energy & chemical change

Energy & Chemical Change

Mr. Solsman

Chapter 15

• Chemistry is the study of matter, its composition and changes, and the energy effects that accompany these changes.

• Chapter 15 deals with energy and how energy changes relate to chemical reactions.

Why are energy factors important?

• Can we power cars using water as fuel?

• Energy factors are critical for the safe operation of chemical production facilities (and lab experiments).

• Energy factors are important in determining if a process can be operated at a profit.

The Nature of Energy

• Energy is the ability to do work or produce heat.

• Two Classifications - Kinetic & Potential• Kinetic Energy is the Energy of Motion• Potential Energy is Stored Energy

– can be a result of position or composition.– we will be most concerned with potential

energy in the form of composition.

The Law of Conservation of Energy

• Energy can neither be created or destroyed.

• Energy can only be converted from one form to another.

Kinetic Energy – the energy of motion

a. Depends on velocity

b. Depends on mass

Kinetic energy = ½ mv2

Positional Potential Energy

a. Depends on mass

b. Depends on distance

c. Depends on an attractive force

Potential Energy ↔ mad

Compositional Potential Energy

• Depends on the chemical bonds in a compound.

• A chemical reaction can store energy or release energy.

• Burning wood is an example of releasing energy.

• Plant growth is an example of storing energy.

Energy is a State Function

• State functions are those which depend only on the current state of a system and not on it’s history.

• This concept is critical to the content of this chapter.

Temperature and Heat

• Temperature is a measure of how hot something is.– (Well, duh!)

• Temperature is related to how fast the particles of a system are moving.

Temperature

• Temperature is a measure of the average kinetic energy of particles in the system.

• As kinetic energy is added to a system, the molecules move faster and the average kinetic energy increases – we say the temperature goes up.

HEAT

• Heat is NOT the same as temperature.

• Heat is the FLOW of energy from one object to another.

• Heat ALWAYS flows from hot to cold.

• Heat flow does not always raise or lower the temperature, it can change the kinetic OR the potential energy.

TYPES OF ENERGY CHANGES

A. Exothermic changes – changes where energy is released.

B. Endothermic changes – changes where energy is absorbed.

Terminology

• The SYSTEM is what we are focusing our attention on.

• The SURROUNDINGS is the rest of the universe.

Units of Heat

• The calorie is defined as the amount of heat needed to raise the temperature of one gram of water by one ºC.

• The Calorie (food calorie) is equal to 1000 calories.

Units of Energy

• The SI unit of energy is the joule.

• 1 calorie = 4.184 joules

• How many joules is 65 calories?

• How many calories is 457 joules?

Heat Calculations

• How much energy (in joules) does it take to heat 25 grams of water from 22 ºC to 45 ºC?

Heat Calculations

• Q = c x m x ∆T

(where ∆ means “change in”)

• ∆T = TFinal – TInitial

• m is the mass

• c is called the specific heat capacity. The units of c are J / g ºC

Typical Calculation

• How many joules are required to heat 1.50 kg of iron from 15 ºC to 37 ºC?

• Q = c x m x ∆T

= 0.45 J/g ºC x 1.50 kg x 1000 g/kg x 22 ºC

= 15000 J

WOW!!!!

• How many joules are required to heat 1.50 kg of water from 15 ºC to 37 ºC?

• Q = c x m x ∆T

= 4.2 J/g ºC x 1.50 kg x 1000 g/kg x 22 ºC

= 140000 J

• Compare this to the value of 15,000 J for iron. Why are these values so different?

• Specific Heat Capacity

Specific Heat Capacity

Problem

• If it required 54 J to heat 15 grams of a sample from 45 ºC to 60. ºC, what is the specific heat capacity of the material?

• Which of the materials on the specific heat capacity slide would the material most likely be?

Specific Heat Capacity

Chemical Energy and the Universe

• Thermochemistry is the study of heat changes that accompany chemical reactions and phase changes.

4Fe(s) + 3O2(g) → 2FeO3(s) + 1625 kJ

• The system is the specific part of the universe that contains the reaction or process you wish to study.

The surroundings are everything else other than the system in the universe.

• The universe is defined as the system plus the surroundings.

• Enthalpy is the heat content of a system at constant pressure.

