New chm 152 unit 6 power points sp13

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THERMODYNAMICS

• Internal Energy• Enthalpy• Entropy• Free Energy• Chapter 17 (McM)• Chapter 20 Silberberg

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Chapter 20

Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions

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Goals & Objectives

• See the following Learning Objectives on page 914.

• Understand these Concepts:• 20.1-22.

• Master the Skills:• 20.1-10.

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Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions

20.1 The Second Law of Thermodynamics: Predicting Spontaneous Change

20.2 Calculating Entropy Change of a Reaction

20.3 Entropy, Free Energy, and Work

20.4 Free Energy, Equilibrium, and Reaction Direction

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Thermodynamics• the study of the changes in energy and the

transfers of energy that accompany chemical and physical processes.

• Addresses three fundamental questions– Will 2 or more substances react when they are

mixed under specified conditions?– If they do react, what energy changes and

transfers are associated with their reaction?– If a reaction occurs, to what extent does it

occur?

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Thermodynamics

• Used to determine if a reaction can occur under specified conditions.– spontaneous reaction--can occur under the

specified conditions– nonspontaneous reaction--do not occur to a

significant extent under the specified conditions

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First Law of Thermodynamics

• The internal energy of an isolated system is constant.

• The total amount of energy in the universe is constant.

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Some Thermodynamic Terms

• System--the substances involved in the chemical and physical changes under investigation

• Surroundings--the rest of the universe• Universe--the system and its

surroundings

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Types of Thermodynamic Systems

• Open system--can exchange both matter and energy with its surroundings

• Closed system--has a fixed amount of matter but can exchange energy with its surroundings

• Isolated system--has no contact with its surroundings

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Thermodynamic State of a System

• defined by a set of conditions that completely specifies all the properties of the system

• State functions--the properties of a system( pressure, temperature, energy, e.g.) whose values depend only on the state of the system

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Spontaneous Change

A spontaneous change is one that occurs without a continuous input of energy from outside the system.

All chemical processes require energy (activation energy) to take place, but once a spontaneous process has begun, no further input of energy is needed.

A nonspontaneous change occurs only if the surroundings continuously supply energy to the system.

If a change is spontaneous in one direction, it will be nonspontaneous in the reverse direction.

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The First Law of Thermodynamics Does Not Predict Spontaneous Change

Energy is conserved. It is neither created nor destroyed, but is transferred in the form of heat and/or work.

DE = q + w

The total energy of the universe is constant:

DEsys = -DEsurr or DEsys + DEsurr = DEuniv = 0

The law of conservation of energy applies to all changes, and does not allow us to predict the direction of a spontaneous change.

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DH Does Not Predict Spontaneous Change

A spontaneous change may be exothermic or endothermic.

Spontaneous exothermic processes include:• freezing and condensation at low temperatures,• combustion reactions,• oxidation of iron and other metals.

Spontaneous endothermic processes include:• melting and vaporization at higher temperatures,• dissolving of most soluble salts.

The sign of DH does not by itself predict the direction of a spontaneous change.

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Figure 20.1 A spontaneous endothermic chemical reaction.

water

Ba(OH)2 ·8H2O(s) + 2NH4NO3(s) → Ba2+(aq) + 2NO3-(aq) + 2NH3(aq) + 10H2O(l)

DH°rxn = +62.3 kJ

This reaction occurs spontaneously when the solids are mixed. The reaction mixture absorbs heat from the surroundings so quickly that the beaker freezes to a wet block.

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Enthalpy Change

= DH SDHfo (products)

SDHfo(reactants)

• where DHfo is the standard molar

enthalpy of formation and DH is the enthalpy change for the reaction.

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Enthalpy Change

• Calculate the enthalpy change for the following reaction at 298K.

• C3H8(g) + 5O2(g) ----> 3CO2(g) + 4H2O(l)

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The Second Law of Thermodynamics

• In spontaneous changes the universe tends toward a state of greater disorder.

