Unit 5: Spontaneity of Reaction You need to bring your ...cpeachee.weebly.com/uploads/5/9/9/2/5992251/spontaneitynotes... · You need to bring your textbooks everyday of this unit

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March 05, 2013

Unit 5: Spontaneity of Reaction

You need to bring your textbooks everyday of this unit.

THE LAWS OF THERMODYNAMICS

1st Law of Thermodynamics à Energy is conserved ΔE = q + w

2nd Law of Thermodynamics à A system NOT at equilibrium will move toward it.

3rd Law of Thermodynamics à The entropy of a pure crystalline substance at absolute zero


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Processes that are spontaneous in one direction are NOT spontaneous in the other direction.

*Spontaneous does NOT mean RAPID. Be careful not to mix up thermodynamics with kinetics.

Thermodynamics tells us about direction and extent of movement, but we need KINETICS to tell us about rate.

Together, kinetics and thermodynamics describe a reaction.

Standard Heat of Formation: ΔHf° compounds formed from elements in their standard states.

Exothermic heats of formation are (), endothermic are (+).

Heat of formation of an element in its standard state = 0.

Enthalpy and SpontaneityA spontaneous change involves going to a lower energy state.

Endothermic and Exothermic changes can be spontaneous.

Spontaneous changes often occur for combustion reactions, neutralization reactions, melting, dissolving.


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Here are some specific examples:

Combustion: CH4(g) + 2O2(g) à CO2(g) + 2H2O(g) ΔHf = 890.4 kJ

Neutralization: H+(aq) + OH(aq) à H2O(l) ΔHf = 55.9 kJ

Melting: H2O(s) à H2O(l) ΔHf = +6.01 kJ

Dissolving: NH4NO3(s) àNH4+(aq) + NO3(aq) ΔHf = +25 kJ

An exothermic enthalpy change favors spontaneity, but another thermodynamic

property needs to be considered, entropy

Entropy: Symbol is S Units of S = J/mol K (vs. kJ/mol for ΔH)

It is the measure of order/disorder, the higher the degree of disorder, the greater the entropy.

Second Law of Thermodynamics restated: When a spontaneous change occurs, the entropy of the universe as a whole increases. The driving force of any spontaneous change is an increase in the entropy of the universe. There is a drive toward the states that have the highest probability of existing.


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Systems will move toward the state with the most possible arrangements or the highest probability of existing.

Third Law of Thermodynamics: the entropy of a pure crystalline substance at absolute zero is zero. ie: S (at 0 K) = 0

Entropy changes:

ΔS = Sfinal Sinitial (Δ is always final initial)

ΔS > 0 (ΔS = +)

disorder is increasing, system becoming less organized, more random.

ΔS < 0 (ΔS = )

disorder is decreasing, system getting more organized.

ΔS > 0 is thermodynamically favorable

disorder can be reduced by energy input.


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Entropy Changes for Physical Phase Changes

For a given substance: S(s) < S(l) < S(g)

For freezing (l to s) molecules are more ordered so ΔS is negative.

Entropy Changes for Chemical Reactions

1. NH3(g) + HCl(g) à NH4Cl(s), ΔS is negative gas to solid more order

2. 2O3(g) à 3O2(g), ΔS positive more moles of gas after reaction

3. H2(g) à 2H(g), ΔS is positive more moles of gas after reaction

4. Ag+(aq) + Cl(aq) à AgCl(s), ΔS negative solid more order than aqueous solutions ions


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S increases as …

volume increases

particle mobility increases (s à l à g)

temperature increases

particles dissolve

moles of gas increases

Example 17.1 à Which of the following involve an increase in the entropy?

a. melting a solid

b. sublimation

c. freezing

d. mixing

e. separation

f. boiling


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Example 17.2 à Predict the sign of ΔS° for each of the following changes

a. Na (s) + ½ Cl2 (g) à NaCl (s)

b. N2 (g) + 3H2 (g) à 2NH3 (g)

c. NaCl (s) à Na1+ (aq) + Cl1 (aq)

d. NaCl (s) à NaCl (l)

e. CaCO3 (s) à CaO (s) + CO2 (g)

f. 2SO2 (g) + O2 (g) à 2SO3 (g)

Practice Problems 1- 10 in textbook.