• Enthalpy (heat) of reaction is the change in enthalpy during a reaction symbolized as ΔHrxn.

ΔHrxn = Hfinal – Hinitial

ΔHrxn = Hproducts – Hreactants

• Enthalpy changes for exothermic reactions are always negative.

• Enthalpy changes for endothermic reactions are always positive.

• Cold-pack reaction:

• 27 kJ + NH4NO3(s) → NH4+

(aq) + NO3-

(aq)

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STOICHIOMETRY

THERMOCHEMICAL REACTION – A REACTION THAT INCLUDES THE ENERGY TERM.

1. 136 kcal + 2H2O(l) 2H2(g) + O2(g)

2. C(s) + O2(g) CO2(g) + 94.0 kcal3. How much heat is absorbed when 30.0

grams of water decomposes?4. What amount of heat is released when 4.4

grams of CO2 form?

5. 3H2(g) + N2(g) 2NH3

H = -11.0kcal/mole NH3

a. What is the thermochemical reaction? b. What amount of heat is released if 1.6 grams

of H2 reacts with nitrogen?

6. N2(g) + 2O2(g) 2NO2(g)

H = 8.1kcal/mole NO2

a. What is the thermochemical reaction? b. What amount of heat is needed when 3.50

liters of O2 react with nitrogen?

Changes of State

• Molar enthalpy (heat) of vaporization refers to the heat required to vaporize one mole of a liquid substance.

• Molar enthalpy (heat) of fusion is the amount of heat required to melt one mole of a solid substance.

• Molar enthalpy (heat) of vaporization refers to the heat required to vaporize one mole of a liquid substance.

• Molar enthalpy (heat) of fusion is the amount of heat required to melt one mole of a solid substance.

Phase Change Diagram

Phase Changes are physical changes.

Na+Cl-(s) Na+Cl-(l) - MELTING

H2O(l) H2O(g) -EVAPORATION

CO2(s) CO2(g) -SUBLIMATION

Ne(g) Ne(l) - CONDENSATION

H2O(g) H2O(s) -DEPOSITION

Cu(l) Cu(s) -FREEZING

• How much heat does it take to completely vaporize 100. grams of room temperature water?

• The total heat required is the heat to raise the temperature of the water to its boiling point PLUS the heat to vaporize it at its boiling point.

• Q = m c ΔT

• H = n (ΔHpc )

• n is in moles

• ΔH is in joules/mole or kcal/mol

• How much heat does it take to completely vaporize 100. grams of room temperature(25 ºC) water?

• Q = 100. g * 4.18 J/g ºC * (100-25) ºC

• ΔHvap=(100. g/18.0 g/mol)*40.7 kJ/mol

• Total Heat Needed = Q + ΔH

= 31,400 J + 224,000 J = 255,000 J

• How much heat is given off when 10.0 grams of water vapor at 100.0 ºC is converted into ice at –15.0 ºC ?

• c is 4.18 J/g ºC for liquid water

• c is 2.0 J/g ºC for water vapor

• c is 2.03 J/g ºC for solid water

HESS’S LAW It makes no difference how reactants are

converted to products, the net energy change will always be the same.

EXAMPLES

1. Combustion of Carbon H

C(s) + O2(g) CO2(g) -94.0kcal

Possible Mechanism:

C(s) + .5O2(g) CO(g) -26.4kcal

CO(g) + .5O2(g) CO2(g) -67.6kcal

H

C(s) + O2(g)

CO2(g)

H=-94.0 kcal

CO(g) + .5O2(g)H=-26.4 kcal

H=-67.6 kcal

Reaction Coordinate

Water Gas Reaction

H2O(g) + C(s) CO(g) + H2(g) +31.4kcal

CO(g) + .5O2(g) CO2(g) -67.6kcal

H2(g) + .5O2(g) H2O(g) -57.8kcal

H

C(s) + O2(g)

CO2(g)

H= -94.0kcal

C(s) + H2O(g)

CO(g) + H2(g) + O2(g)

H = +31.4kcal

H= -125.4kcal

Reaction Coordinate

CO2(g) + H2O(g)

H= -94.0kcal

Using HESS’S LAW1. The reactants are theoretically decomposed

into elements in their standard states.

