and S, all other energy terms can be determined. The most important ones in relation to chemical reactions are H and G.

Atkins & de Paula: Elements of Physical Chemistry 6e Entropy of Reaction System

Entropy definition dS = dQ/T meaning: at T, change of S is proportional to change of heat. Entropy calculation Basis of S calculation: 3rd Law of thermodynamics: S = 0 at absolute T = 0. Assign standard entropy of a substance (1 mole, at 1 atm. 25°C) as = 298 0 𝑝𝑝 298 𝐶𝐶 (S°298 value of common substances are 𝑆𝑆available� in most chemistry𝑑𝑑𝑑𝑑 / chem. Eng. handbooks.) 0 𝑇𝑇 The entropy change in a reaction at T, p This term becomes zero if there is no phase change during the reaction reaction:vAA + vBB D vCC + vDD = , = + + , , , 𝑇𝑇 , 0 0 0 0 0 𝑑𝑑𝑑𝑑 𝑄𝑄𝑝𝑝푝𝑝𝑝𝑝𝑝𝑝𝑝 ∆𝑆𝑆𝑇𝑇 � 𝜈𝜈𝑖𝑖𝑆𝑆𝑖𝑖 𝑇𝑇 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 − � 𝜈𝜈𝑖𝑖𝑆𝑆𝑖𝑖 𝑇𝑇 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 Δ𝑆𝑆𝑖𝑖 𝑇𝑇 𝑆𝑆298 � Δ𝐶𝐶𝑝𝑝 𝑖𝑖 If a reaction is adiabatic, it can only proceed when S>0 (reaction298 𝑇𝑇reaches𝑇𝑇𝑗𝑗 equilibrium when S=0). Δ Usually S analysis of a reaction system is complicated and inconvenient to use. Δ Atkins & de Paula: Elements of Physical Chemistry 6e Enthalpy of Reaction System

Enthalpy definition H = U+ pV ⇒ dH = dU + pdV + Vdp = S dT + Vdp ΔH = ΔU + Δn RT (gas, g-l, g-s, g-s,l phase at constant p and T) ΔH = ΔU (liquid, or solid (dV = 0)) Enthalpy calculation 0 Standard enthalpy of formation, Hf

Assign H°f of all stable substances (O2, N2, CO2 H2O etc.) at 298K & 1atm as 0 (not at 0K as for S) (H°298 values of common substances are available in most chemistry / chem. eng. handbooks.)

Enthalpy of combustion, Hc -Often used for combustion process, similar to enthalpy of formation Enthalpy change when 1 mole of substance combusted completely

(reacted completely with O2).

Atkins & de Paula: Elements of Physical Chemistry 6e Enthalpy of Reaction System

The enthalpy change in a reaction at T, p

reaction:vAA + vBB vCC + vDD

=⇄ , , 0 0 0 ∆𝐻𝐻𝑇𝑇 � 𝜈𝜈𝑖𝑖𝐻𝐻𝑖𝑖 𝑇𝑇 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 − �The𝜈𝜈𝑖𝑖 𝐻𝐻ΔH𝑖𝑖 𝑇𝑇of 𝑟𝑟phase𝑟𝑟𝑟𝑟𝑟𝑟 transformations including In which g-l-s or crystalline phase change = + + , 𝑇𝑇 , 0 0 𝑖𝑖 𝑇𝑇 298 𝑝𝑝 𝑖𝑖 𝑝𝑝푝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝푝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 ΔH is a direct measureΔ𝐻𝐻 of𝐻𝐻 the reaction�298Δ𝐶𝐶 heat𝑑𝑑𝑑𝑑 associated� Δ𝐻𝐻 with a reaction (if only chemical energy change exists). When ΔH < 0 - the reaction is exothermic and when ΔH > 0 is the reaction is endothermic. Reaction heat is of critical importance in reaction engineering (reactor and process design).

Atkins & de Paula: Elements of Physical Chemistry 6e Gibbs Free Energy of Reaction System

Gibbs Free Energy definition G = U + pV TS = H TS ⇒ ΔG = ΔH ΔS = ΔT + VΔp Gibbs free energy calculation − − − T − S 0 Standard Gibbs free energy, G298 Similar to S0 and H0, the standard Gibbs free energy of formation of a substance, G0, is defined at reference state of T=298 K and p = 1 atm

(G°298 values of common substances are available in most chemistry / chem. eng. handbooks.) The enthalpy change in a reaction at T, p

reaction:vAA + vBB vCC + vDD

=⇄ , , 0 0 0 ∆𝐺𝐺𝑇𝑇 � 𝜈𝜈𝑖𝑖𝐺𝐺𝑖𝑖 𝑇𝑇 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 − � 𝜈𝜈𝑖𝑖𝐺𝐺𝐻𝐻𝑖𝑖 𝑇𝑇 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 In which , = , , (value of each individual substance) 0 0 0 0 = or, GT can Δbe𝐺𝐺𝑖𝑖 determined𝑇𝑇 Δ𝐻𝐻𝑖𝑖 𝑇𝑇 − 𝑇𝑇directlyΔ𝑆𝑆𝑖𝑖 𝑇𝑇 from (value from overall reaction) 0 0 0 Δ Δ𝐺𝐺𝑇𝑇 Δ𝐻𝐻𝑇𝑇 − 𝑇𝑇Δ𝑆𝑆𝑇𝑇 Atkins & de Paula: Elements of Physical Chemistry 6e Gibbs Free Energy of Reaction System

ΔG° is one of the most important thermodynamics properties for a chemical reaction system. - It determines the direction of reaction to proceed ∆ ° < reaction can proceed (we don’t know how fast it will be!) for a reaction at  GT 0  constant T, p, ∆GT ° = 0 reaction at equilibrium (no further change in system) we have  ∆GT ° > 0 reaction will NOT proceed (or can proceed backward!) • ΔG0 < 0 is the pre-condition which MUST be met for any process (not limited to chemical reaction systems) to occur (spontaneous process). • ΔG0 < 0 indicates a specified reaction has tendency to proceed; however, it CANNOT tell how fast that reaction will occur - reaction kinetics tell the reaction rate. • A process/reaction proceeds always in the direction of MINIMISING Gibbs free energy. This is a very important concept. • A process/reaction will stop at ΔG0 = 0, this is called equilibrium state.

