chapter 5

Chapter 5Thermochemistry

Definitions• Thermochemistry - the study of how energy in the

form of heat is consumed and produced by chemical reactions.

• Thermochemical Reaction: ExampleH2(g) + O2(g) 2 H2O (l) + heat

• Energy is anything having the capacity to do work or to transfer heat.

• Work is Force X Distance• Thermodynamics The study of energy and its

transformation from one form to another.

Energy ExamplesDefinition-Energy is anything with the capacity to do

work, or create heat.• Food• Gasoline• Electricity• An apple on a tree• A baseball moving

Energy UnitsUnits of Energy

Joules = kg(m/s)2

Calories, an older unit; the energy to increase one gram of water one deg. C

Calories, unit of food energy = kcal

Definitions Continued• Heat is energy transferred between

objects because of a difference in their temperatures.

• Thermodynamics is the study of relationship between chemical reactions and changes in heat energy.

• Heat transfer is the process of heat energy flowing from one object into another

Potential: due to position or composition - can be converted to work PE = mgh (m = mass, kg, g = force of gravity(9.8 m/s2), and h = vertical distance in meters)• These units multiplied together = joule• (chemical energy is a form of potential energy)

• Kinetic: due to motion of the object KE = 1/2 mv 2 (m = mass(kg), v = velocity(m/s))

• These units multiplied together also equal a joule also.

• Joule = kg(m/s)2

Two Types of Energy

Kinetic vs. Potential Energy

• A State Function depends only on the present state of the system - not how it arrived there.

• It is independent of pathway.

Potential Energy: A State Function

Other State Functions?

Heat (a form of energy)? Work? Altitude? Altitude?

Other State Functions?

Heat (a form of energy)? Yes, not path dependent Work? Altitude?

Other State Functions?

Heat (a form of energy)? Yes, not path dependent Work? No, work depends on path Altitude?

Other State Functions?

Heat (a form of energy)? Yes, not path dependent Work? No, work depends on path Altitude? Yes, does not depend on path Enthalpy ΔH,?

Other State Functions?

Heat (a form of energy)? Yes, not path dependent Work? No, work depends on path Altitude? Yes, does not depend on path Enthalpy ΔH,? Yes, does not depend on path

• Energy is conserved: the First Law of thermodynamics is called the Law of Conservation of Energy which states that energy cannot be created nor destroyed, but can be converted from one form to another.

The Nature of Energy

Energy at the Molecular LevelKinetic energy at the molecular level depends on the mass and velocity of the particle. Since velocity depends on temperature, then kinetic energy does too, because matter cannot be created nor destroyed. Another way to say this is that absolute temperature and kinetic energy are directly proportional to each other.Another important form of energy is potential energy at the atomic-molecular level arises from electrostatic interactions.Potential Energy can be converted into kinetic, for example an apple falling from a tree.

Electrostatic Potential EnergyCoulombic attraction, not gravitational force, determines the potential energy of matter at the atomic level. The energy is determined by multiplying the charge in coulombs of the two particles and dividing by the distance squared in meters.

Eel Q1 x Q2

d• Eel is the electrostatic potential energy• Q is the charge in coulombs• d is the distance between particles in meters

2

Electrostatic Potential Energy

Energy of an ionic bond

Bond length

Bonds contain potential energyEnergy required to break bondsEnergy released when bonds are created

Ionic compounds do not exist as discrete molecules. Instead they exist as crystals where ions of opposite charges occupy positions known as lattice sites.

In an ionic compound the ions organize in such a way as to minimize repulsive and maximize attractive forces. This is an example of Coulomb’s Law in action.

Crystal Lattice of NaCl

+-

+-

The arrangement of charged particles in a covalent bond

organized in such a way as to minimize repulsive and maximize

attractive forces to give the lowest potential energy.

The Covalent BondThe covalent bond in an example of a Columbic attraction.

• System: the part of the universe that is the focus of a thermodynamic study. Example a beaker or test tube in the lab

• Surroundings: everything in the universe that is not part of the system.

• Universe = System + Surroundings

• An isolated system exchanges neither energy nor matter with the surroundings.

Terms Describing Energy Transfer

Examples

Heat Flow• In an exothermic process, heat flows from a

system into its surroundings.• In an endothermic process, heat flows from

the surroundings into the system

Phase Changes

Internal Energy

• The internal energy of a system is the sum of all the KE and PE of all of the components of the system.

