Chapter 6.1-6.2 Thermochemistry PHALL not integrated into presentation.

Chapter 6.1-6.2

Thermochemistry

PHALL not integrated into presentation

THERMODYNAMICSTHERMODYNAMICS

Courtesy of lab-initio.com

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Chapter 6

Table of Contents

Copyright © Cengage Learning. All rights reserved

6.1 The Nature of Energy

6.2 Enthalpy and Calorimetry

6.3 Hess’s Law

6.4 Standard Enthalpies of Formation

6.5 Present Sources of Energy

6.6New Energy Sources

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Chapter 6

Table of Contents


• Finishing ch. 5 test - on computer• Notes with computer for section 6.1-6.2 including

problems (note to me: assigned as HW last year - incorporated into notes this year)

• HW: Pre-lab Calorimetry • Turn in part 2 ch. 5 take home Test• Turn in previous lab reports• Turn in ch. 5 Interactive Notes packet for grade.• Turn in any missing or late assignments - check

Pinnacle for grades

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Section 6.1

The Nature of Energy


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• Capacity to do work or to produce heat.• That which is needed to oppose natural

attractions.• Law of conservation of energy – energy

can be converted from one form to another but can be neither created nor destroyed.

• The First Law of Thermodynamics The total energy content of the

universe is constant.

Energy

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Thermodynamics

• Thermodynamics is the study of energy and its interconversions.

Definitions #1Definitions #1

Energy: The capacity to do work or produce heat

Potential Energy: Energy due to position or composition

Kinetic Energy: Energy due to the motion of the object

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• Potential energy – energy due to position or composition.

• Kinetic energy – energy due to motion of the object and depends on the mass of the object and its velocity.

Energy

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• In the initial position, ball A has a higher potential energy than ball B.

Initial Position

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• After A has rolled down the hill, the potential energy lost by A has been converted to random motions of the components of the hill (frictional heating) and to the increase in the potential energy of B.

Final Position

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Types of Energy

1. Kinetic energy

2. Potential energy

3. Chemical energy

4. Heat energy

5. Electric energy

6. Radiant energy

Energy Transformations

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What are 2 ways energy is transferred?

1) By doing work2) by Heat transfer - thermal energy is transferred

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Energy Transfer by Doing work

• Energy can be transferred from one object to another by doing work. When work is done on an object, it results in a change in the object's motion (more specifically, a change in the object's kinetic energy).

Energy is often defined as the ability to do work. Work equals force multiplied by distance.

An illustration of how doing work is an example of energy transfer

Suppose that a person exerts a force on the wheelbarrow that is initially at rest, causing it to move over a certain distance. Recall that the work done on the wheelbarrow by the person is equal to the product of the person's force multiplied by the distance traveled by the wheelbarrow. Notice that when the force is exerted on the wheelbarrow, there's a change in its motion. Its kinetic energy increases. But where did the wheelbarrow get its kinetic energy? It came from the person exerting the force, who used chemical energy stored in the food they ate to move the wheelbarrow. In other words, when the person did work on the wheelbarrow, they transferred a certain amount of chemical energy stored in the person was transferred to the wheelbarrow, causing its kinetic energy to increase. As a result, the person's store of chemical energy decreases and the wheelbarrow's kinetic energy increases.

Wherever you look, you can see examples of energy transfers. When you turn on a light, you see result of energy being transferred from the sun to the plants to the coal to electricity and finally to light you see. During each of these transfers, energy changes form. There are two main forms of energy, kinetic energy (motion) and potential energy (position). To further classify energy, these forms are sometimes further described as thermal (heat), elastic, electromagnetic (light, electrical, magnetic), gravitational, chemical (food), and nuclear energy.

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• Heat involves the transfer of energy between two objects due to a temperature difference.

• Work – force acting over a distance.• Energy is a state function; work and heat are

not: State Function – property that does not

depend in any way on the system’s past or future (only depends on present state).

Energy

State Functions State Functions depend ONLY on the present state of the system

ENERGY IS A STATE FUNCTION

A person standing at the top of Mt. Everest has the same potential energy whether they got there by hiking up, or by falling down from a plane

WORK IS NOT A STATE FUNCTION

WHY NOT???

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• System – part of the universe on which we wish to focus attention.

• Surroundings – include everything else in the universe.

Chemical Energy

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• Exothermic Reaction: Energy flows out of the system.

• Energy gained by the surroundings must be equal to the energy lost by the system.

Chemical Energy

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• Endothermic Reaction: Heat flow is into a system. Absorb energy from the surroundings.

Chemical Energy

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Concept Check

Is the freezing of water an endothermic or exothermic process? Explain.

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Concept Check

Classify each process as exothermic or endothermic. Explain. The system is underlined in each example.

a) Your hand gets cold when you touch ice.

b) The ice gets warmer when you touch it.

c) Water boils in a kettle being heated on a stove.