Atkins & de Paula: Elements of Physical Chemistry 6e Important Notes about S, H and G

Δ Δ Δ S, H and G are widely used in analysing various systems. Our current discussion of these function is limited to the application to a reaction system. Δ Δ Δ A clear definition of the reaction conditions is necessary before start calculation of these properties. There are many ways these thermodynamic properties can be determined. Only most commonly used ones in relation to a chemical reaction are given. It is very important to understand the study a chemical reaction by means of thermodynamics tells only state - a ‘snapshot’ of system, and how a process proceeds from one state to another (reversible-irreversible, with or without work done / exchange heat with surrounding). There is no factor of time involved.

Atkins & de Paula: Elements of Physical Chemistry 6e Gibbs energy minimum

Extent of reaction ( ): The amount of reactants being converted to products. Its unit is mole. ξ In a very general way, the extent of reaction is calculated as

d = dnA/vA (where v is the stoichiometric number of the reactant A, which is ξ A negative for the reactant!!) Why do we need a new quantity? Consider a generic reaction: 2A ↔ 3B The amount of A consumed is different from the amount of B produced. Which number should be used in the report? Consider: A ↔ B Assume an infinitesimal amount d of A turns into B, dn -d A ξ On the other hand, dn = d = ξ B Atkins & de Paula: Elements of Physical Chemistry 6e ξ Gibbs energy minimum

Example 1: N2(g) + 3H2(g) ↔ 2NH3(g) When the extent of reaction changes from to mole, what are the changes of each reagent? ξ = 0 ξ = 1.0 Solution: identify vj:

v(N2) = -1; v(H2) = -3; v(NH3) = 2. since d = 1.0 mole, dn(N ) = -1x1.0 mole = -1.0 mole, ξ 2 dn(H2) = -3x1.0 mole = -3.0 moles,

dn(NH3) = 2x1.0 mole = 2.0 moles,

Example 2: CH4(g) + Cl2(g) ↔ CHCl3(l) + HCl(g),

in which the amount of reactant Cl2(g) decreases by 2 moles. What is the extent of the reaction? Solution:

Atkins & de Paula: Elements of Physical Chemistry 6e Atkins & de Paula: Elements of Physical Chemistry 6e Atkins & de Paula: Elements of Physical Chemistry 6e Gibbs energy

Variation of Gibbs energy during a reaction process

Atkins & de Paula: Elements of Physical Chemistry 6e The reaction Gibbs energy: ΔrG

The slope of the Gibbs energy plotted against the extent of reaction: = 𝜕𝜕𝐺𝐺 , ∆𝑟𝑟𝐺𝐺 here r signifies a derivative𝜕𝜕𝜕𝜕 𝑝𝑝 𝑇𝑇 • A reaction for which rG is called exergonic. Δ • A reaction for which G is called endergonic. Δr < 0 • G , the forward reaction is spontaneous. r Δ > 0 • G , the reverse reaction is spontaneous. Δr < 0 G • Δ r > 0 , the reaction is at equilibrium!!! Δ = 0

Atkins & de Paula: Elements of Physical Chemistry 6e The reaction Gibbs energy

Molecular interpretation of the minimum in the reaction Gibbs energy

Gibbs energy of the system decreases as the reaction progress

Gibbs energy of a system consisting of different portions of reactants and products

Atkins & de Paula: Elements of Physical Chemistry 6e Calculation of reaction Gibbs energy

Consider the reaction : A ↔ B initial amount: n n

final amount: nAfA0 nBfB0

Ginitial Bn An G n n final = μB BfB0 + μA AfA0 G - G n n – n n ) =final μ initial+ μ B Bf A Af B A n - n n - n - ΔG = B Bf = (μA Af + μ ) B (μ B0 A+ μ A0 - ) = μ (B B0) + μ ( A0) = μ Δξ + μ ( Δξ) = (μ μA Δξ = = 𝜕𝜕𝐺𝐺 , ∆𝑟𝑟𝐺𝐺 𝜇𝜇𝐵𝐵 − 𝜇𝜇𝐴𝐴 • When A B, the reaction (A →𝜕𝜕𝜕𝜕 B) 𝑝𝑝is𝑇𝑇 spontaneous. • When , the reverse reaction (B → A) is spontaneous. μB > μA • When , the reaction is spontaneous in neither direction (equilibrium μB > μA condition). μ = μ Atkins & de Paula: Elements of Physical Chemistry 6e Description of equilibrium

Perfect gas equilibrium : A(g) ↔ B(g)

rG B – A

Δ = μ +μ ln + ln 𝜃𝜃 𝜃𝜃 Δ𝑟𝑟𝐺𝐺 = 𝜇𝜇𝐵𝐵 +𝑅𝑅𝑅𝑅 ln𝑝𝑝𝐵𝐵 − 𝜇𝜇=𝐴𝐴 𝑅𝑅𝑅𝑅 + 𝑝𝑝𝐴𝐴 ln 𝜃𝜃 𝑝𝑝𝐵𝐵 𝜃𝜃 Q ( pB /pΔA𝑟𝑟𝐺𝐺) is a 𝑅𝑅reaction𝑅𝑅 quotientΔ𝑟𝑟𝐺𝐺 𝑅𝑅𝑅𝑅 𝑄𝑄 𝑝𝑝𝐴𝐴 At equilibrium, rG , therefore = + ln ln = Δ = 0 (K is used to denote𝜃𝜃 the ratio of partial pressures at 𝜃𝜃equilibrium) 0 Δ𝑟𝑟𝐺𝐺 𝑅𝑅𝑅𝑅 𝐾𝐾 → 𝑅𝑅𝑅𝑅 𝐾𝐾 − ∆𝑟𝑟𝐺𝐺 Note: The difference in standard molar Gibbs energies of the products and reactants is equal to the difference in their standard Gibbs energies of formation, thus, θ θ θ Δr G = Δf G (B) - Δf G (A)

