Chemical Thermodynamics lecture 3 mod 2

• Module 2, Lecture 3 is about Chemical Thermodynamics Part 3

Learning Objectives:

• Calculate the change in Gibbs Free Energy at standard Conditions from values and from the equation: ∆G° = ∆H° - TAS°
• Understanding concept of coupled reactions
• When ∆G = 0 the reaction is at equilibrium
• Appreciate the difference between ∆G and ∆G° in relation to equilibrium.
• Perform calculations relating ∆G, ∆G°and K.

Textbook Reference

• Chapter 19

Gibbs Energy

• Gibbs energy (G) is defined as: G = H-TS
• H is the enthalpy of the system
• S is the entropy of the system
• T is the temperature in Kelvin (0 °C = 273 K)
• Absolute values of G cannot be measured, therefore focus is on the change in G for a process, ∆G, where: ∆G = ∆H – TAS
• For a chemical reaction, the terminology used is: ∆G = ∆H – T∆S

Gibbs Free Energy and Spontaneity

• ∆G indicates whether a process is spontaneous
• If ∆G is negative, the process is spontaneous
• If ∆G is zero, the system is at equilibrium
• If ∆G is positive, the process is non-spontaneous
• A non-spontaneous process will have a spontaneous reverse reaction.
• The magnitude of ∆G indicates how far a spontaneous process will proceed towards completion, but not spontaneity.
• If ∆H is positive and ∆S is negative, the reaction is never spontaneous at any temperature T
• If ∆H is negative and ∆S is positive, the reaction is spontaneous at all temperatures, T
• When ∆H and ∆S have the same sign, spontaneity of the reaction relies on T
• ∆G = ∆H – T∆S

Examples

• The equation for Gibbs Free energy is: ∆G = ∆H – T∆S
• Melting of ice: H2O(s) → H2O(I)
• Because ∆H is positive and ∆S is positive, the reaction is spontaneous only at high temperatures (above 0 °C)
• Freezing of water: H2O(l) → H2O(s)
• Because ∆H is negative and ∆S is negative, the reaction is spontaneous only at low temperatures (below 0 °C)

Entropy of Dissolution

• Equation of Dissolution: KCI(s) → K+(aq) + Cl-(aq)
• S°(KCI, s) = 83 J mol-1 K-1
• S°(K+, aq) = 103 J mol-1 K-1
• S°(Cl-, aq) = 57 J mol-1 K-1
• △S° = ((103 + 57) – 83) J mol-1 K-1 = 77 J mol-1 K-1
• Entropy change for dissolution of any ionic salt is favourable because it forms a less ordered system

Enthalpy of Dissolution

• Equation of Dissolution: KCI(s) → K+(aq) + Cl-(aq)
• ∆H°(KCI, s) = -436 kJ mol-1
• ∆H°(K+, aq) = -252 kJ mol-1
• ∆H°(Cl-, aq) = -167 kJ mol-1
• ∆H° = [(-252 + -167) – (-436)] kJ mol-1 = +17 kJ mol-1
• ∆G° = ∆H° − T∆S° = 17 × 103 J mol-1 – (298 K × 77 J mol-1 K-1) = -5.9 kJ mol-1 (Spontaneous)

Gibbs Free Energy at Standard Conditions

• Thermodynamic data is typically tabulated under standard conditions
• Use ∆H° and ∆S° values to obtain ∆G° values
• ∆G° = ∆H° - TAS°
• ∆G° can only be used as a criterion of spontaneity under standard conditions
• ∆Gº will determine the position of equilibrium in a chemical system

Calculating Gibbs Free Energy

• ∆G for a reaction can be calculated by:
• Using ∆G = ∆H – T∆S
• G measures energy including enthalpy and entropy
• Using ∆G data
• The standard Gibbs energy of formation, ∆Gº is analogous to the standard enthalpy of formation, ∆H°
• The ∆G° of any pure element is defined as zero

Gibbs Free Energy Calculation Example

• Reaction: CO(NH2)2(aq) + H2O(l) → CO2(g) + 2NH3(g)
• Temperature: 25.0 °C
• Given ∆H° = +119.2 kJ mol-1 and ∆S° = +354.8 J K-1 mol-1
• ∆G° = ∆H° – TAS° = +119.2 × 103 J mol-1- (298 K × 354.8 J K-1 mol-1) = +13.2 × 103 J mol-1 = +13.2 kJ mol-1
• The is non-spontaneous under standard conditions, therefore the opposite reaction is spontaneous

Gibbs Free Energy Calculation Example with Hess's Law

• Values of ∆fG° are tabulated and can be used in Hess's Law-type calculations
• △rG° = ∑[△fG° (products)] - ∑[△fG° (reactants)](taking account of stoichiometry)
• The Gibbs energy change for a reaction equals the sum of the Gibbs energies of formation of the products minus the sum of the Gibbs energies of formation of the reactants.