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Calculating ΔS°rxn (° = standard thermodynamic conditions, 1 atm, 25°C, 1 M solution)

ΔS°rxn = Σ S°products Σ S°reaction

S° = absolute entropy of a substance at 25°C (can be calculated based on the fact that S° = 0 at 0 K)

*S° for an element in its standard state is NOT zero.

(No element has a standard state defined where T = 0 K.)

Example 17.3 à Predict the sign of ΔS° then calculate ΔS° for each of the following

reactions.

a. 2SO3 (g) à 2SO2 (g) + O2 (g)

b. Fe2O3 (s) + 3H2 (g) à 2Fe (s) + 3H2O (g)

c. H2 (g) + ½ O2 (g) à H2O (l)


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Example 17.3 à Predict the sign of ΔS° then calculate ΔS° for each of the following

reactions. (continued)

d. N2 (g) + 3H2 (g) à 2NH3 (g)

e. HCl (g) à H1+ (aq) + Cl1 (aq)


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Example 17.4 à For the reaction: C2H2 (g) + 4F2 (g) à 2CF4 (g) + H2 (g) ΔS° is equal to 358 J/K. Use this value and data from Thermodynamic Data Sheet to calculate S° for CF4 (g)

Practice Problems 12-16 evens in

textbook.


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The Two Factors Affecting Spontaneity

1. Tendency to go to a lower enthalpy.

2. Tendency to go to a higher entropy.

Gibb's Free Energy: ΔG = ΔH TΔS

ΔG = energy free to do useful work

ΔH = change in heat content (enthalpy)

TΔS = energy associated with changing order/disorder

ΔG depends on the nature of products, reactants, and the conditions (P, T, and concentration) not on the path for which the the rx. is carried out.

Two Ways to Calculate ΔGorxn

1. Use ΔG°rxn = ΔH° TΔS° where temperature is in Kelvin. (K = oC + 273.16)

Calculate à ΔH°rxn = Σ ΔH°prod Σ ΔH°react and

Calculate à ΔS°rxn = Σ S°prod Σ S°reactSubstitute and calculate à ΔG°rxn = ΔH° TΔS°

2. Use ΔG°rxn = Σ ΔG°f prod Σ ΔG°f reactΔG°f for an element in its standard state is 0 kJ/mol.


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Relationship between ΔG° and spontaneity:

ΔG° = (), spontaneous

ΔG° = 0, equilibrium

ΔG° = (+), nonspontaneous, spontaneous in the reverse rx.

What Does Spontaneous or Nonspontaneous Mean?

1. Physical change: if spontaneous, these go to completion, 100% change.

There is no tendency for an ice cube to stop melting at 10°C.

H2O(s) à H2O(l) at 10°C

2. Chemical reactions: Spontaneous means a reaction that gives predominantly products.

The system may reach equilibrium, the right side will be favored.


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Nonspontaneous does not mean "no reaction." Here, spontaneous, nonspontaneous does not mean "go or no go." Nonspontaneous means the reaction mixture is mostly reactants

Nonspontaneous reactions can be made to occur by use of an outside energy source.

An example, electrolysis of water and other electrolytic reactions.

Qualitative Relationship Between ΔG° and Position at Equilibrium

1. When ΔG° is ()

The more () the ΔG°, the higher the % reaction.

Large ΔG° rxns go more toward completion than low ΔG° reactions.

Reactants are always higher than products, but the bottom of the curve skews

left or right between 50 and 100%.

2. When ΔG° is (+)

The more (+) the ΔG°, the lower the % reaction.