Atkins & de Paula: Elements of Physical Chemistry 6e Equilibrium of a general reaction

Example: 2A + B ↔ C + 3D

The reaction Gibbs energy, rG, is defined in the same way as discussed earlier : Δ = + ln where the reaction quotient, Q, has the form:𝜃𝜃 Δ𝑟𝑟 𝐺𝐺 Δ𝑟𝑟𝐺𝐺 𝑅𝑅𝑅𝑅 𝑄𝑄 Q = activities of products/activities of reactants in a compact expression = v j are the𝑄𝑄 corresponding∏ 𝛼𝛼𝑗𝑗𝑣𝑣𝑗𝑗 stoichiometric numbers; positive for products and negative for reactants.

vA = -2; vB = -1; vC = 1; vD = 3

Atkins & de Paula: Elements of Physical Chemistry 6e Justification θ ∆rG = ∆rG + RT lnQ

Justification of the equation : dG jdnj Assuming that the extent of reaction equals dξ, = ∑μ one gets dnj vjd then dG dn v d = ξj j j j dG/d u v  G v j j = r∑μ =j ∑j μ ξ because ln a ) ( ξ) = j∑ j Δ =j ∑μ G θ v ln a ))} μr= μ + jRTj ( j v θ v ln a )) Δ = ∑{ j (μj + RT j ( j G θ ln a )vJ = ∑(r μ ) + ∑ (RTj ( Gθ ln a )vj G ln = Δr + RT ∑ ( j r θ θ =θ Δ + RT θ (∏( ) = Δ θ + RT (Q ) ∆rG = ∑v∆ f G − ∑v∆ f G Products Reac tants

Atkins & de Paula: Elements of Physical Chemistry 6e reaction quotient

Again, we use K to denote the reaction quotient at an equilibrium point, vj K Qequilibrium aj )equilibrium is called thermodynamic equilibrium constant. Note that until now is = = (∏ expressed in terms of activities) K K

Example: calculate the quotient for the reaction: A + 2B ↔ 3C + 4D Solution: first, identify the stoichiometric number of each reactant:

vA - vB - vC and vD . = = 1, = 2, = 3,3 4 = 4 𝑎𝑎𝐶𝐶𝑎𝑎𝐷𝐷 At the equilibrium condition:𝑄𝑄 2 𝑎𝑎𝐴𝐴𝑎𝑎𝐵𝐵 = 3 4 𝑎𝑎𝐶𝐶𝑎𝑎𝐷𝐷 𝐾𝐾 2 𝑎𝑎𝐴𝐴𝑎𝑎𝐵𝐵

Atkins & de Paula: Elements of Physical Chemistry 6e Equilibrium Constants

Example 1. Consider a hypothetical equilibrium reaction A(g) + B(g) ↔ C(g) + D(g) While all gases may be considered ideal. The following data are available for this reaction: Compound μθ(kJ mol-1) A(g) -55.00 B(g) -44.00 C(g) -54.00 D(g) -47.00

Calculate the value of the equilibrium constant Kp for the reaction at 298.15K.

Solution:

Atkins & de Paula: Elements of Physical Chemistry 6e Equilibrium Constants

Example 2: Using the data provided in the data section, calculate the standard Gibbs energy and the equilibrium constant at 25oC for the following reaction

CH4(g) + Cl2(g) ↔ CHCl3(g) + HCl(g)

Solution: (chalkboard) -1 fG (CHCl3, g) = -73.66 kJ mol Gθ(HCl, g) = - 95.30 kJ mol-1 Δf Gθ(CH , g) = - 50.72 kJ mol-1 Δf 4 θ Δ

Atkins & de Paula: Elements of Physical Chemistry 6e Atkins & de Paula: Elements of Physical Chemistry 6e Thermodynamic equilibrium constant K

• K is a dimensionless quantity. • K can be expressed in terms of fugacity for gas phase reactions or activities for aqueous phase reactions. • Gas phase: Fugacity is a dimensionless quantity and equals to the numerical value of partial pressure expressed in bar, i.e. j pj/p where p bar). • Aqueous phase: θ θ f = = 1 1. Neutral solution: the activity, a, is equal to the numerical value of the -1 molality, i.e. bj/b where b mol kg . 2. Electrolyte solution:θ The activityθ shall now be calculated as *b /b , = 1 j j j where the activity coefficient, , denotes distance from the ideal systemθ where there is no ion-interactions among constituents. α = γ γ

Atkins & de Paula: Elements of Physical Chemistry 6e Chemical Reaction - The Equilibrium

For a chemical reaction vAA + vBB vCC + vDD when = , we say the reaction is in chemical reaction equilibrium ⇄ • What0 is a chemical reaction equilibrium? 𝐺𝐺𝑇𝑇 0 + - NH4 Example 1: NH3(aq)+H2O(l) NH4 (l)+OH (aq) At constant T and p, when t → ∞ ⇄ NH3 = + − 𝑁𝑁𝐻𝐻4 𝑂𝑂𝐻𝐻 t Example 2: 2NO(g) + O2(g3) 22NO2(g𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) 𝑐𝑐𝑐𝑐𝑐𝑐 𝑁𝑁𝑁𝑁 𝐻𝐻 0 reaction quotient At constant T and p, when t → ∞ ⇄ NO2 NO = 2 O2 𝑝𝑝𝑁𝑁𝑁𝑁2 2 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 t • The concentration of a gas 𝑝𝑝is𝑁𝑁𝑁𝑁 usually𝑃𝑃𝑂𝑂2 measured as partial pressure • At an equilibrium the reaction quotient becomes constant

Atkins & de Paula: Elements of Physical Chemistry 6e Chemical Reaction - The Equilibrium

Chemical reaction equilibrium

vAA + vBB vCC + vDD – An equilibrium is a state at which no further change is possible under that ⇄ specific set of reaction system parameters – An equilibrium is a dynamic process, meaning that the rate of forward reaction is equal to that of reverse – At equilibrium = , i.e 0 = , 𝐺𝐺𝑇𝑇 0 , = , = , 0 0 0 0 0 𝑇𝑇 𝑖𝑖 𝑖𝑖 𝑇𝑇 𝑖𝑖 𝑖𝑖 𝑇𝑇 𝑖𝑖 𝑖𝑖 𝑇𝑇 𝑖𝑖 𝑖𝑖 𝑇𝑇 –∆𝐺𝐺Any change� 𝜈𝜈 𝐺𝐺 of reaction𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 − � parameters𝜈𝜈 𝐺𝐺 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 will redefine0 ⇒ � a𝜈𝜈 system𝐺𝐺 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 leading� to𝜈𝜈 𝐺𝐺a new𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 equilibrium, which may not be the same as that before the change

– Kp indicates only the relation between the partial pressures of reactants and products at equilibrium, how fast a reaction proceed is controlled by reaction kinetics.

