Copyright © Tyna L. Heise 2001 - 2002 Chapter Nineteen Copyright © Tyna L. Heise 2001 - 2002 All Rights Reserved

Spontaneous Processes Understanding and designing chemical reactions: How rapidly does the reaction proceed? - reaction rates - controlled by a factor related to energy - the lower the activation energy, the faster the reaction proceeds

Spontaneous Processes How far toward completion will the reaction go? - equilibrium constants - depends on rates of forward and reverse reactions - equilibrium should also be dependent on energy in some way due to dependence on reaction rates

Spontaneous Processes Chemical thermodynamics is the relationship between equilibrium and energy. First Law of Thermodynamics: for a reaction that occurs at constant pressure, the enthalpy change equals the heat transferred between the system and its surroundings Energy is conserved!! * Enthalpy is important in helping us determine if a reaction will proceed!

Spontaneous Processes energy is neither created nor destroyed in any process, energy can only be transferred or converted from one form to another DE = q + w a spontaneous process occurs without any outside intervention as energy is conserved

Spontaneous Processes

Spontaneous Processes temperature is going to effect the spontaneity of a process if discussing a phase change, at the temperature of the phase change, the phases compete for spontaniety, and neither is said to win over the other

Spontaneous Processes Sample exercise: Under 1 atm pressure, CO2(s) sublimes at -78°C. Is the transformation of CO2(s) to CO2(g) a spontaneous process at -100°C?

Spontaneous Processes Sample exercise: Under 1 atm pressure, CO2(s) sublimes at -78°C. Is the transformation of CO2(s) to CO2(g) a spontaneous process at -100°C? *No, if the temp had been higher it would change phase spontaneously, but lower than the sublimation point favors the reverse, so the solid remains a solid.

Spontaneous Processes Reversible and Irreversible Processes: State functions: define a state and do not depend upon the pathway temperature internal energy enthalpy

Spontaneous Processes Reversible and Irreversible Processes: Reversible processes is a unique way for a system to change its state, than go back to its state by following the exact same path but in the opposite direction phase changes at constant temp only one specific value of q (heat) system in equilibrium

Spontaneous Processes Reversible and Irreversible Processes: Irreversible processes cannot be simply restored to their original state using the same path, it may be forced to go back, but by a different pathway phase changes at different temps two q values need to be established qforward and qreverse any spontaneous reaction

Spontaneous Processes Thermodynamics can tell us direction of reaction extent of reaction NOT speed of reaction

Entropy and the 2nd Law Spontaniety depends upon two factors Enthalpy (DH): heat of reaction exothermic normally spontaneous endothermic normally NOT spontaneous Entropy (DS): disorder of the system natural law indicates reactions go in the direction that leads to more disorder

Entropy and the 2nd Law The Spontaneous Expansion of a Gas: When the stopcock is opened, the gas will spontaneously flow to fill the empty half, but it WILL NOT flow backward without work being done on the system.

Entropy and the 2nd Law The Spontaneous Expansion of a Gas: Gas expands because of the tendency for the molecules to ‘spread out’ among the different arrangements that they can take.

Entropy and the 2nd Law Entropy: measurement of randomness or chaos melting ice dissolving salts The more disordered or random a system, the larger its entropy

Entropy and the 2nd Law Sample exercise: Indicate whether each of the following reactions produces an increase or decrease in the entropy of the system: a) CO2(s)  CO2(g)

Entropy and the 2nd Law Sample exercise: Indicate whether each of the following reactions produces an increase or decrease in the entropy of the system: a) CO2(s)  CO2(g) Entropy increases

Entropy and the 2nd Law Sample exercise: Indicate whether each of the following reactions produces an increase or decrease in the entropy of the system: b) CaO(s) + CO2(g)  CaCO3(s)

Entropy and the 2nd Law Sample exercise: Indicate whether each of the following reactions produces an increase or decrease in the entropy of the system: b) CaO(s) + CO2(g)  CaCO3(s) Entropy decreases

Entropy and the 2nd Law Entropy: measurement of randomness or chaos state function for a process that occurs at constant temperature, the entropy change is dependent on the heat transferred during the reverse of the reaction (qrev) DS = qrev/T

Entropy and the 2nd Law Sample exercise: The normal boiling point of ethanol, C2H5OH, is 78.3°C, and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the change in entropy when 25.8 g of C2H5OH(g) at 1 atm pressure condenses to liquid at the normal boiling point?