First Law of Thermodynamics• The first law of thermodynamics states that

the energy gained or lost by a system must equal the energy lost or gained by surroundings. ΔE = q + w (mathematical statement)

• The calorie (cal) is the amount of heat necessary to raise the temperature of 1 g of water 1oC.

• The joule (J) is the SI unit of energy; 4.184 J = 1 cal.

Energy Flow Diagrams

E = q + w E = change in system’s internal energy q = heat

Endothermic +qExothermic -qExpansion –w (since system is losing energy to do work)Compression +w

Change in Internal Energy

First Law ProblemFind the change in energy of a system when 12 j of energy flows into the system while the system is doing 8 j of work on the surroundings.

First Law ProblemFind the change in energy of a system when 12 j of energy flows into the system while the system is doing 8 j of work on the surroundings.

ΔE = q + w ↔ ΔE = + 12 – 8 = 4 j

PV Work

h

ΔV = Vfinal – Vinitial

ΔV = A x d A

W = F x dP = F/A

Facts

F = PA

Expansion

A

Atm = 14 lb/in2

W = PA x d

W = P ΔV

PV Work

h

ΔV = Vfinal – Vinitial

ΔV = A x d A W = F x d

P = F/A

Facts

F = PA

SubstitutingW = PxAxdW = PΔV

Since expansion is defined as negative work and ΔV is positive for expansion, then we change the sign of PΔV to –PΔV,ΔE = q - PΔV

Expansion

PV Work

h

ΔV = Vfinal – Vinitial

ΔV = A x d A W = F x d

P = F/A

Facts

F = PA

SubstitutingW = PxAxdW = PΔV

Since expansion is defined as negative work and ΔV is positive for expansion, then we change the sign of PΔV to –PΔV,ΔE = q - PΔV (L-atm = 101.3 j)

Expansion

Calculation of WorkCalculate the work in L•atm and joules associated with the expansion of a gas in a cylinder from 54 L to 72 L at a constant external pressure of 18 atm(Note that as the gas expands, it does work on its surroundings. Energy flows out of the gas, so it is a negative quantity)

Calculation of WorkCalculate the work in L•atm and joules associated with the expansion of a gas in a cylinder from 54 L to 72 L at a constant external pressure of 18 atm(Note that as the gas expands, it does work on its surroundings. Energy flows out of the gas, so it is a negative quantity)

W = -pΔVW = -18atm(72-54)L = -320L-atm

-320L-atm101.3 L-atmj

= -3.2 j

Calculation of WorkCalculate the work in L•atm and joules associated with the expansion of a gas in a cylinder from 54 L to 72 L at a constant external pressure of 18 atm(Note that as the gas expands, it does work on its surroundings. Energy flows out of the gas, so it is a negative quantity)

W = -pΔVW = -18atm(72-54)L = -320L-atm

-320L-atm101.3 L-atmj

= -3.2 j

Note: The result is negative, which we would predict relative to expansion. This is the reason that work is -pΔV, to give the correct sign for expansion.

• Enthalpy (H) = E + PV ( mathematical definition)

• Change in Enthalpy (H) = E + PV

• At constant P, qP = E + PV, therefore qP = H

• At Constant V, qv = ΔE, since ΔV = 0

• H = change in enthalpy: an energy flow as heat (at constant pressure)

• H > 0, Endothermic; H < 0, Exothermic

Enthalpy and Change in Enthalpy

Heating Curves

Heat Capacities• Molar heat capacity (cp) is the heat required to

raise the temperature of 1 mole of a substance by 1oC at constant pressure.

• q = ncpT

• Specific heat (cs) is the heat required to raise the temperature of 1 gram of a substance by 1oC at constant pressure.

• Heat capacity (Cp) is the quantity of heat needed to raise the temperature of some specific object by 1oC at constant pressure.

Phase Change and Energy• Molar heat of fusion (Hfus) - the heat

required to convert 1 mole of a solid substance at its melting point to 1 mole of liquid.

• q = nHfus

• Molar heat of vaporization (Hvap) - the heat required to convert 1 mole of a substance at its boiling point to 1 mole of vapor.

• q = nHvap

WATER THERMO VALUES• Ice H2O (s) 2.06 j/g-°C

• Water H2O (l) 4.184 j/g-°C

• Steam H2O (g) 1.86 j/g-°C• Heat of Fusion (melting) 334.0 j/g• Heat of vaporization (evaporation) 2257 j/g

Practice During a strenuous workout, a student

generates 2000 kJ of heat energy. What mass of water would have to evaporate from the student’s skin to dissipate this much heat?