Atkins & de Paula: Elements of Physical Chemistry 6e Equilibrium Constant

Definition of equilibrium constant, K The equilibrium constant is the reaction quotient at = . 0 Expressions of equilibrium constant for various reactions𝐺𝐺𝑇𝑇 0 = – gas phase 2NO(g) + O2(g) 2NO2(g) 2 𝑝𝑝𝑁𝑁𝑁𝑁2 𝑝𝑝 2 ⇄ 𝐾𝐾 = 𝑂𝑂2 – gas-solid phase CaCO3(s) CaO (s)+CO2(g) 𝑝𝑝𝑁𝑁𝑁𝑁𝑃𝑃 𝐾𝐾𝑝𝑝 𝑃𝑃𝐶𝐶𝑂𝑂2 ⇄ + - = – liquid phase NH3(aq)+H2O(l) NH4 (l)+OH (aq) + − 𝑁𝑁𝐻𝐻4 𝑂𝑂𝐻𝐻 𝐾𝐾𝑐𝑐 ⇄2+ - = 3 2 – liquid-solid Cu(OH)2(s) Cu (aq)+2OH (aq) 𝑁𝑁𝑁𝑁 𝐻𝐻 0 2+ − 2 𝑐𝑐 ⇄ 𝐾𝐾 = 𝐶𝐶𝐶𝐶/ 𝑂𝑂𝐻𝐻 – gas-liquid NH3(g)+H2O(l) NH4OH(aq) 𝐾𝐾𝑝𝑝 1 𝑃𝑃𝑁𝑁𝑁𝑁3 ⇄ = = – general vAA + vBB vCC + vDD 𝜈𝜈𝐶𝐶 𝜈𝜈𝐷𝐷 𝜈𝜈𝐶𝐶 𝜈𝜈𝐷𝐷 𝐶𝐶 𝐷𝐷 𝑝𝑝𝐶𝐶 𝑝𝑝𝐷𝐷 𝑐𝑐 𝐴𝐴 𝐵𝐵 𝜈𝜈𝐴𝐴 𝜈𝜈𝐵𝐵 Atkins & de Paula: Elements of Physical Chemistry⇄ 6e 𝐾𝐾 𝜈𝜈 𝜈𝜈 𝐴𝐴 𝐵𝐵 𝑝𝑝𝐴𝐴 𝑝𝑝𝐵𝐵 Equilibrium Constant

Equilibrium constant, Kp, and Gibbs free energy DG

A gas phase reaction vAA + vBB vCC + vDD At equilibrium, assuming all the gases follow the ideal gas law at p = 1 atm, ⇄ ΔG°= (viG°i)prod− (viG°i)reac=0 and at any p(≠1) ΔG= (viGi)prod− (viGi)reac=0 When reaction occurs at constant temperature (isothermal reaction) ∑ ∑ ∑ ∑ T=const PV=RT Intergration dG=Vdp −SdT dG=Vdp dG=RTdp/p ΔG=RT ln(p/p0)

G° is defined at p0=1 atm, so that ΔG=G −G°=RT ln(p/1) Gi=Gi°+RT lnpi ∑(v (G° +RT ln(p p )) − ∑(v (G° +RT ln(p p )) =0 i i C D prod i i A B reac→

∑(viG°i)prod − ∑(vi G°i)reac= −[∑(viRT ln(pC pD))prod − ∑(vi RT ln(pA pB))reac]

= ln = ln 𝜈𝜈𝑐𝑐 𝜈𝜈𝐷𝐷 𝜈𝜈𝐶𝐶 𝜈𝜈𝐷𝐷 0 𝑝𝑝𝑐𝑐⁄𝑎𝑎𝑎𝑎𝑎𝑎 𝑝𝑝𝐷𝐷⁄𝑎𝑎𝑎𝑎𝑎𝑎 𝑝𝑝𝑐𝑐 𝑝𝑝𝐷𝐷 Δ𝐺𝐺 −𝑅𝑅𝑅𝑅 𝜈𝜈𝐴𝐴 𝜈𝜈𝐵𝐵 − 𝑅𝑅𝑅𝑅 𝜈𝜈𝐴𝐴 𝜈𝜈𝐵𝐵 𝑝𝑝𝐴𝐴⁄𝑎𝑎𝑎𝑎𝑎𝑎 𝑝𝑝𝐵𝐵⁄𝑎𝑎𝑎𝑎𝑎𝑎 𝐴𝐴 𝐵𝐵 = 0 𝑝𝑝 𝑝𝑝 – By definition: 𝜈𝜈𝐶𝐶 𝜈𝜈𝐷𝐷 ΔG = − RT ln Kp 𝑝𝑝𝑐𝑐 𝑝𝑝𝐷𝐷 𝑝𝑝 𝜈𝜈𝐴𝐴 𝜈𝜈𝐵𝐵 𝐾𝐾 𝑝𝑝𝐴𝐴 𝑝𝑝𝐵𝐵 Atkins & de Paula: Elements of Physical Chemistry 6e Equilibrium Constant