Entropy and the 2nd Law Sample exercise: The normal boiling point of ethanol, C2H5OH, is 78.3°C, and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the change in entropy when 25.8 g of C2H5OH(g) at 1 atm pressure condenses to liquid at the normal boiling point? DS = qrev/T

Entropy and the 2nd Law Sample exercise: The normal boiling point of ethanol, C2H5OH, is 78.3°C, and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the change in entropy when 25.8 g of C2H5OH(g) at 1 atm pressure condenses to liquid at the normal boiling point? DS = qrev/T qrev = DHvap = 38.56kJ/mol T = 351.45 K

Entropy and the 2nd Law Sample exercise: The normal boiling point of ethanol, C2H5OH, is 78.3°C, and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the change in entropy when 25.8 g of C2H5OH(g) at 1 atm pressure condenses to liquid at the normal boiling point? DS = qrev/T 38.56kJ 1000 J 1 mol mol 1 kJ 46 g T = 351.45 K

Entropy and the 2nd Law Sample exercise: The normal boiling point of ethanol, C2H5OH, is 78.3°C, and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the change in entropy when 25.8 g of C2H5OH(g) at 1 atm pressure condenses to liquid at the normal boiling point? DS = qrev/T 838.26 J/g * 25.8 g = -21627 J T = 351.45 K

Entropy and the 2nd Law Sample exercise: The normal boiling point of ethanol, C2H5OH, is 78.3°C, and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the change in entropy when 25.8 g of C2H5OH(g) at 1 atm pressure condenses to liquid at the normal boiling point? DS = qrev/T = -21627 J/351.45 K = -61.5 J/K

Entropy and the 2nd Law Second law of Thermodynamics: In any reversible process, DSuniv = 0. In any irreversible reaction, DSuniv >0. DSuniv = DSsys + DSsurr In an isolated system, just the entropy of the system is considered.

Molecular Interpretation On the microscopic level, the number of gas molecules can be directly related to the amount of entropy in a system. The more gas molecules present, the higher the entropy value a phase change that increases the number of gas molecules would increase entropy a phase change that decreases the number of gas molecules would decrease entropy

Molecular Interpretation Three moles of gas combine to form two moles of gas, thus decreasing the number of molecules.

Molecular Interpretation Degrees of freedom creating new bonds decreases the freedom of movement atoms may have had. 3 degrees motion in one direction, translational movement vibrational movement spinning, rotational movement

Molecular Interpretation Figure 19.12

Molecular Interpretation Sample exercise: Choose the substance with the greatest entropy in each case: 1 mol of H2(g) at STP or 1 mol of H2(g) at 100°C and 0.5 atm.

Molecular Interpretation Sample exercise: Choose the substance with the greatest entropy in each case: 1 mol of H2O(s) at 0°C or 1 mol of H2O(l) at 25°C.

Molecular Interpretation Sample exercise: Choose the substance with the greatest entropy in each case: 1 mol of H2(g) at STP or 1 mol of SO2(g) at STP.

Molecular Interpretation Sample exercise: Choose the substance with the greatest entropy in each case: 1 mol of N2O4(g) at STP or 2 mol of NO2(g) at STP.

Molecular Interpretation Sample exercise: Predict whether is DS is positive or negative in each of the following processes: HCl(g) + NH3(g)  NH4Cl(s)

Molecular Interpretation Sample exercise: Predict whether is DS is positive or negative in each of the following processes: 2SO2(g) + O2(g)  2SO3(s)

Molecular Interpretation Sample exercise: Predict whether is DS is positive or negative in each of the following processes: cooling of nitrogen gas from 20°C to -50°C

Calculation of Entropy Changes Entropy Calculations: no easy method for measuring entropy experimental measurements on the variation of heat capacity with temperature can give a value known as absolute entropy zero point of reference for perfect crystalline solids tabulated as molar quantities, J/mol-K

Calculation of Entropy Changes Entropy differs from enthalpy standard molar entropies are not 0 entropies of gases are greater than those of liquids and solids entropies increase with molar mass entropies increase with number of atoms in formula DS° = nS(products) - mS(reactants)

Gibbs Free Energy Spontaneity involves both thermodynamic concepts: entropy and enthalpy DG = DH – TDS G = Gibbs Free Energy H = Enthalpy T = Temperature (K) S = Entropy

Gibbs Free Energy Gibbs Free Enegry If DG is negative, the reaction is spontaneous and proceeds in the forward direction If DG is zero, the reaction is at equilibrium If DG is positive, the reaction is nonspontaneous and proceeds in the reverse direction