Practice During a strenuous workout, a student

generates 2000 kJ of heat energy. What mass of water would have to evaporate from the student’s skin to dissipate this much heat?

g2257 j

Practice During a strenuous workout, a student

generates 2000 kJ of heat energy. What mass of water would have to evaporate from the student’s skin to dissipate this much heat?

g2257 j

10 3 jkj

Practice During a strenuous workout, a student

generates 2000 kJ of heat energy. What mass of water would have to evaporate from the student’s skin to dissipate this much heat?

g2257 j

10 3 jkj

2000 kj

Practice During a strenuous workout, a student

generates 2000 kJ of heat energy. What mass of water would have to evaporate from the student’s skin to dissipate this much heat?

g2257 j

10 3 jkj

2000 kj = 886 g water

Practice5.53 From Text

Exactly 10 mL of water at 25oC was added to a hot iron skillet. All of the water was converted into steam at 100oC. If the mass of the pan was 1.20 kg and the molar heat capacity of iron is 25.19 J/mol•oC, what was the temperature change of the skillet?

Sample Problem Solution

25.19 jmole-°C

Sample Problem Solution

25.19 jmole-°C

mole55.85 g

Sample Problem Solution

25.19 jmole-°C

mole55.85 g

103 g

kg

Sample Problem Solution

25.19 jmole-°C

mole55.85 g

103 g

kg1.20 kg

Sample Problem Solution

25.19 jmole-°C

mole55.85 g

103 g

kg1.20 kg

Sample Problem Solution

25.19 jmole-°C

mole55.85 g

103 g

kg1.20 kg

Now the energy required to heat 10mL of water from 25°C to 100°C and then to vaporize the water is outlined below.

Sample Problem Solution

25.19 jmole-°C

mole55.85 g

103 g

kg1.20 kg

Now the energy required to heat 10mL of water from 25°C to 100°C and then to vaporize the water is outlined below.

Heating from 25°C to 100°C 4.184 j

g-°C

Sample Problem Solution

25.19 jmole-°C

mole55.85 g

103 g

kg1.20 kg

Now the energy required to heat 10mL of water from 25°C to 100°C and then to vaporize the water is outlined below.

Heating from 25°C to 100°C 4.184 j

g-°C10.0g

Sample Problem Solution

25.19 jmole-°C

mole55.85 g

103 g

kg1.20 kg

Now the energy required to heat 10mL of water from 25°C to 100°C and then to vaporize the water is outlined below.

Heating from 25°C to 100°C 4.184 j

g-°C10.0g 75 °C

Sample Problem Solution

25.19 jmole-°C

mole55.85 g

103 g

kg1.20 kg

Now the energy required to heat 10mL of water from 25°C to 100°C and then to vaporize the water is outlined below.

Heating from 25°C to 100°C 4.184 j

g-°C10.0g 75 °C

= 3138 j

Evaporating 10.0 mL of water

Sample Problem Solution

25.19 jmole-°C

mole55.85 g

103 g

kg1.20 kg

Now the energy required to heat 10mL of water from 25°C to 100°C and then to vaporize the water is outlined below.

Heating from 25°C to 100°C 4.184 j

g-°C10.0g 75 °C

= 3138 j

Evaporating 10.0 mL of water2257 j

g

Sample Problem Solution

25.19 jmole-°C

mole55.85 g

103 g

kg1.20 kg

Now the energy required to heat 10mL of water from 25°C to 100°C and then to vaporize the water is outlined below.

Heating from 25°C to 100°C 4.184 j

g-°C10.0g 75 °C

= 3138 j

Evaporating 10.0 mL of water2257 j

g10.0 g

Sample Problem Solution

25.19 jmole-°C

mole55.85 g

103 g

kg1.20 kg

Now the energy required to heat 10mL of water from 25°C to 100°C and then to vaporize the water is outlined below.

Heating from 25°C to 100°C 4.184 j

g-°C10.0g 75 °C

= 3138 j

Evaporating 10.0 mL of water2257 j

g10.0 g =22570 j

Sample Problem Solution

25.19 jmole-°C

mole55.85 g

103 g

kg1.20 kg

Now the energy required to heat 10mL of water from 25°C to 100°C and then to vaporize the water is outlined below.