Equilibrium constant, Kc, and Gibbs free energy ΔG Ideal solution (liquid and solid) Pi - vapor pressure of component i x - mole fraction of component i in solution Raoult’s law pi=xipi* or xi=pi / pi* where i Pi* - equil. vapor pressure of pure component i Thus for a solution we can also write Gi = (Gi° + RT ln xi) (compare to gas Gi=Gi°+RT lnpi) which lead to, in a similar way, the relation between ΔG and Kc, = ln = ln = ln 𝜈𝜈𝐶𝐶 𝜈𝜈𝐷𝐷 𝜈𝜈𝐶𝐶 𝜈𝜈𝐷𝐷 0 𝑥𝑥𝑐𝑐 𝑥𝑥𝐷𝐷 𝐶𝐶 𝐷𝐷 Δ𝐺𝐺 −𝑅𝑅𝑅𝑅 𝜈𝜈𝐴𝐴 𝜈𝜈𝐵𝐵 −𝑅𝑅𝑅𝑅 𝜈𝜈𝐴𝐴 𝜈𝜈𝐵𝐵 −𝑅𝑅𝑅𝑅 𝐾𝐾𝑐𝑐 Summary of G calculation𝑥𝑥𝐴𝐴 for𝑥𝑥𝐵𝐵 ideal gas and𝐴𝐴 solution𝐵𝐵 • For a pure substance at constant T and p, G = G°+ RT ln p (1)

• For a mixture of ideal gas at const T and p, G = Gi = (Gi°+ RT ln pi) (2-1) • For a mixture of ideal solution at const T and p, G = G = (G °+ RT ln x ) (2-2) ∑ i ∑ i i  For a situation that a mixture (gas or solution) under concern is not ideal, ∑ ∑ the pi or xi cannot be related to G by expressions (2-1) & (2-2). How do you calculate G?

Atkins & de Paula: Elements of Physical Chemistry 6e Equilibrium Constant

Real mixture (Non ideal substances)

For a real gas: pi,real ≠ pi,ideal, we define the effective pressure (called fugacity) f, , = , or pi,real = fi pi,ideal. 𝑝𝑝𝑖𝑖 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 For𝑖𝑖 a real solution:𝑝𝑝𝑖𝑖 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 : pi,vap,real ≠ pi,vap,ideal, we define the effective concentration, ai, 𝑓𝑓 � , , a is called activity : = , , or pi,vap,real = ai pi,vap,ideal. 𝑝𝑝𝑖𝑖 𝑣𝑣𝑣𝑣𝑣𝑣 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑖𝑖 𝑣𝑣𝑣𝑣𝑣𝑣 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 – Chemical potential,𝑎𝑎 𝑖𝑖μ �𝑝𝑝 Using the same relation of G with pi and xi for ideal gas and solution, but G is replaced by a new term, μ, called chemical potential , representing the non- ideal situation

ideal gas Gi= Gi°+RT lnPi real gas: μi = μi°+ RT ln pi,real = μi°+RT ln fipi,ideal

ideal solution Gi=Gi°+RT ln xi real solution: μi= μi°+RT lnxi,real = μi°+RT ln aixi,ideal  Values of f and a for real gases and solutions can be found from literature  Chemical potential can also be used to describe an ideal gas or solution, i.e. G = μ. i.e. for ideal gas μi = μi°+ RT ln pi and for ideal solution μi= μi°+RT ln xi

Atkins & de Paula: Elements of Physical Chemistry 6e Equilibrium Constant

A chemical reaction, if in equilibrium (Δ = 0) v A + v B v C + v D A B C μ D = ln ⇄ = ln = ideal gas or where 𝜈𝜈𝐶𝐶 𝜈𝜈𝐷𝐷 0 0 𝑝𝑝𝑐𝑐 𝑝𝑝𝐷𝐷 𝑝𝑝 𝑝𝑝 𝑝𝑝 𝜈𝜈𝐴𝐴 𝜈𝜈𝐵𝐵 Δ𝐺𝐺 −𝑅𝑅𝑅𝑅 𝐾𝐾 Δ𝜇𝜇 −𝑅𝑅𝑅𝑅 𝐾𝐾 𝐾𝐾 𝑝𝑝𝐴𝐴 𝑝𝑝𝐵𝐵 = ln = = real gas where 𝜈𝜈𝐶𝐶 𝜈𝜈𝐷𝐷 𝜈𝜈𝐶𝐶 𝜈𝜈𝐷𝐷 𝜈𝜈𝐶𝐶 𝜈𝜈𝐷𝐷 0 𝑓𝑓𝑐𝑐 𝑓𝑓𝐷𝐷 𝑝𝑝𝑐𝑐 𝑝𝑝𝐷𝐷 𝑓𝑓𝑐𝑐 𝑓𝑓𝐷𝐷 𝑓𝑓 𝑓𝑓 𝜈𝜈𝐴𝐴 𝜈𝜈𝐵𝐵 𝜈𝜈𝐴𝐴 𝜈𝜈𝐵𝐵 𝜈𝜈𝐴𝐴 𝜈𝜈𝐵𝐵 𝑝𝑝 Δ𝜇𝜇 = −𝑅𝑅𝑅𝑅 ln𝐾𝐾 =𝐾𝐾 𝑓𝑓𝐴𝐴ln𝑓𝑓𝐵𝐵 𝑝𝑝𝐴𝐴 𝑝𝑝𝐵𝐵 𝑓𝑓𝐴𝐴=𝑓𝑓𝐵𝐵 𝐾𝐾 ideal solution or where 𝜈𝜈 𝜈𝜈 𝐶𝐶 𝐶𝐶 𝐷𝐷 𝐷𝐷 0 0 𝜈𝜈 𝜈𝜈 Δ𝐺𝐺 −𝑅𝑅𝑅𝑅 𝐾𝐾𝑐𝑐 Δ𝜇𝜇 −𝑅𝑅𝑅𝑅 𝐾𝐾𝑐𝑐 𝐾𝐾𝑝𝑝 𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐵𝐵 real solution = ln where = 𝜈𝜈 𝜈𝜈 = 𝜈𝜈 𝜈𝜈 𝐶𝐶 𝐷𝐷 𝜈𝜈𝐶𝐶 𝜈𝜈𝐷𝐷 𝐶𝐶 𝐷𝐷 0 𝑎𝑎𝑐𝑐 𝑎𝑎𝐷𝐷 𝐶𝐶 𝐷𝐷 𝑎𝑎𝑐𝑐 𝑎𝑎𝐷𝐷 𝑎𝑎 𝑎𝑎 𝜈𝜈𝐴𝐴 𝜈𝜈𝐵𝐵 𝜈𝜈𝐴𝐴 𝜈𝜈𝐵𝐵 𝜈𝜈𝐴𝐴 𝜈𝜈𝐵𝐵 𝑐𝑐 When solids or liquidsΔ𝜇𝜇 are− 𝑅𝑅present𝑅𝑅 𝐾𝐾 either as𝐾𝐾 reactants𝑎𝑎𝐴𝐴 𝑎𝑎𝐵𝐵 or𝐴𝐴 products,𝐵𝐵 their𝑎𝑎𝐴𝐴 𝑎𝑎𝐵𝐵 vapour𝐾𝐾 pressures, at constant T, are usually constant - thus can be included in the Kp without appearing in the expression.