• In thermodynamics, entropy is a measure of the degree of disorder.

• Entropy tends to increase.

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• Likely

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• Unlikely

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The Concept of Entropy (S)

Entropy refers to the state of order.

A change in order is a change in the number of ways of arranging the particles, and it is a key factor in determining the direction of a spontaneous process.

solid liquid gasmore order less order

crystal + liquid ions in solution

more order less order

more order less order

crystal + crystal gases + ions in solution

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The sign of ΔS for a reaction does not, by itself, predict the direction of a spontaneous reaction.

If we consider both the system and the surroundings, we find that all real processes occur spontaneously in the direction that increases the entropy of the universe.

DSuniv = DSsys + DSsurr > 0

For a process to be spontaneous, a decrease in the entropy of the system must be offset by a larger increase in the entropy of the surroundings.

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Comparing Energy and Entropy

The total energy of the universe remains constant.

DEsys + DEsurr =DEuniv = 0

DH is often used to approximateDE.

For enthalpy there is no zero point; we can only measure changes in enthalpy.

The total entropy of the universe increases.

DSuniv =DSsys +DSsurr > 0

For entropy there is a zero point, and we can determine absolute entropy values.

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Entropy

• Entropy can be indirectly measured.• Absolute standard molar entropy values

can be found in the textbook.• An increase in entropy corresponds to an

increase in disorder.• When DS is _______, disorder increases.• When DS is _______, disorder

decreases.

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The Third Law of Thermodynamics

A perfect crystal has zero entropy at absolute zero.

Ssys = 0 at 0 K

A “perfect” crystal has flawless alignment of all its particles. At absolute zero, the particles have minimum energy, so there is only one microstate.

S = k lnW = k ln 1 = 0

To find the entropy of a substance at a given temperature, we cool it as close to 0 K as possible. We then heat in small increments, measure q and T, and calculate DS for each increment. The sum of these DS values gives the absolute entropy at the temperature of interest.

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Standard Molar Entropies

S° is the standard molar entropy of a substance, measured for a substance in its standard state in units of J/mol·K.

The conventions for defining a standard state include:• 1 atm for gases• 1 M for solutions, and• the pure substance in its most stable form for

solids and liquids.

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Factors Affecting Entropy

• Entropy depends on temperature.– For any substance, S° increases as temperature increases.

• Entropy depends on the physical state of a substance.– S° increases as the phase changes from solid to liquid to gas.

• The formation of a solution affects entropy.

• Entropy is related to atomic size and molecular complexity.– Remember to compare substances in the same physical state.

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Figure 20.6A Visualizing the effect of temperature on entropy.

Computer simulations show each particle in a crystal moving about its lattice position. Adding heat increases T and the total energy, so the particles have greater freedom of motion, and their energy is more dispersed. S therefore increases.

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Figure 20.6B Visualizing the effect of temperature on entropy.

At any T, there is a range of occupied energy levels and therefore a certain number of microstates. Adding heat increases the total energy (area under the curve), so the range of occupied energy levels becomes greater, as does the number of microstates.

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Figure 20.6C Visualizing the effect of temperature on entropy.

A system of 21 particles occupy energy levels (lines) in a box whose height represents the total energy. When heat is added, the total energy increases and becomes more dispersed, so S increases.

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Figure 20.7 The increase in entropy during phase changes from solid to liquid to gas.

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Figure 20.8 The entropy change accompanying the dissolution of a salt.

pure solid

pure liquidsolution

MIX

The entropy of a salt solution is usually greater than that of the solid and of water, but it is affected by the organization of the water molecules around each ion.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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Figure 20.9 The small increase in entropy when ethanol dissolves in water.

Ethanol (A) and water (B) each have many H bonds between their own molecules. In solution (C) they form H bonds to each other, so their freedom of motion does not change significantly.

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Figure 20.10 The entropy of a gas dissolved in a liquid.

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Entropy and Atomic Size

S° is higher for larger atoms or molecules of the same type.