Gibbs Free Energy Calculation Example with Hess's Law and Reaction Stoichiometry

• Calculate ∆G° for the reaction: C2H5OH(I) + 3O2(g) → 2CO2(g) + 3H2O(g) at 25.0 °C
• Given: ∆G°(C2H5OH, I) = -174.8 kJ mol-1, ∆G°(CO2, g) = -394.4 kJ mol-1, ∆G°(H2O, g) = -228.6 kJ mol-1
• ∆rG° =[2 × -394.4 kJ mol-1 + 3 × -228.6 kJ mol-1] - [1 × -174.8 kJ mol-1 + 3 x 0 kJ mol-1] = -1299.8 kJ mol-1
• Result: The reaction is spontaneous under standard conditions

Coupled Reactions

• A spontaneous reaction (∆G -ve) can be used to drive a non-spontaneous reaction (∆G +ve) by coupling them together
• In biological systems, the hydrolysis reaction of adenosine triphosphate, ATP, is often involved

Example of Coupled Reactions

• Glucose metabolism: reaction of glucose with hydrogen phosphate to give glucose-6-phosphate
• ∆G = +13.4 kJ mol-1 at pH = 7
• Reactions: Glu → Glu-6-P, AG = +13.4 kJ mol-1 and ATP → ADP, AG = -30.5 kJ mol-1
• Coupled Version of Reactions will be spontaneous at pH = 7: Glu + ATP → Glu-6-P + ADP, AG = -17.1 kJ mol-1

Gibbs Free Energy at Equilibrium

• ∆G = 0
• There is no driving force for chemical or physical change, when ∆G = 0
• The system is at equilibrium when ∆G = 0
• Example Reaction: N2(g) + 3H2(g) ⇄ 2NH3(g)
• The equilibrium composition is the same, regardless of whether starting with pure reactants, N2 and H2, or pure product, NH3
• Forward and reverse reactions occur at equilibrium with identical rates, showing no net change in composition
• Equilibrium minimizes Gibbs Free Energy, therefore neither forward or reverse reactions proceed to completion

K, Gibbs Free Energy, and Equilibrium

• ∆G provides the criterion for spontaneity and has a relationship with ∆G°, which is obtained from tables
• The equation showing this relationship is: △rG = △rG° + RT InQ
• Within this equation, Gibbs energy available point at a single point in the reaction is directly related to products/reactants
• Previously shown:
• If ∆G = 0, the system is at equilibrium
• Q = K at equilibrium
• 0 = △rG° + RT InK
• Gibbs Free Energy Equation: △rG° = -RT InK

Example Problem

• Lugol's Iodine is a solution of triiodide, to treat iodine deficiency and as a disinfectant
• I3- is prepared by the reaction of I2 with I-: I2(aq) + I-(aq) ⇄ I3-(aq) (K = 1.26x10-3 at 298 K)

Calculate the value of ∆rGº for this reaction?

• Solve for standard Gibbs Free Energy: ∆G° = -RTInK
• ∆G° = - (8.314 J K-1 mol¯¹ x 298 K x ln(1.26x10-3)) = 16542 J mol-1 = 16.54 kJ mol-1

Given Specific Quantities, will the reaction at equilibrium?

• Solution contains [I2] = 0.65 mol L-1, [I-] = 1.30 mol L-1 and [I3¯] = 0.02 mol L-1
• Determine if ∆rG = 0 by finding Q (the reaction quotient).
• Q = [I3]/[I2][I] = (0.02) / (0.65)(1.30) = 0.0237
• Solve for Gibbs Free Energy: ∆G = ∆rG° + RT InQ = 16.54x103 J + (8.314 J K¯¹ mol-1 x 298 K x In (0.0237)) = 7270 J mol-1 = 72.7 kJ mol-1
• ∆G ≠ 0, therefore the system is not at equilibrium.

Le Chatelier's Principle

• Since the value of Gibbs Free Energy is positive, the position of equilibrium will move to the left in order to achieve equilibrium, causing the [I3¯] to decrease.