e.g. CaCO3(s) = CaO(s) + CO2 (g) Kp = pCaO pCO2/pCaCO3 ⇒ Kp’=PCO2

Atkins & de Paula: Elements of Physical Chemistry 6e Equilibrium Constant

The equilibrium constant, K, is a function of temperature = ln = exp = ln = exp 0 or 0 0 Δ𝐺𝐺 0 Δ𝜇𝜇 TheΔ𝐺𝐺 effect− of𝑅𝑅𝑅𝑅 inert𝐾𝐾 on⇒ 𝐾𝐾the equilibrium− 𝑅𝑅𝑅𝑅 compositionΔ𝜇𝜇 − -𝑅𝑅depending𝑅𝑅 𝐾𝐾 ⇒ 𝐾𝐾 on the reaction− 𝑅𝑅𝑅𝑅 stoichiometry : Δv = vC + vD - vA - vB Δv < 0 : adding an inert cause the equilibrium to shift to more reactants Δv = 0 : adding an inert has no effect on the equilibrium composition Δv > 0 : adding an inert push the equilibrium to shift to more products When Δv > 0, a high p push the equilibrium to the direction of more completion, and visa verse The expression of equilibrium constant depends on the way reaction equation is written – For a reverse reaction its equilibrium constant

vAA + vBB vCC + vDD vCC + vDD vAA + vBB – For reaction equation written with different stoichiometry coefficient ⇄ ⇄ 2NO(g) + O2(g) 2NO2(g) NO(g) + ½ O2(g) NO2(g) Atkins & de Paula: Elements of Physical Chemistry 6e ⇄ ⇄ Equilibrium Constant

The expression of equilibrium constant depends on the way reaction equation is written – For a reverse reaction its equilibrium constant

vAA + vBB vCC + vDD vCC + vDD vAA + vBB

⇄ ⇄ , = 𝜈𝜈𝐶𝐶 𝜈𝜈𝐷𝐷 , , = 𝜈𝜈𝐴𝐴 𝜈𝜈𝐵𝐵 = 𝜈𝜈𝐶𝐶 𝜈𝜈𝐷𝐷 −1 = , 𝑝𝑝𝐶𝐶 𝑝𝑝𝐷𝐷 𝑝𝑝𝐴𝐴 𝑝𝑝𝐵𝐵 𝑝𝑝𝐶𝐶 𝑝𝑝𝐷𝐷 −1 𝐾𝐾𝑝𝑝 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝜈𝜈𝐴𝐴 𝜈𝜈𝐵𝐵 𝐾𝐾𝑝𝑝 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝜈𝜈𝐶𝐶 𝜈𝜈𝐷𝐷 𝜈𝜈𝐴𝐴 𝜈𝜈𝐵𝐵 𝐾𝐾𝑝𝑝 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 –For reaction equation𝑝𝑝𝐴𝐴 𝑝𝑝𝐵𝐵 written with different𝑝𝑝𝐶𝐶 stoichiometry𝑝𝑝𝐷𝐷 𝑝𝑝𝐴𝐴 𝑝𝑝coefficient𝐵𝐵 2NO(g) + O2(g) 2NO2(g) NO(g) + ½ O2(g) NO2(g) = , = = / ⇄ 2 / ⇄ 𝑝𝑝𝑁𝑁𝑁𝑁2 𝑝𝑝𝑁𝑁𝑁𝑁2 1 2 𝑝𝑝 2 𝑝𝑝 1 2 𝑝𝑝 𝐾𝐾 𝑂𝑂2 𝐾𝐾′ 𝑃𝑃 𝑝𝑝𝑁𝑁𝑁𝑁𝑃𝑃 𝑃𝑃𝑁𝑁𝑁𝑁𝑝𝑝𝑂𝑂2

Atkins & de Paula: Elements of Physical Chemistry 6e Activities of Solids and Pure Liquids

and (!!!) • Illustration: Express the equilibrium constant for the heterogeneous reaction • α(solid) = 1 α(pure liquid) = 1 NH4Cl(s) ↔ NH3(g) + HCl(g) • Answer: In term of fugacity (i.e. thermodynamic equilibrium constant):

Kp = In term of molar fraction (this one is not a thermodynamic equilibrium

constant): Kx =

Atkins & de Paula: Elements of Physical Chemistry 6e Estimate reaction compositions at equilibrium

• Example 1: Given the standard Gibbs energy of reaction H2O(g) → H2(g) + -1 1/2O2(g) at 2000 K is + 135.2 kJ mol , 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. • Solution: (details will be discussed in class) lnK - x - - - x - 3 1 1 1 = (135.2 10 J mol )/(8.3145 JK mol 2000K) K x - = 8.13037 / 4 / == 2.9446 10 𝜃𝜃 1 2 𝜃𝜃 𝑝𝑝𝑂𝑂2 𝑝𝑝 𝑝𝑝𝐻𝐻2𝑝𝑝 𝜃𝜃 p𝐾𝐾 k𝑝𝑝𝐻𝐻pa2𝑂𝑂𝑝𝑝 assuming the mole fraction of O equals x total = 200 2 pO2 x p p x p ) H2 = total, p p – p – p - x) p H2O= 2( total O2 H2 Atkins & de Paula: Elements= total of Physical Chemistry= (1 36e total Equilibrium in biological systems