Li Na K Rb Cs

Atomic radius (pm) 152 186 227 248 265

Molar mass (g/mol)

6.941 22.99 39.10 85.47 132.9

S°(s) 29.1 51.4 64.7 69.5 85.2

HF HCl HBr HI

Molar mass (g/mol) 20.01 36.46 80.91 127.9

S°(s) 173.7 186.8 198.6 206.3

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Entropy and Structure

NaCl AlCl3 P4O10 NO NO2 N2O4

S°(s) 72.1 167 229

S°(g) 211 240 304

For allotropes, S° is higher in the form that allows the atoms more freedom of motion.S° of graphite is 5.69 J/mol·K, whereas S° of diamond is 2.44 J/mol·K.

For compounds, S° increases with chemical complexity.

These trends only hold for substances in the same physical state.

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Entropy Changes in the System

The standard entropy of reaction, DS°rxn, is the entropy change that occurs when all reactants and products are in their standard states.

DS°rxn = SmS°products - SnS°reactants

where m and n are the amounts (mol) of products and reactants, given by the coefficients in the balanced equation.

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Predicting the Sign of DS°rxn

We can often predict the sign of DS°rxn for processes that involve a change in the number of moles of gas.

• DS°rxn is positive if the amount of gas increases;

– e.g., H2(g) + I2(s) → 2HI(g).

• DS°rxn is negative if the amount of gas decreases; – e.g.,

• DS° is likely to be positive if a new structure forms that has more freedom of motion.

N2(g) + 3H2(g) 2NH3(g)

CH2

H2C CH2 (g)

CH3 CH CH2 (g) S > 0

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Sample Problem 20.2 Calculating the Standard Entropy of Reaction, DS°rxn

PROBLEM: Predict the sign of DS°rxn and calculate its value for the combustion of 1 mol of propane at 25°C.

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)

PLAN: From the change in the number of moles of gas (6 mol yields 3 mol), the entropy should decrease (DS°rxn < 0). To find DS°rxn, we apply Equation 20.4, using the S° values from Appendix B.

SOLUTION:

ΔS°rxn = [(3 mol CO2)(S° of CO2) + (4 mol H2O)(S° of H2O)] - [(1 mol C3H8)(S° of C3H8) + (5 mol O2)(S° of O2)

= [(3mol)(213.7J/K·mol) + (4 mol)(69.9J/K·mol)] – [(1mol)(269.9 J/K·mol) + (5 mol)(205.0 J/K·mol) = -374 J/K

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Changes in Entropy

• Calculate the entropy change for the following reaction at 298K. Indicate whether disorder increases or decreases.

• 2NO2(g) -----> N2O4(g)

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Free Energy Change, DG

• If heat is released in a chemical reaction, some of the heat may be converted to useful work. Some of it may be used to increase the order of the universe. If the system becomes more disordered, then more energy becomes available than indicated by enthalpy alone.

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Gibbs Free Energy

The Gibbs free energy (G) combines the enthalpy and entropy of a system; G = H – TS.

The free energy change (DG) is a measure of the spontaneity of a process and of the useful energy available from it.

DGsys = DHsys - T DSsys

DG < 0 for a spontaneous process

DG > 0 for a nonspontaneous process

DG = 0 for a process at equilibrium

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Calculating DG°

DG°sys = DH°sys - T DS°sys

DG°rxn can be calculated using the Gibbs equation:

DG°rxn can also be calculated using values for the standard free energy of formation of the components.

DG°rxn = SmDG°products - SnDG°reactants

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Gibbs Free Energy Change

• The following relationship exists for standard molar Gibbs free energy Gibbs Free Energy Change changes:

DGo = SDGfo(products) -SDGf

o(reactants)

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Sample Problem 20.4 Calculating DG°rxn from Enthalpy and Entropy Values

PROBLEM: Potassium chlorate, a common oxidizing agent in fireworks and matchheads, undergoes a solid-state disproportionation reaction when heated.