– Water vapor condenses on a cold pipe.– Ice cream melts.

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Concept Check

Classify each process as exothermic or endothermic. Explain. The system is underlined in each example.

a) Your hand gets cold when you touch ice.

b) The ice gets warmer when you touch it.

c) Water boils in a kettle being heated on a stove.

– Water vapor condenses on a cold pipe.– Ice cream melts.

Exo

Endo

Endo

Exo

Endo

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Concept Check

For each of the following, define a system and its surroundings and give the direction of energy transfer.

a) Methane is burning in a Bunsen burner in a laboratory.

b) Water drops, sitting on your skin after swimming, evaporate.

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Concept Check

Hydrogen gas and oxygen gas react violently to form water. Explain.

Which is lower in energy: a mixture of hydrogen and oxygen gases, or water?

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• Law of conservation of energy is often called the first law of thermodynamics.

• Internal energy E of a system is the sum of the kinetic and potential energies of all the “particles” in the system.

• To change the internal energy of a system:ΔE = q + w

q represents heat

w represents work

Internal Energy

Copyright © Houghton Mifflin Company. All rights reserved. 6–

Work vs. Energy Flow

Click here to watch visualization.

http://college.cengage.com/chemistry/zumdahl/chemistry/7e/assets/students/protected/fae/index.html?layer=act&src=qtiworkflowflash_6_6.xml&w=750&h=434

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• Sign reflects the system’s point of view.• Endothermic Process:

q is positive (heat enters the system)• Exothermic Process:

q is negative (heat exits the system)

Internal Energy

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• Sign reflects the system’s point of view.• Surroundings do work on the system:

w is positive• System does work on surroundings:

w is negative

Internal Energy

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• Work = P × A × Δh = PΔV P is pressure. A is area. Δh is the piston moving a

distance. ΔV is the change in

volume.

Work = F x d (from 9th grade) - applying to a gas

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• For an expanding gas, ΔV is a positive quantity because the volume is increasing. Thus ΔV and w must have opposite signs:

w = –PΔV • To convert between L·atm and Joules, use

1 L·atm = 101.3 J.

Work

Work, Pressure, and VolumeWork, Pressure, and Volume

Expansion Compression+ΔV (increase) -ΔV (decrease)

-w results +w results

Esystem decreases

Work has been done by the system on the surroundings

Esystem increases

Work has been done on the system by the surroundings


Figure 6.4 The Piston, Moving a Distance Against a Pressure P, Does Work On the Surroundings

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Exercise

Which of the following performs more work?

a) A gas expanding against a pressure of 2 atm from 1.0 L to 4.0 L.

b) A gas expanding against a pressure of 3 atm from 1.0 L to 3.0 L.

w = –PΔV

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Exercise

Which of the following performs more work?

a) A gas expanding against a pressure of 2 atm from 1.0 L to 4.0 L.

b) A gas expanding against a pressure of 3 atm from 1.0 L to 3.0 L.

They perform the same amount of work.

(-6 L*atm)

w = –PΔV

ΔΔE = q + wE = q + wΔE = change in internal energy of a system

q = heat flowing into or out of the system

-q if energy is leaving to the surroundings+q if energy is entering from the surroundings

w = work done by, or on, the system

-w if work is done by the system on the surroundings

+w if work is done on the system by the surroundings

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Thermodynamic Quantities

Energy Change in Chemical Energy Change in Chemical ProcessesProcesses

Endothermic:

Reactions in which energy flows into the system as the reaction proceeds.

+ qsystem - qsurroundings

Energy Change in Chemical Energy Change in Chemical ProcessesProcesses

Exothermic: Reactions in which energy flows out of the system as the reaction proceeds.

- qsystem + qsurroundings

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Exercise 1

• Exercise 1 Internal Energy Calculate ΔE for a system undergoing an endothermic process in

which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system.

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Exercise 1

• Exercise 1 Internal Energy Calculate ΔE for a system undergoing an endothermic process in

which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system.

ANSWER: 17.0 kJ

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Exercise 2

Exercise 2 PV Work Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external pressure of 15 atm.

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Exercise 2

Exercise 2 PV Work Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external pressure of 15 atm.

ANSWER: ‒270 L•atm

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Exercise 3 Internal Energy, Heat, and Work

Exercise 3 A balloon is being inflated to its full extent by heating the air inside it. In the final stages of this process, the volume of the balloon changes from 4.00 • 106 L to 4.50 • 106 L by the addition of 1.3 • 108 J of energy as heat. Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate ΔE for the process. (To convert between L atm and J, use 1 L ⋅ ⋅atm = 101.3 J.)

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Exercise 3 Internal Energy, Heat, and Work

Exercise 3 A balloon is being inflated to its full extent by heating the air inside it. In the final stages of this process, the volume of the balloon changes from 4.00 • 106 L to 4.50 • 106 L by the addition of 1.3 • 108 J of energy as heat. Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate ΔE for the process. (To convert between L atm and J, use 1 L ⋅ ⋅atm = 101.3 J.)