Biological standard state: pH = 7. + For a reaction: A + vH (aq) ↔ P = + ln 𝜃𝜃 𝑏𝑏𝑃𝑃 Δ𝑟𝑟𝐺𝐺 Δ𝑟𝑟𝐺𝐺 𝑅𝑅𝑅𝑅 𝜈𝜈 𝑏𝑏𝐴𝐴𝑏𝑏𝐻𝐻+ = + ln + ln 𝜃𝜃 1 𝑏𝑏𝑃𝑃 Δ𝑟𝑟𝐺𝐺 Δ𝑟𝑟𝐺𝐺 𝑅𝑅𝑅𝑅 + 𝜈𝜈 𝑅𝑅𝑅𝑅 G‡ the first two terms of the above𝐻𝐻 equation form𝑏𝑏𝐴𝐴 r G‡ G v , r r Δ Δ G‡ is defined as Standardθ reaction Gibbs energy for biochemical systems r Δ = Δ + 7 RT ln10 A better practice is to transfer the above equation into ‡ rG rG 7v ln10, and then recognizesθ v is the stoichiometric number of H+. Δ = Δ − RT

Atkins & de Paula: Elements of Physical Chemistry 6e Equilibrium in biological systems

Example: For a particular reaction of the form A → B + 2H+ in aqueous - o ‡ solution, it was found that rG at 28 C. Estimate the value of rG . Solution: θ 1 Δ = 20kJ mol Δ ‡ rG rG v here the stoichiometricθ number of H+ is 2, i.e. v = 2 Δ = Δ − 7 RT ln10 ‡ - - - - rG -1 - 3 1 1 Δ = 20 kJ mol − 7(2)(8.3145x10 kJ K mol )x(273+ 28K) ln10 1 - 1 = 20 kJ mol − 80.676 kJ mol (Notably, when measured with1 the biological standard, the standard Gibbs = − 61 kJ mol energy of reaction becomes negative.! A transition from endergonic to exergonic process. )

Two factors affect the thermodynamic equilibrium constant: (1) Enthalpy, and (2) Entropy. Boltzmann distribution is independent of the nature of the particle.

Atkins & de Paula: Elements of Physical Chemistry 6e Response of equilibria to reaction conditions

The response of equilibria to reaction conditions • Equilibria may respond to changes in pressure, temperature, and concentrations of reactants and products. • The equilibrium constant is not affected by the presence of a catalyst. How equilibria respond to pressure? • The thermodynamic equilibrium constant is a function of the standard reaction Gibbs energy, rG . K • Standard reaction Gibbs energyθ Δ Gθ is defined at a single standard Δ r pressure and thus is independent of the pressure used in a specific reaction. • The thermodynamic equilibrium constant is therefore independent of reaction pressure. Such a relationship can be expressed as:

= 𝜕𝜕𝐾𝐾 0 𝜕𝜕𝑝𝑝 𝑇𝑇

Atkins & de Paula: Elements of Physical Chemistry 6e Response of equilibria to reaction conditions

• Although the thermodynamics equilibrium constant K is independent of pressure, it does not mean that the equilibrium composition is independent of the pressure!!! • Example: Consider a gas phase reaction 2A(g) ↔ B(g)

assuming that the mole fraction of A equals xA at equilibrium, then xB = 1.0 – xA, / ) = = / 𝜃𝜃 𝜃𝜃 1.0 − 𝑥𝑥𝐴𝐴 𝑝𝑝𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑝𝑝 (1.0 − 𝑥𝑥𝐴𝐴 𝑝𝑝 because K 𝐾𝐾does not change, 𝜃𝜃x 2must change2 in response to any 𝑥𝑥𝐴𝐴𝑝𝑝𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑝𝑝 A 𝑥𝑥𝐴𝐴𝑝𝑝𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 variation in ptotal!!!

Atkins & de Paula: Elements of Physical Chemistry 6e Le Chatelier’s Principle

A system at equilibrium, when subject to a disturbance, responds in a way that tends to minimize the effect of the disturbance. The above statement suggests that if the total pressure of a system is increased, the system will shift to the direction that will have smaller number of molecules, i.e. smaller pressure.

3H2(g) + N2(g) ↔ 2NH3(g).

Atkins & de Paula: Elements of Physical Chemistry 6e Le Chatelier’s Principle

Example: Predict the effect of an increase in pressure on the Haber reaction, 3H2(g) + N2(g) ↔ 2NH3(g). Solution:According to Le Chatelier’s Principle, an increase in pressure will favor the product. prove: = = = = 2 2 2 2 𝑝𝑝𝑁𝑁𝐻𝐻3 𝑥𝑥𝑁𝑁𝐻𝐻3𝑝𝑝𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑥𝑥𝑁𝑁𝐻𝐻3 𝐾𝐾𝑥𝑥 𝐾𝐾 3 3 3 3 2 2 Therefore, to keep𝑝𝑝𝑁𝑁 the2𝑝𝑝𝐻𝐻 2thermodynamic𝑥𝑥𝑁𝑁2𝑝𝑝𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑥𝑥𝐻𝐻2 𝑝𝑝equilibrium𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑥𝑥𝑁𝑁2 𝑥𝑥constant𝐻𝐻2𝑝𝑝𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 K 𝑝𝑝unchanged,𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 the equilibrium mole fractions Kx must change by a factor of 4 if the pressure ptotal is doubled.