4KClO3(s) 3KClO4(s) + KCl(s)+7 -1+5 Δ

PLAN: To solve for DG°, we need values from Appendix B. We use DHf° to calculate DH°rxn, use S° values to calculate DS°rxn and then apply the Gibbs equation.

SOLUTION:DH°rxn = SmDHf°(products) - SnDHf°(reactants)

= [(3 mol KClO4)(DHf° of KClO4) + (1 mol KCl)(DHf° of KCl)] ‒ [(4 mol KClO3)(DHf° of KClO3)]

= [(3 mol)(-432.8 kJ/mol) + (1 mol)(-436.7 kJ/mol) – [(4 mol)(-397.7 kJ/mol)] = 144 kJ

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Sample Problem 20.4

DS°rxn = SmS°(products) - SS°(reactants)

= [(3 mol of KClO4)(S° of KClO4) + (1 mol KCl)(S° of KCl)] ‒ [(4 mol KClO3)(S° of KClO3)] = [(3 mol)(151.0 J/mol·K) + (1 mol)(82.6 J/mol·K) – [(4 mol)(-397.7 kJ/mol)] = -36.8 J/K

DG°sys = DH°sys - T DS°sys = 144 kJ – (298 K)(-36.8 J/K) 1000 J

1 kJ

= -133 kJ

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Sample Problem 20.5 Calculating DG°rxn from DG°f Values

PLAN: We apply Equation 20.8 and use the values from Appendix B to calculate DG°rxn.

SOLUTION:

PROBLEM: Use DG°f values to calculate DG°rxn for the reaction in Sample Problem 20.4:

4KClO3(s) 3KClO4(s) + KCl(s)Δ

DG°rxn = SmDG°products - SnDG°reactants

= [(3 mol KClO4)(DGf° of KClO4) + (1 mol KCl)(DGf° of KCl)] ‒ [(4 mol KClO3)(DGf° of KClO3)]

= [(3 mol)(-303.2 kJ/mol) + (1 mol)(-409.2 kJ/mol)] ‒ [(4 mol)(-296.3 kJ/mol)]

= -134 kJ

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Gibbs Free Energy Change

• Calculate the Gibbs free energy change for the following reaction at 298K. Indicate whether the reaction is spontaneous or nonspontaneous under these conditions.

• C3H8(g) + 5O2(g) ----> 3CO2(g) + 4H2O(l)

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Effect of Temperature on Reaction Spontaneity

Reaction is spontaneous at all temperatures

If DH < 0 and DS > 0

DG < 0 for all T

Reaction is nonspontaneous at all temperatures

If DH > 0 and DS < 0

DG > 0 for all T

DGsys = DHsys - T DSsys

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Effect of Temperature on Reaction Spontaneity

Reaction becomes spontaneous as T increases

If DH > 0 and DS > 0

DG becomes more negative as T increases.

Reaction becomes spontaneous as T decreases

If DH < 0 and DS < 0

DG becomes more negative as T decreases.

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Table 20.1 Reaction Spontaneity and the Signs of DH, DS, and DG

DH DS -TDS DG Description

– + – – Spontaneous at all T

+ – + + Nonspontaneous at all T

+ + – + or – Spontaneous at higher T;

nonspontaneous at lower T

– – + + or – Spontaneous at lower T;

nonspontaneous at higher T

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Sample Problem 20.7 Determining the Effect of Temperature on ΔG

PROBLEM: A key step in the production of sulfuric acid is the oxidation of SO2(g) to SO3(g):

2SO2(g) + O2(g) → 2SO3(g)

At 298 K, DG = -141.6 kJ; DH = -198.4 kJ; and DS = -187.9 J/K

(a) Use the data to decide if this reaction is spontaneous at 25°C, and predict how DG will change with increasing T.

(b) Assuming DH and DS are constant with increasing T, is the reaction spontaneous at 900.° C?

