Thermodynamics of Reactions Spontaneity, Entropy, and Free Energy Chapter 16

1 st Law of Thermodynamics The energy of the universe is constant (conservation of energy)

Spontaneous Processes Can be fast OR slow. Occurs without outside intervention

True, or Not True? Ball rolls downhill spontaneously. Ball rolls uphill spontaneously. Wood burns spontaneously in O 2. CO 2 and H 2 O spontaneously form wood. Steel rusts spontaneously. Rust spontaneously turns into iron and water.

Its All About Entropy Entropy is… All about the ability of energy to spread out Related to probability. Over-simplified to be about disorder.

Positional Entropy When a substance has more opportunities to exist, (ie, particles can have more than one arrangement) it is also said to have more positional entropy. Increasing moles of gas Increasing volume Decreasing pressure Changing state of matter ( s  l  g ) Exothermic rxns **

Try Me Conceptual Determine which of the following pairs has the most positional entropy: − 1 mol H 2 at STP or 1 Mol H 2 at 100 o C, 0.5 atm − 1 mol N 2 at STP or 1 mol N 2 at 100 K 2atm − 1 mol H 2 O (s) at 0 o C or 1 mole H 2 O (l) at 20 o C

2 nd Law of Thermodynamics Spontaneous processes increase the entropy of the universe.  S universe =  S system +  S surroundings

The Value of S Sign is determined by enthalpy: − exothermic − endothermic Magnitude is determined by temperature

Try Me Calculation The melting point of tungsten (W) is the second highest among the elements, at 3680 K. The enthalpy of fusion for this metal is 35.2 kJ/mol. What is the entropy of fusion?

3 rd Law of Thermodynamics The entropy of a perfect crystal at 0 Kelvin is zero. In Phase Out of Phase Angle Bending Rotation Translation

Entropy for a chemical reaction  S o rxn =  nS o p -  nS o r

Try Me Calculate the change in entropy at 25 o C for the reaction: 2 NiS + 3 O 2  2 SO NiO Given Entropy Values: SO 2 = 248 J/Kmol NiO = 38 J/Kmol O 2 = 205 J/Kmol NiS = 53 J/Kmol

Free Energy Gibbs Free Energy- another method for determining spontaneity. Also indicates the amount of available energy that is capable of doing work. As an energy source is used, the energy is not destroyed, only converted to a non-usable form.

 G =  H - T  S Defines Gibbs energy in terms of enthalpy and entropy. All three factors will contribute to reaction spontaneity. When  G is negative, the reaction is spontaneous.

Rearrange that formula Lets start with the following 3 formulae:  G =  H –T  S  S surr = -  H/T  S univ =  S surr +  S syst  S universe = -  G /T

Fancy Pants Charts HHHH SSSS GGGG Spontaneous? __+ Always __ +/-When temp is low +++/-When temp is high +__+Never

Free Energy in Chemical Reactions Standard Free energy is used so that we can compare the relative tendency to occur.  G o =  H o – T  S o

2SO 2(g) + O 2(g)  2SO 3(g) The above rxn. Occurs at 25 o C and 1 atm. Calculate  H o,  S o, and  G o using the following data: substance  H o f S o kJ/molJ/Kmol SO 2(g) SO 3(g) O 2(g) 0205

It also works like Hess’s Law problems: What is the  G o for the reaction if the mechanism is: C di(s) + O 2(g)  CO 2(g)  G o = -397kJ C gr(s) + O 2(g)  CO 2(g)  G o = -394 kJ

Dependence on Pressure Enthalpy does not depend on pressure. Entropy does depend on temperature. S low pressure > S high pressure

 G =  G o + RT ln(P) This can be adapted to reflect partial pressures for the reaction it describes. When that occurs, the formula can be re-written:  G o = -RT ln(Q)

@ Equilibrium K = Q The free energy is the lowest possible it will ever be for the system.

Try Me Out The overall rxn for rusting iron by oxygen is 4Fe (s) + 3O 2(g)  2 Fe 2 O o C, find the equilibrium constant given: Substance  H o S o Fe2O3(s) Fe(s) 027 O2(g) 0 205

Free Energy and Work Achieving the maximum amount of work from a process is highly unlikely because of transfers of energy.  G represents the maximum possible quantity of work a system is capable of doing.