Gibbs Free Energy Gibbs Free Enegry State function Table 19.3

Gibbs Free Energy Sample exercise: By using data from Appendix C, Calculate DG° at 298 K for the combustion of methane: CH4(g) + 2O2(g) ®CO2(g) + 2H2O(g)

Gibbs Free Energy Sample exercise: By using data from Appendix C, Calculate DG° at 298 K for the combustion of methane: CH4(g) + 2O2(g) ®CO2(g) + 2H2O(g) -50.8 0 -394.4 2(-228.57)

Gibbs Free Energy Sample exercise: By using data from Appendix C, Calculate DG° at 298 K for the combustion of methane: CH4(g) + 2O2(g) ®CO2(g) + 2H2O(g) -50.8 0 -394.4 2(-228.57) DG° = products - reactants

Gibbs Free Energy Sample exercise: By using data from Appendix C, Calculate DG° at 298 K for the combustion of methane: CH4(g) + 2O2(g) ®CO2(g) + 2H2O(g) -50.8 0 -394.4 2(-228.57) DG° = products – reactants (-394.4 + 2(-228.57)) – (-50.8 + 0)

Gibbs Free Energy Sample exercise: By using data from Appendix C, Calculate DG° at 298 K for the combustion of methane: CH4(g) + 2O2(g) ®CO2(g) + 2H2O(g) -50.8 0 -394.4 2(-228.57) DG° = products – reactants (-394.4 + 2(-228.57)) – (-50.8 + 0) -800.7

Gibbs Free Energy Sample exercise: Consider the combustion of propane to form CO2(g) and H2O(g) at 298K. Would you expect DG° to be more negative or less negative than DH°?

Gibbs Free Energy Sample exercise: Consider the combustion of propane to form CO2(g) and H2O(g) at 298K. Would you expect DG° to be more negative or less negative than DH°? more negative, using Gibbs Free Enegry formula, more moles of gas being produced would be increasing entropy, +DS,

Free Energy and Temperature Table 19.4

Free Energy and Temperature Sample exercise: Using standard enthalpies of formation and standard entropies in Appendix C, calculate DH° and DS° at 298 K for the following reaction: 2SO2(g) + O2(g) ® 2SO3(g)

Free Energy and Temperature Sample exercise: Using standard enthalpies of formation and standard entropies in Appendix C, calculate DH° and DS° at 298 K for the following reaction: 2SO2(g) + O2(g) ® 2SO3(g) 2(-296.9) 0 2(-395.2) DH = 2(-395.2) - 2(-296.9) = -196.6 kJ

Free Energy and Temperature Sample exercise: Using standard enthalpies of formation and standard entropies in Appendix C, calculate DH° and DS° at 298 K for the following reaction: 2SO2(g) + O2(g) ® 2SO3(g) 2(248.5) 205.0 2(256.2) DS = 2(256.2) – (2(248.5)+205.0) = -189.6 J

Free Energy and Temperature Sample exercise: Using the values obtained, estimate DG° at 400 K DG = DH – TDS

Free Energy and Temperature Sample exercise: Using the values obtained, estimate DG° at 400 K DG = DH – TDS = -196.6 kJ – 400(-0.1896 kJ) = -120.8

Free Energy and the Equilibrium Constant 2 other important ways free energy is a powerful tool Tabulate free energy under nonstandard conditions Directly relate free energy to equilibrium constants

Free Energy and the Equilibrium Constant Tabulate free energy under nonstandard conditions DG = DG° + RTlnQ R = 8.314 J/mol-K T = temperature K Q = reaction quotient

Free Energy and the Equilibrium Constant Sample exercise: Calculate DG at 298 K for the reaction of nitrogen and hydrogen to form ammonia if the reaction mixture consists of 0.50 atm N2, 0.75 atm H2, and 2.0 atm NH3. DG = DG° + RTlnQ

Free Energy and the Equilibrium Constant Sample exercise: Calculate DG at 298 K for the reaction of nitrogen and hydrogen to form ammonia if the reaction mixture consists of 0.50 atm N2, 0.75 atm H2, and 2.0 atm NH3. DG = DG° + RTlnQ 2(-16.6) + 0.008314(298)ln(2.02/0.50(0.753)) -26.0 kJ

Free Energy and the Equilibrium Constant At equilibrium, DG is equal to 0 so… DG° = -RTlnKeq DG° negative: K > 1 DG° zero : K = 1 DG° positive: K < 1 Keq = e-DG°/RT