Heating from 25°C to 100°C 4.184 j

g-°C10.0g 75 °C

= 3138 j

Evaporating 10.0 mL of water2257 j

g10.0 g =22570 j

Now Combine 3138 j + 22570 j = 25708j

Sample Problem Solution

25.19 jmole-°C

mole55.85 g

103 g

kg1.20 kg

25708 j

= 47.5 °C

Heat of ReactionHeat of reaction is also known as enthalpy of reaction (Hrxn) is the heat absorbed or released by a chemical reaction.

A Specific Enthalpy• The standard enthalpy of formation (Hf

o) is also called the standard heat of formation and is the enthalpy change of the a formation reaction.

• A formation reaction is the process of forming 1 mole of a substance in its standard state from its component elements in their standard states.

• H2(g) + 1/2 O2(g) ---> H2O(l) Hfo for water

• The standard state of a substance is its most stable form under 1 bar pressure and 25oC.

Methods of Determining Hrxn

1. from calorimetry experiments

2. from enthalpies of formation

3. using Hess’s Law

Calorimetry• Calorimetry is the measurement of the change in

heat that occurs during a physical change or chemical process.

• A calorimeter is the device used to measure the absorption or release of heat by a physical or chemical process.

• There are two types of calorimeters, styrofoam calorimeter where P is constant and a bomb calorimeter where V is constant.

• Bomb calorimeters measure ΔE, while Styrofoam calorimeters measure ΔH.

Measuring Heat CapacityA Styrofoam calorimeter is open to the atmosphere with pressure being constant.

Here the First Law of Thermodynamics is used. The heat lost by the aluminum beads is absorbed by the water in the calorimeter. Using the specific heat of water and its mass the heat lost by the aluminum beads can be easily calculated by unit analysis.

Measuring Heat CapacityAssuming there were 750.0 g of Al present in 333 g of boiling water at 100°C, the specific heat of aluminum can be calculated by using the final temperature of the aluminum and water which we will assume to be 44.0°C and the initial temperature of the water to be 21.0°C . This calculation is shown below:

4.184 jg-°C

333 g H2O 44.0-21.0°C750.0 g Al100-44.0°C

= 0.763 j/g-°C

Calorimetry: Bomb Calorimeter• E = -qcal = -CcalT

• A bomb calorimeter is a constant-volume device used to measure the heat of a combustion reaction.

Practice Write the standard enthalpy of formation

reaction for nitric acid.

Practice Write the standard enthalpy of formation

reaction for nitric acid.Standard enthalpy, ΔH°, is the change in energy when elements, in their standard state (25°C and 1 atm) combine to make 1 mole of products at their standard state. Standard State date is found in the appendix of your textbook.

• Hrxn° = npHf(products) nrHf(reactants)

Calculating Hrxn°

• Hrxn° = npHf(products) nrHf(reactants)

Calculating Hrxn°

½ N2 (g) + ½ H2 (g) + 3/2 O2 (g) HNO3 (l)

• Hrxn° = npHf(products) nrHf(reactants)

Calculating Hrxn°

½ N2 (g) + ½ H2 (g) + 3/2 O2 (g) HNO3 (l) 1/2(0.00) ½(0.00) 3/2(0.0) 1(-135.1 kj)

• Hrxn° = npHf(products) nrHf(reactants)

Calculating Hrxn°

½ N2 (g) + ½ H2 (g) + 3/2 O2 (g) HNO3 (l) 1/2(0.00) ½(0.00) 3/2(0.0) 1(-135.1 kj)

ΔH°rxn = [1(-135.1 kj)] – [1/2(0.00) + 1/2(0.00) + 3/2(0.0)]

ΔH°rxn = - 135.1 kj

These values are found in the appendix of your text

Use Table 5.2 to calculate an approximate enthalpy of reaction for CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(l)

Example

One step in the production of nitric acid is the combustion of ammonia. Using the data in the appendix to calculate the enthalpy of this reaction.4NH3(g) + 5O2(g) ---> 4NO(g) + 6H2O(g)

PracticePractice

Fuel ValuesCompound Molecular

FormulaFuel Value

(kJ/g)*Methane CH4 50.0

Ethane C2H6 47.6

Propane C3H8 46.3

Butane C4H10 45.8

* Based on the formation of H2O (g)