Atkins & de Paula: Elements of Physical Chemistry 6e Response of equilibria to reaction conditions

The response of equilibria to temperature According to Le Chatelier’s Principle: Exothermic reactions: increased temperature favors the reactants. Endothermic reactions: increased temperature favors the products. The van’t Hoff equation: ln = 𝜃𝜃 𝑟𝑟 ln𝑑𝑑 𝐾𝐾 ∆ 𝐻𝐻 𝑎𝑎 = 2 / 𝑑𝑑)𝑑𝑑 𝑅𝑅𝑇𝑇 𝜃𝜃 𝑑𝑑 𝐾𝐾 ∆𝑟𝑟𝐻𝐻 𝑏𝑏 − 𝑑𝑑(1 𝑇𝑇 𝑅𝑅

Atkins & de Paula: Elements of Physical Chemistry 6e Derivation of the van’t Hoff equation

ln = 𝜃𝜃 ∆𝑟𝑟𝐺𝐺 • Differentiate ln K with respect to 𝐾𝐾temperature− ln 𝑅𝑅𝑅𝑅 / ) = 𝜃𝜃 𝑑𝑑 𝐾𝐾 1 𝑑𝑑(∆𝑟𝑟𝐺𝐺 𝑇𝑇 • Using Gibbs-Helmholtz equation − 𝑑𝑑𝑑𝑑 / 𝑅𝑅) 𝑑𝑑𝑑𝑑 = 𝜃𝜃 𝜃𝜃 𝑑𝑑(∆𝑟𝑟𝐺𝐺 𝑇𝑇 ∆𝑟𝑟𝐻𝐻 thus − 2 𝑑𝑑ln𝑑𝑑 𝑇𝑇 = 𝜃𝜃 𝑑𝑑 𝐾𝐾 ∆𝑟𝑟𝐻𝐻 • Because d(1/T)/dT = 1/T2: 2 ln𝑑𝑑𝑑𝑑 𝑅𝑅𝑇𝑇 = − / ) 𝜃𝜃 𝑑𝑑 𝐾𝐾 ∆𝑟𝑟𝐻𝐻 − 𝑑𝑑(1 𝑇𝑇 𝑅𝑅 Atkins & de Paula: Elements of Physical Chemistry 6e Atkins & de Paula: Elements of Physical Chemistry 6e van’t Hoff equation

θ For an exothermic reaction, ΔrH < 0, thus < , suggesting that increasing𝑑𝑑 ln 𝐾𝐾 the reaction temperature 𝑑𝑑𝑑𝑑 0 will reduce the equilibrium constant.

Atkins & de Paula: Elements of Physical Chemistry 6e Applications of the van’t Hoff equation

• Provided the reaction enthalpy, rH , can be assumed to be independent of temperature, eqn. 7.23b (or 9.26b inθ 7th edition) illustrates that a plot of – lnK against 1/T should yield a straightΔ line of slope rH / . • Example: The data below show the equilibrium constantθ measured at Δ R different temperatures. Calculate the standard reaction enthalpy for the system. /K 350 400 450 500 K 3.94x10-4 1.41x10-2 1.86x10-1 1.48 T Solution: 2.86x10-3 2.50x10-3 2.22x10-3 2.00x10-3 ln K 7.83 4.26 1.68 -0.39 1/T −

Atkins & de Paula: Elements of Physical Chemistry 6e Applications of the van’t Hoff equation

Self-test : The equilibrium constant of the reaction

2SO2(g) + O2(g) ↔ 2SO3(g) is 4.0x1024 at 300K, 2.5x1010 at 500K, and 2.0x104 at 700K. Estimate the reaction enthalpy at 500K. Solution : discussion: 1. Do we need a balanced reaction equation here? 2. What can be learned about the reaction based on the information provided? 3. Will the enthalpy become different at 300K or 700K?

Atkins & de Paula: Elements of Physical Chemistry 6e K at different temperatures

The equilibrium constant at temperature 2 can be obtained in terms of the known equilibrium constant K at . T Since the standard reaction enthalpy is also a function of temperature, when 1 T1 integrating the equation 9.26b from to 2, we need to assume that rH is constant within that interval. ө 1 / T T Δ ln = 𝐾𝐾2 1 𝑇𝑇2 𝜃𝜃 / ∆𝑟𝑟𝐻𝐻 1 � 𝑑𝑑 𝐾𝐾 − � 𝑑𝑑 𝐾𝐾1 1 𝑇𝑇1 𝑅𝑅 𝑇𝑇 so ln ln = (7.24) 𝜃𝜃 𝑟𝑟 2 1 ∆ 𝐻𝐻 1 1 𝐾𝐾 − 𝐾𝐾 − 2 − 1 Equation 7.24 provides a non-calorimetric𝑅𝑅 𝑇𝑇 method𝑇𝑇 of determining standard reaction enthalpy. (Must keep in mind that the reaction enthalpy is actually temperature-dependent!)

Atkins & de Paula: Elements of Physical Chemistry 6e K at different temperatures

Example, The Haber reaction

N2(g) + 3H2(g) ↔ 2NH3(g) At 298 K, the equilibrium constant K = 6.1x105. The standard enthalpy of -1 formation for NH3 equals -46.1 kJ mol . What is the equilibrium constant at 500K?

Answer: First, calculate the standard reaction enthalpy, rH , H H H ) ө r f f 2 f 2 Δ ө ө – ө ө Δ = 2Δ H (NH3) − 3 Δ (H ) − Δ (N kJ mol-1 = 2(− 46.1) 3(0) − 1(0) then ln K – ln - – = 2− 92.2 ln K 5 3 1 ( 2) (6.1x10 ) = (−1/8.3145)(− 92.2x10 J mol ) (1/500 1/298) K 2( ) = − 1.71 Despite the decrease in equilibrium constant as a result of temperature = 0.18 increase, yet in industrial production it is still operated at an elevated temperature (kinetics vs thermodynamics)

Atkins & de Paula: Elements of Physical Chemistry 6e temperature dependence of the equilibrium constant

Practical Applications of the Knowledge of the temperature dependence of the equilibrium constant

(i) M(s) + 1/2O2(g) → MO(s)

(ii) 1/2C(s) + 1/2O2(g) → 1/2CO2(g)

(iii) C(s) + 1/2O2(g) → CO(g)

(iv) CO(g) + 1/2O2(g) → CO2(g) This is carried out based on two criteria (1) Gibbs energy is a state quantity and thus can be added or subtracted directly. (2) When the standard reaction Gibbs energy is negative, the forward reaction is favored (i.e. K > 1). Standard reaction Gibbs energy is sometimes referred as Free Gibbs energy.

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