chapter 19: chemical thermodynamics

Chapter 19: Chemical Thermodynamics

Tyler BrownHailey Messenger

Shiv PatelAgil Jose

19.1 Spontaneous Processes


Spontaneous Process:• A process that proceeds on its own without any

outside assistance• Occurs in a definite direction

“Processes that are spontaneous in one direction are nonspontaneous in the other”

(Pg 805 Brown and LeMay)


Reversible Process:• A process in which the system is changed in a way

that the system and surroundings can be restored to their original state by exactly reversing the change (can be completely restored to original condition)

• Reversible processes reverse direction whenever an infinitesimal change is made in some property of the system


Irreversible Process:• Cannot be reversed to return system and

surroundings to original state


Isothermal Process:• A process with a constant temperature


• First Law of Thermodynamics: Energy is conserved

• “Physical and chemical processes have a directional character” (Pg 804)

• I.e. Sodium and Chlorine come together on their own to make NaCl, but it does not decompose of its own accord.


• Experimental conditions help to determine if a process is spontaneous

• I.e. When the temperature > 0°C in ordinary atmospheric pressure, ice melting is spontaneous.

» In these conditions, liquid water turning in to ice is not spontaneous


• Entropy: The ratio of heat delivered to the temperature at which it is delivered

• “A reversible change produces the maximum amount of work that can be achieved by the system on the surroundings (wrev= wmax)” (Page 807)

• All spontaneous processes are IRREVERSIBLE


Tell whether or not the following process is spontaneous

Ice melts at -1 degree Celsius


Tell whether or not the following process is spontaneous

Ice melts at -1 degree Celsius

This would not be spontaneous because Ice does not melt under 0 degrees Celsius. This reaction would be spontaneous in the reverse reaction


• 19.1 Problems: 7-18

19.2 Entropy and the Second Law of Thermodynamics

19.3 The Molecular Interpretation of Entropy

Molecular motion

• Translational Motion – entire molecule moves in one direction (like throwing a baseball)

• Vibrational Motion – atoms in the molecule periodically move toward and away from one another

• Rotational Motion – molecules spin like a top

What is Statistical Thermodynamics?

• It’s the use of tools of statistics and probability to provide the link between macroscopic and microscopic worlds

• We molecules in bulk in a microstate: a single possible arrangement of the positions and kinetic energies of gas molecules in a specific thermodynamic state

So what equation do we use?

• Boltzmann created this beautiful equation:

S = k lnW

K is Boltzmann’s constant which is: 1.38 x 10-23J/K

-This means entropy is the measure of how many microstates are associated with a particular macroscopic state

Relationship between microstates and entropy

• The more microstates, the more entropy• Increasing volume, temperature, and

number of molecules increases entropy due to larger number of microstates

GENERALLY SPEAKING

Entropy increases when:- Gases are formed from either solids or

liquids- Liquids or solutions are formed from solids- The number of gas molecules increases

during a chemical reaction

The Third Law of Thermodynamics

• The entropy of a pure crystalline substance at absolute zero is zero

S(0 K) = 0

- Makes sense because when there is no molecular motion (temp is 0 K), there is only ONE microstate

Trends

• Sharp increase of entropy at melting points• Sharp increase of entropy at boiling points

Entropy of solids < Entropy of liquids < Entropy of gas

Example

• What will the sign of change of entropy be for the following reactions?

2Na(s) + Cl2 (g) 2NaCl (s)

2H2 (g) + O2 (g) 2H2O (l)

Answers • Because of Boltzmann’s equation, we know that

the more particles there are in a system, the more microstates there are, so the more entropy there is, so:

• The reaction to form NaCl has a negative sign for change in entropy, since there are less particles after the reaction

• The reaction to form water also has a negative sign for change in entropy since there are less particles

Example

• In a chemical reaction two gases combine to form a solid. What do you expect for the sign of change in S? For which of the processes does the entropy of the system increase? A)The melting of ice cubes at -5 degrees Celsius and 1 atm pressure B) dissolution of sugar in a cup of hot coffee C) the reaction of nitrogen atoms to form N2

Answer• Entropy decreases when two gases form a

solid, so negative sign• A) Positive because the ice is melting from a

structured solid to unstructured liquid• B) Positive because the sugar went from

being structured solids to numerous separated particles

• C) Negative, there is a huge decrease in the number of gas particles (by 1/2 to be exact)

Practice, practice, practice!

• Page 838 #’s 19.27 – 19.40• A closer look on page 810

19.4 Entropy Changes in Chemical Reactions

How do we do it?

• We can’t measure change in entropy for a reaction, but we CAN find the absolute value of the entropy for many substances at any temperature

• Standard molar entropies are denoted by S° (pronounced S knot) for any pure substance in their standard state at 1 atm

Observations on Standard Entropy

• Standard molar entropies (SME) of elements at reference temperature 298 K are not zero

• SME of gases greater than liquids• SME generally increase with molar mass• SME generally increase with number of

atoms in molecular formula

So what’s the equation we use?