Hess’s Law• Hess’s law states that the enthalpy change of a reaction

that is the sum of two or more reactions is equal to the sum of the enthalpy changes of the constituent reactions. Due to the fact than enthalpy is a State Function

1. If a reaction is reversed, H sign changes. N2(g) + O2(g) 2NO(g) H = 180 kJ 2NO(g) N2(g) + O2(g) H = 180 kJ

2. If the coefficients of a reaction are multiplied by an integer, H is multiplied by that same integer.

6NO(g) 3N2(g) + 3O2(g) H = 540 kJ

Calculations via Hess’s Law

H2(g) + 1/2O2(g) H2O(l) -285.8 kJC2H4(g) + 3O2(g) 2H2O(l) + 2CO2(g) -1411 kJC2H6(g) + 7/2O2(g) 3H2O(l) + 2CO2(g) -1560 kJ

Calculate the enthalpy change for C2H4(g) + H2(g) C2H6(g) using the following data.

Example

H2(g) + 1/2O2(g) H2O(l) -285.8 kJC2H4(g) + 3O2(g) 2H2O(l) + 2CO2(g) -1411 kJ

Calculate the enthalpy change for C2H4(g) + H2(g) C2H6(g) using the following data.

Example

3H2O(l) + 2CO2(g) C2H6(g) + 7/2O2(g) +1560 kJ

H2(g) +1/2O2(g)+C2H4(g)+3O2(g)+3H2O(l)+2CO2(g) H2O(l)+ 2H2O(l) + 2CO2(g)+ C2H6(g) + 7/2O2(g) -136.8 simplifyC2H4(g) +H2(g) C2H6(g) -136.8 kj

Review

ChemTour: State Functions and Path Functions

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This ChemTour defines and explores the difference between state and path functions using a travel analogy that leads into a discussion of energy, enthalpy, heat, and work.

ChemTour: Internal Energy

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This ChemTour explores how energy is exchanged between a system and its surroundings as heat and/or work, and how this transfer in turn affects the internal energy (E) of a system.

ChemTour: Pressure-Volume Work

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An animated ChemTour of an internal combustion engine shows how a system undergoing an exothermic reaction can do work on its surroundings; students can explore the relationship among pressure, volume, and work.

ChemTour: Heating Curves

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In this ChemTour, students use interactive heating curve diagrams to explore phase changes, heat of fusion, and heat of vaporization. Macroscopic views of ice melting and water boiling are shown in sync with the appropriate sections of the heating curve.

ChemTour: Calorimetry

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This ChemTour demonstrates how a bomb calorimeter works, and walks students through the equations used to solve calorimetry problems. Includes an interactive experiment.

ChemTour: Hess’s Law

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This ChemTour explains Hess’s law of constant heat of summation using animated sample problems and step-by-step descriptions.

Isothermal Expansion of Ideal Gas

An ideal gas in a sealed piston is allowed to expand isothermally (at a constant temperature) against a pressure of 1 atm. In what direction, if at all, does heat flow for this process?

A) into the system B) out of the system C) heat does not flow

Isothermal Expansion of Ideal Gas

Consider the following arguments for each answer and

vote again:

A. When the gas expands isothermally, it does work without a decrease in its energy, so heat must flow into the system.

B. During the expansion, the gas pressure decreases, thereby releasing heat to the surroundings.

C. The fact that the process is isothermal means that heat does not flow.

When air is released from a tire does it get warmer or cooler?

When air is released from a tire does it get warmer or cooler?

Cooler

Isothermal means that the temperature is unchanged, so what must happen for the temperature to remain the same?

When air is released from a tire does it get warmer or cooler?

Cooler

Isothermal means that the temperature is unchanged, so what must happen for the temperature to remain the same?

Heat must flow into the tire

Adiabatic Compression of an Ideal Gas

An ideal gas in an insulated piston is compressed adiabatically (q = 0) by its surroundings. What can be said of the change in the temperature (ΔT) of the gas for this process?

A) ΔT > 0 B) ΔT = 0 C) ΔT < 0

Adiabatic Compression of an Ideal Gas

Consider the following arguments for each answer

and vote again:

A. The surroundings are doing work on the system, and no heat is flowing. Therefore, ΔE > 0 and so ΔT > 0.

B. The volume of the gas decreases, but the pressure increases to keep the product of the pressure and volume constant. Therefore, the temperature is also constant.