Reactants are always lower than products, but the bottom of the curve skews

left or right between 0 and 50%.

The more ()ΔG°rxn , the higher the % completion of the reaction at 25°C.

The more (+)ΔG°rxn , the lower the % completion of the reaction at 25°C.

Reactions that are nonspontaneous in one direction will be spontaneous in the other.

Also, remember that conditions will often not be standard.


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Example 17.5 à Consider the reaction 2SO2 (g) + O2 (g) à 2SO3 (g) carried out at

25 °C and 1 atm. Calculate ΔH°, ΔS°, and ΔG°.

Example 17.6 à Methanol, CH3OH (l), is a highoctane fuel used in highperformance

racing engines. Calculate ΔG°f for it’s combustion reaction. Note water vapor is a product.


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March 05, 2013

Example 17.7 à A chemical engineer wants to determine the feasibility of making ethanol,

C2H5OH by reacting water ethylene C2H4, according to the equation:

C2H4 (g) + H2O (l) à C2H5OH (l)

Is this reaction spontaneous under standard conditions?

Example 17.8 à Given the following data

S (s) + 3/2 O2 (g) à SO3 (g) ΔG° = 371 kJ

2SO2 (g) + O2 (g) à 2SO3 (g) ΔG° = 142 kJ

Calculate ΔG° for the reaction: S (s) + O2 (g) à SO2 (g)


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Ex 17.5 pg. 456 Calculate ΔG at 230 °C for the reduction of one mole of Fe2O3 with hydrogen; the products are iron metal and water vapor

Homework #3pgs. 468-469

18a, 18b, 20a, 20b, 22a, 24 all, 26 all, 28 all, 55 all, and 56 all


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March 05, 2013

Example 17.9 à #29 for our textbook: The alcohol in most liqueurs is ethanol,

C2H2OH. It is produced by the fermentation of the glucose in fruit or grain.

C6H12O6 (aq) à 2C2H5OH (l) + 2CO2 (g)

ΔH° = 82.4 kJ ΔG° = 219.8 kJ at 25 °C

a. Calculate ΔS° for this reaction. Is the sign reasonable

b. Calculate S° for C6H12O6 (aq)

c. Calculate ΔHf° for C6H12O6 (aq)

c. Calculate ΔHf° for C6H12O6 (aq)

Example 17.9 à #29 for our textbook: The alcohol in most liqueurs is ethanol,

C2H2OH. It is produced by the fermentation of the glucose in fruit or grain.

C6H12O6 (aq) à 2C2H5OH (l) + 2CO2 (g)

ΔH° = 82.4 kJ ΔG° = 219.8 kJ at 25 °C


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March 05, 2013

The Effect of Temperature on Spontaneity

ΔH° and ΔS° are virtually temperature independent.

So, ΔG = ΔH° T ΔS° can be used to calculate ΔG at any temp. Temperature is considered in T ΔS°.

1. ΔG = ΔH° T ΔS° ΔH° and + ΔS°

2. ΔG = ΔH° T ΔS° + ΔH° and ΔS°

3. ΔG = ΔH° T ΔS° + ΔH° and + ΔS°

4. ΔG = ΔH° T ΔS° ΔH° and ΔS°

Situations 2 and 3 where the signs for ΔH° and ΔS° are ALIKE are temperature dependent.

Situations 1 and 4, where the signs are OPPOSITE are not temperature dependent.


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Calculating the Temperature at Which a Reaction Becomes Spontaneous/Nonspontaneous

ΔG = ΔH° T ΔS°

Example 17.11 à #33a from textbook

Discuss the effects of temperature change on the spontaneity of the

following reactions at 1 atm.