2. The products are theoretically formed from elements in their standard states.

3. The reactions and their enthalpy values are then added together.

Problem Set Predict the enthalpy changes for the following

reactions:

CH4(g) + 2 O2(g) CO2(g) + 2H2O(l)

2SO2(g) + O2(g) 2SO3(g)

2NH3(g) + 3.5 O2(g) 2NO2(g) + 3H2O(g)

P4O10(s) + 6H2O(l) 4H3PO4(aq)

H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l)

MORE OF THE SAME

Using ΔHR = ΣΔHf of products - ΣΔHf of reactants, find ΔHR for:

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

2. C3H8(g) + 5 O2(g) 3CO2(g) + 4H2O(l)

3. 2NH3(g) + 2.5 O2(g) 2NO(g) +3H2O(l)

4. Mg3N2(s) +6H2O(l) 3Mg(OH)2 (s) + 2NH3(g)

THE SAME OLD SAME OLD5. Given: C6H6(l) +7.5 O2(g) 6CO2(g) +

3H2O(l) ΔHR = -780.9 kcal

Find the Heat of Formation of C6H6(l)6.How much heat is released in the reaction of

4.50 g H2 with N2?

3H2(g) + N2(g) 2NH3(g) + 22.0 kcal

7. What is Δ HR for the reaction in #6?

8. How much heat is needed in the reaction of 45.9 grams of carbon and excess sulfur?

84.0 kcal + 4C(s) + S8(s) 4CS2(l)

9. What mass of water is formed if 95.6 kcal of heat were released when H2 and O2 react?

2H2(g) + O2(g) 2H2O(l) + 136.6 kcal

Use Hess’s Law to calculate ΔHR for:

C4H10(g) + 6.5 O2(g) 4CO2(g) + 5H2O(l)

given: C(s) + O2(g) CO2(g) ΔH = -393.5 kJ H2(g) + .5 O2(g) H2O(l) ΔH = -286 kJ

4C(s) + 5H2(g) C4H10(g) ΔH = -125 kJ

Given: CaO+H2O Ca(OH)2 ΔH = -15.9 kcal CaC2+2H2O Ca(OH)2+ C2H2 ΔH = -31.0 kcal Find ΔHR for CaC2 + H2O CaO + C2H2

Find ΔHR: C8H18 8C + 9H2

Given: ΔHR

C8H18+12.5O2 8CO2+9H2O -1307.5 kcal C + O2 CO2 -94.0 kcal H2 + .5 O2 H2O

-68.3 kcal

Find ΔHR for CO(g) +.5O2(g) CO2(g) given: ΔHR

C(s) +.5O2(g) CO(g) -26.4kcal C(s) + O2(g) CO2(g) -94.0kcal

Spontaneous Processes

• A spontaneous process is a physical or chemical change that once begun, occurs with no outside intervention.

• (Many spontaneous processes require some energy from the surroundings to start the process.)

• Entropy is a measure of the possible ways that the energy of a system can be distributed, and this is related to the freedom of the system’s particles to move and the number of ways they can be arranged.

• The second law of thermodynamics states that spontaneous processes always proceed in such a way that the entropy of the universe increases.

• Entropy is sometimes considered a measure of disorder or randomness of the particles in a system.

• The more spread out the particles are, the more disorder.

• Entropy changes associated with changes in state can be predicted.

• Entropy increases as a substance changes from a solid to a liquid and from a liquid to a gas.

• Dissolving a gas in a solvent always results in a decrease in entropy.

• Assuming no change in physical state occurs, the entropy of a system usually increases when the number of gaseous product particles is greater than the number of gaseous reactant particles.

• With some exceptions, entropy increases when a solid or liquid dissolves in a solvent.

• The random motion of particles of a substance increases as its temperature increases.

• In nature, the change in entropy tends to be positive when:

− The reaction or process is exothermic, which raises the temperature of the surroundings

− The entropy of the system increases.

• Predict the entropy change:

• H2O(l) → H2O(g)

• 2SO3(g) → 2SO2(g) + O2(g)

• NaCl(s) → Na+ (aq) + Cl-

(aq)

• ClF(g) + F2(g) → ClF3(g)

• N2(g) + 3H2(g) → 2NH3(g)