ΔS° = ΣnS(products) – ΣnS(reactants)

In englishChange in entropy equals sum of entropies

of products minus the sum of entropies of reactants

Entropy Changes in the Surroundings

• The change in entropy of the surroundings will depend on how much heat is absorbed or given off by the system

• Isothermal process equation:

ΔSsurr=-qsys/T

- At constant pressure, qsys is the enthalpy change of the system, ΔH

Example

• Calculate change in standard entropy for the following reaction

• 2Na(s) + Cl2 (g) 2NaCl (s)

Answer• 1. The standard entropy change is equal to the standard entropy of

products minus the reactants

2. So we must first find the standard entropy of the products(using a standard entropy table), and multiply them by the number of moles in the reaction

Standard entropy of NaCl = 72.1 J/K-mol, so 2(72.1)= 144.2

3. Then we find the standard entropy of reactantsStandard entropy of Na = 51.2 J/K-mol, so 2(51.2)= 102.4 J/K-molStandard entropy of Cl2 = 223.1 J/K-mol, 223.1 J/K-mol

4. Now we subtract product and reactants144.2 J/K-mol – (102.4 J/K-mol + 223.1 J/K-mol) = - 181.3 J/K-mol

Example

• Cyclopropane and propylene isomers both have the formula C3H6. Based on the molecular structures shown, which of these isomers would you expect to have the higher standard molar entropy at 25 degrees Celsius?

Cyclopropane Propylene

Answer

• Propylene because it is much less organized compared to Cyclopropane

We love to practice!

• Page 839 #’s 19.41 – 19.48• Example on page 821, figure 19.5

19.5 Gibbs Free Energy


• Standard Free Energies of Formation: Used to calculate standard free-energy change for chemical processes


• Equation for Standard Free-Energy Change:

ΔG° = ΣnΔG°f (products) - ΣmΔG°f

(reactants)


• G = H – TS

• T is the absolute temperature.

• The change in free energy of a system (ΔG) is given by: ΔG= ΔH – TΔS


• If ΔG is negative, the reaction is spontaneous

• If ΔG is zero, then reaction is at equilibrium

• If ΔG is positive, the forward reaction is not spontaneous, but the reverse reaction is.


• For a certain chemical reaction, ΔH = -35.4 kJ and ΔS = - 85.5 J/K. Calculate ΔG for the reaction at 298 K.


• For a certain chemical reaction, ΔH = -35.4 kJ and ΔS = - 85.5 J/K. Calculate ΔG for the reaction at 298 K (19.51 C)

ΔG = ΔH –TΔSΔG = - 35.4 kJ – (298K)(-85.5 J/K)ΔG = - 35.4 kJ + 25500 JΔG = -35.4 kJ + 25.5 kJΔG = -9.9 kJ


• 19.5 Problems: 49 -70

19.6 Free Energy and Temperature


•The equation ΔG=ΔH+(–TΔS) in which ΔH is the enthalpy term and –TΔS is the entropy term, can be used to demonstrate how free energy is affected by the change in temperature.•The value of –TΔS depends directly on the absolute temperature, T, ΔG will vary with the temperature.


• T is a positive number at all temperatures other than absolute zero.

• Both the enthalpy and entropy can be either positive or negative.

• When ΔS is positive, which means greater randomness than the original state, then the TΔS is negative. When the ΔS is negative, then the TΔS is positive.


• ΔG will always be negative when the process is spontaneous, and the opposite is true when the process is not spontaneous.


• H2O(s) → H2O(l) ΔH>0 ΔS>0• This process is endothermic, meaning the

ΔH is positive. Entropy also increases with this process, meaning that ΔS is positive as well, which makes –TΔS a negative value. This means that the –TΔS term dominates, making ΔG negative. A negative ΔG means that this process is spontaneous at T>0ºC


• Calculate ΔG for 2NO2(g) → N2O4(g) at 298K using appendix C.


• ΔH 2NO2(g) → N2O4(g) = -58.02 kJ/mol

• ΔS 2NO2(g) → N2O4(g) = -175.79 J/K•


• ΔGº =• -58.02kJ – (298K)(-175.79J/K)(1kJ/1000J)• -58.02kJ – (298K)(-0.176kJ/K)• -58.02kJ – (-52.45kJ)• ΔGº = -5.57kJ• This means that at 298K, • 2NO2(g) → N2O4(g) is a spontaneous reaction.


• Textbook problems: chapter 19, 65,66,79. 80, and 87

19.7 Free Energy and the Equilibrium Constant

Most reactions occur in nonstandard conditions

• To find the free energy in nonstandard reactions, we use the standard free energy change to calculate

ΔG = ΔG° + RT ln(Q)R is ideal gas constant = 8.314 J/mol-KT is absolute temperatureQ is reaction quotient for that particular

mixture of interest

Some more info

• At equilibrium ΔG = 0• Q = equilibrium constant K

Trends

• The more negative ΔG°, is the larger K is• The more positive ΔG° is, the smaller K is

Example

• 2CO(g) + O2 (g) 2CO2 (g)

• Calculate the change in free energy for the above reaction at 298K, for a reaction mixture that consists of 1.0 atm O2 , 2.0 atm CO , and 0.75 atm CO2

Answer1. First we must find Q in order to use the equation provided

earlier. So we solve for the reaction quotient:Q=(PCO2 )2 / (PCO)2(PO2 )=(.75 atm)2 / (2.0 atm)2 (1.0 atm) = .14atm

2. Now calculate standard change in free energy for reaction using tables

Products – Reactants, so 2(-394.4kJ/mol) – (2(-137.2kJ/mol) + 0)Change in standard free energy is -514.4 kJ/mol3. Now we can finally use ΔG = ΔG° + RT ln(Q) ΔG = (-514.4 kJ/mol) + (8.314 J/mol-K)(298 K) (1 kJ/1000J)ln(.14)

= (-514.4 kJ/mol) – 4.871 kJ/mol = -519.271 = - 520 kJ/mol

Practice Problems

• Page 841• Problems 19.71 – 19.95

• You can find answers on pages A-24 thru A-25

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chapter 19: chemical thermodynamics

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