C. The gas is being compressed to a more ordered state, which corresponds to a lower temperature.

ΔT of a Released Rubber Band

Consider a stretched rubber band that is suddenly released. What can be said of the change in the temperature (ΔT) of the rubber band for this process?

A) ΔT > 0 B) ΔT = 0 C) ΔT < 0

ΔT of a Released Rubber Band

Consider the following arguments for each answer and vote again:

A. The stretched rubber band is at a higher energy state than the unstretched rubber band. Releasing the stretched rubber band causes the energy to be released.

B. Because the recoil of the rubber band is rapid, this process is essentially adiabatic. Therefore, the temperature of the rubber band will not change.

C. As the rubber band contracts, it does work and its energy decreases, resulting in a decrease in its temperature.

Specific Heat Capacity of Al and Fe

A 1.0 gram block of Al (cs = 0.9 J/·°C-1·g) at 100 °C and a 1.0 gram block of Fe (cs = 0.4 J/·°C-1·g) at 0 °C are added to 10 mL of water (cs = 4.2 J·/°C-1·g) at 50 °C. What will be the final temperature of the water?

A) < 50 °C B) 50 °C C) > 50 °C

Specific Heat Capacity of Al and Fe

Consider the following arguments for each answer and vote again:

A. The specific heat capacity of Fe(s) is smaller than that of Al(s), so heat from both the Al(s) and the water will be required to warm the Fe(s).

B. The average initial temperature of the three components is 50 °C. Therefore, the final temperature of the water will be 50 °C.

C. The specific heat capacity of Al(s) is greater than that of Fe(s), so the Al block at 100 °C will heat the water more than the Fe block will cool it.

This is called a first law problem (energy cannot be created nor destroyed), and a mathematical statement would be – q = q. Remember q is measured in joules, so – j = + j

This is called a first law problem (energy cannot be created nor destroyed), and a mathematical statement would be – q = q. Remember q is measured in joules, so – j = + j

Since the heat capacity of aluminum is twice that of carbon, then it will lose more heat than iron will gain. This means more heat is lost by the aluminum than gained by the iron. Therefore, the water will increase in temperature and the resultant temperature will be greater than 50°C.

Equilibium Constant of Cyclooctatetraene

Cyclooctatetraene, C8H8, can undergo a transformation between two possible states, A and B, by rearranging its 4 double bonds.

Which of the following graphs depicts the dependence of the equilibrium constant (K) on temperature for the conversion from state A to state B?

A) B) C)

Equilibium Constant of Cyclooctatetraene

Consider the following arguments for each answer and vote again:

A. The intermediate is at a higher energy state than either states A or B, so at high temperatures, the reaction will favor the intermediate and K will decrease.

B. The enthalpies of formation for states A and B are equal, so ΔH° = 0 and K is not temperature dependent.

C. At high temperatures, the conversion from state A to state B will occur at a much faster rate, thus increasing the value of K.

Enthalpies of N2, NH3, and N2O

The reaction of N2(g) and O2(g) to form N2O(g) is an endothermic process. The reaction of N2(g) and H2(g) to form NH3(g) is an exothermic process. Given this information, which of the following species has the lowest enthalpy of formation, ?

A) N2(g) B) NH3(g) C) N2O(g)

Enthalpies of N2, NH3, and N20

Consider the following arguments for each answer and vote again:

A. N2(g) is the elemental form of nitrogen, which by definition will have   = 0, the lowest possible enthalpy of formation.

B. N2O(g) has a higher than N2(g), O2(g), and H2(g), whereas NH3(g) has a lower than N2(g), O2(g), and H2(g).

C. The formation of an N−N double bond and a N−O double bond, as found in N2O, releases more energy than does the creation of 3 N−H bonds to form NH3(g).

fH

fH fH

Polymerization of Ethylene

Which of the following is true of ΔH° for the polymerization of ethylene to form polyethylene? Note: the C-C single bond enthalpy is ~350 kJ/mole and the C-C double bond enthalpy is ~600 kJ/mole.

A) ΔH° > 0 B) ΔH° = 0 C) ΔH° < 0

Polymerization of Ethylene

Consider the following arguments for each answer

and vote again:

A. The energy required to break double bonds is more than the energy released by forming new single bonds.

B. The total number of C-C bonds (if we count double bonds twice) does not change with polymerization. Therefore, there can be no change in ΔH°.

C. For each C-C double bond that breaks (~600 kJ/mole), two single bonds form, (2×~350 = ~700 kJ/mole).

THE END