PLAN: We note the sign of DG to see if the reaction is spontaneous and the signs of DH and DS to see the effect of T. We can then calculate whether or not the reaction is spontaneous at the higher temperature.

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Sample Problem 20.7

SOLUTION:

(a) DG < 0 at 209 K (= 25°C), so the reaction is spontaneous.

With DS < 0, the term -TDS > 0 and this term will become more positive at higher T.

DG will become less negative, and the reaction less spontaneous, with increasing T.

(b) DG = DH - TDS

DG = -198.4 kJ – [1173 K)(-0.187.9 kJ/K) = 22.0 kJ

Convert T to K: 900. + 273.15 = 1173 KConvert S to kJ/K: -187.9 J/K = 0.1879 kJ/K

DG > 0, so the reaction is nonspontaneous at 900.°C.

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Figure 20.15 The effect of temperature on reaction spontaneity.

DH

ΔST =

The sign of DG switches at

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Sample Problem 20.8 Finding the Temperature at Which a Reaction Becomes Spontaneous

PROBLEM: At 25°C (298 K), the reduction of copper(I) oxide is nonspontaneous DG = 8.9 kJ). Calculate the temperature at which the reaction becomes spontaneous.

PLAN: We need to calculate the temperature at which DG crosses over from a positive to a negative value. We set DG equal to zero, and solve for T, using the values for DH and DS from the text.

SOLUTION:

DH

DST = =

58.1 kJ

0.165 kJ/K= 352 K

At any temperature above 352 K (= 79°C), the reaction becomes spontaneous.

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The Gibbs Helmholtz Equation

• Calculate DSo for the following reaction at 298K.

• C3H8(g) + 5O2(g) ----> 3CO2(g) + 4H2O(l)

• From previous examples we found• DHo = -2219kJ and• DGo = -2107kJ• Indicate whether disorder increases or

decreases

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The Relationship Between DGo and the Equilibrium Constant

• The standard free energy change for a reaction is DGo. This is the free energy change that would accompany the complete conversion of all reactants, initially present in their standard states, to all products in their standard states. DG is the free energy change for other concentrations and pressures.

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• The relationship between DG and DGo is• DG = DGo + RTlnQ where• R = universal gas

constant(8.314J/moleK)• T = temperature in K• Q = reaction quotient

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• When a system is at equilibrium,

DG = 0 and Q = K. Then:• 0 = DGo + RTlnK• Rearranging gives

DGo = -RTlnK

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DG, Q, and K

QK

DG = RT ln = RT lnQ – RT lnK

If Q and K are very different, DG has a very large value (positive or negative). The reaction releases or absorbs a large amount of free energy.

If Q and K are nearly the same, DG has a very small value (positive or negative). The reaction releases or absorbs very little free energy.

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ΔG and the Equlibrium Constant

For standard state conditions, Q = 1 and

As DG° becomes more positive, K becomes smaller.

As DG° becomes more negative, K becomes larger.

A small change in DG° causes a large change in K, due to their logarithmic relationship.

DG = DG° + RT lnQ

To calculate DG for any conditions:

DG° = -RT lnK

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Table 20.2 The Relationship Between DG° and K at 298 K

DG° (kJ) K Significance

200 9x10-36 Essentially no forward reaction;reverse reaction goes to completion100 3x10-18

50 2x10-9

10 2x10-2

1 7x10-1

0 1 Forward and reverse reactions proceed to same extent-1 1.5

-10 5x101

-50 6x108

-100 3x1017 Forward reaction goes to completion; essentially no reverse reaction-200 1x1035

FO

RW

AR

D R

EA

CT

ION

RE

VE

RS

E R

EA

CT

ION

{

{

{

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• Calculate the value for the equilibrium constant, Kp, for the following reaction at 298K.

• N2O4(g) = 2NO2(g)

• At 25oC and 1.00 atmosphere pressure, Kp=4.3x10-13, for the decomposition of NO2. Calculate DGo at 25oC.

New chm 152 unit 6 power points sp13

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