2PbO (s) + 2SO2 (g) à 2PbS (s) + 3O2 (g)

ΔH° = +830.8 kJ ΔS° = +168 J/K


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March 05, 2013

35a from textbook

At what temperature does ΔG° become zero for the above reaction. Explain the significance of your answer

Example 17.11 à #33b from textbook

(continued) Discuss the effects of temperature change on the spontaneity of the


2As (s) + 3F2 (g) à 2AsF3 (l)

ΔH° = 1643 kJ ΔS° = 0.316 kJ/K


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March 05, 2013

35b from textbook


33c from textbook

Discuss the effects of temperature change on the spontaneity of the


CO (g) à C (s) + ½ O2 (g)

ΔH° = +110.5 kJ ΔS° = 89.4 J/K


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March 05, 2013

35c from textbook


Example 17.12 à #43 from textbook

Red phosphorus is formed by heating white phosphorus. Calculate the

temperature at which the two forms are at equilibrium.

White P: ΔHf° = 0.00 kJ/mol S ° = 41.1 J/mol K

Red P: ΔHf° = 17.6 kJ/mol S ° = 22.8 J/mol K


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Homework #4pgs. 469-470

34, 36, 38, 42, 44, 46, 48

ΔG = ΔG° + RTlnQ

ΔG = ΔH TΔS

ΔG = Σproducts Σreactants


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March 05, 2013

Relationship Between ΔG, Equilibrium Expressions, and Reaction Quotients

ΔG = ΔG° + RTlnQ

ΔG° = standard free energy change

(standard conditions: P= 1 atm, [ ] = 1M, T = 25°C or 298K)

ΔG = free energy under any conditions.

T = temp in Kelvins

Q = Reaction quotient, from the equilibrium constant equation. gases represented by pressure aqueous solutions represented by molar concentration, (solids, liquids do not get used, consider them to be the value of 1)


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Reaction Quotient aA(g) + bB (g) ó cC(g) + dD (g)

Q = (PCc) X (PDd)OR Q = [C]c X [D]d (PAa) X (PBb) [A]a X [B]b

R = 8.314 X 103 kJ/mol K (8.314 J/mol K)

ΔG = ΔG° + RTlnQ

If conditions are NOT standard, ΔG must be calculated, NOT ΔG°.

ΔG changes as the reaction proceeds, with changes in Q. Driving forces (H and S) change until equilibrium is achieved.

Example 17.12 à Consider the above reaction: Zn (s) + 2H1+(aq) à Zn2+ (aq) + H2 (g) at 25 °C. Calculate

a. ΔG°


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b. ΔG when PH2 = 750 mm Hg, [Zn2+] = 0.10 M, [H1+] = 1.0 X 104 M


c. [H1+] when ΔG = 100.0 kJ and all other species involved in the reaction are at standard conditions.


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March 05, 2013

The Free Energy Change and the Equilibrium Constant

At equilibrium, ΔG = 0, and Q = K.

ΔG is a true measure of spontaneity under any conditions.

At equilibrium, ΔG = 0 = ΔG° + RTlnK, therefore

ΔG° = RTlnKNote: K is equilibrium constant (Kgas, Kw, Ka, Kb, Ksp, …)

A frequently confusing issue results from the fact that ΔG° implies 25°C (298K), however ΔG° may be calculated for other temperatures.

The position of equilibrium is related to ΔG°.If K > 1, ΔG° is (), mostly product and spontaneousIf K = 1, ΔG° = 0, system at equilibriumIf K < 1, ΔG° = (+), mostly reactants and nonspontaneous.


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Example 17.13 à Using ΔG°f tables in Appendix 1, calculate the solubility product constant, Ksp, for PbCl2 at 25 °C

Operations Involving Logarithms

Multiplication: ln (xy) = ln x + ln y

Division: ln (x/y) = ln x – ln y

Raising to a Power: ln (xn) = n ln x

Extracting a Root: ln(x1/n) = 1 ln x n

Taking a Reciprocal: ln (1/x) = ln x


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Unit 5: Spontaneity of Reaction You need to bring your ...cpeachee.weebly.com/uploads/5/9/9/2/5992251/spontaneitynotes... · You need to bring your textbooks everyday of this unit