ANSWER: 8.0 • 107 J

Stopped here 2-5-13 - need to Stopped here 2-5-13 - need to make ch.3-4 test shorter.make ch.3-4 test shorter.

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Section 6.2

Temperature and Heat


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Temperature - Check out this simulation regarding the effect temperature on molecular motion.

• Click on the following link and • http://phet.colorado.edu/en/simulation/states-of-matter-b

asics• Click on Run Now. Open with Java. • Adjust the temperatures, elements, and states of matter

to see what happens from the solid, liquid, gas tab.

http://phet.colorado.edu/en/simulation/states-of-matter-basics

http://phet.colorado.edu/en/simulation/states-of-matter-basics

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• A measure of the random motions of the components of a substance.

Temperature

Temperature (T)—is proportional to the average kinetic energy of the molecules, KEave . “Heat ‘em up and speed ‘em up”.

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• A flow of energy between two objects due to a temperature difference between the objects. Heat is the way in which thermal energy is

transferred from a hot object to a colder object.

Heat

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Heat

Two systems with different temperatures that are in thermal contact will exchange thermal energy, the quantity of which is call heat. This transfer of energy in a process (flows from a warmer object to a cooler one, transfers heat because of temperature difference but, remember, temperature is not a measure of energy—it just reflects the motion of particles)

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Section 6.2

Enthalpy and Calorimetry


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Enthalpy

Enthalpy (H)– flow of energy (heat exchange) at constant pressure when two systems are in contact.

Enthalpy of reaction (ΔHrxn) – amount of heat released (negative values) or absorbed (positive values) by a chemical reaction at constant pressure in kJ/molrxn Enthalpy of combustion (ΔHcomb)—heat absorbed or released by burning (usually with O2) in kJ/molrxn; note that combustion reactions yield oxides of that which is combusted

Enthalpy of formation (ΔHf) – heat absorbed or released when ONE mole of compound is formed from elements in their standard states in kJ/molrxn Enthalpy of fusion (ΔHfus)—heat absorbed to melt (overcome IMFs) 1 mole of solid to liquid @ MP expressed in kJ/molrxn Enthalpy of vaporization (ΔHvap)—heat absorbed to vaporize or boil (overcome IMFs) 1 mole liquid to vapor @BP in kJ/molrxn

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• State function• ΔH = q at constant pressure (ex. atmospheric pressure)

• ΔH = Hproducts – Hreactants

Measure only the change in enthalpy, ΔH ( the difference between the potential energies of the products and the reactants)

• A system that releases heat to the surroundings, an exothermic reaction, creates a negative ΔH because the enthalpy of the products is lower than the enthalpy of the reactants of the system (Figure 1).

• A system that absorbs heat from the surroundings, an endothermic reaction, creates a positive ΔH, because the enthalpy of the products is higher than the enthalpy of the reactants of the system.

• A positive ΔH is endothermic rxn• A negative ΔH is exothermic rxn

Change in Enthalpy (ΔH )

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Enthalpy Calculations

Enthalpy can be calculated from several

sources including:

Stoichiometry

Calorimetry

From tables of standard values

Hess’s Law

Bond energies

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Stoichiometrically:

• Exercise 4 • Upon adding solid potassium hydroxide pellets to water the

following reaction takes place:

• KOH(s) → KOH(aq) + 43 kJ/mol

Answer the following questions regarding the addition of 14.0 g of KOH to water:

a) Does the beaker get warmer or colder? How do you know?

b) Is the reaction endothermic or exothermic?

c) What is the enthalpy change for the dissolution of the

14.0 grams of KOH?

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Stoichiometrically:

• Exercise 4 • Upon adding solid potassium hydroxide pellets to water the following

reaction takes place: • KOH(s) → KOH(aq) + 43 kJ/mol

Answer the following questions regarding the addition of 14.0 g of KOH to water:

a) Does the beaker get warmer or colder?

b) Is the reaction endothermic or exothermic?

c) What is the enthalpy change for the dissolution of the

14.0 grams of KOH?

ANSWER: (a) warmer (the +43 kJ/mole is heat on the product side) (b) exothermic (c) −10.7 kJ/molrxn (take 14 g divide by the molar mass of KOH - to get # moles - 1 to 1 ratios and thus if 1 mole produces 43 kJ, then approx. .25 mol would be (.25 mol x 43 kJ)

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Exercise 5

Consider the combustion of propane:

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)

ΔH = –2221 kJ

Assume that all of the heat comes from the combustion of propane. Calculate ΔH in which 5.00 g of propane is burned in excess oxygen at constant pressure.

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Exercise 5

Consider the combustion of propane:

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)

ΔH = –2221 kJ

Assume that all of the heat comes from the combustion of propane. Calculate ΔH in which 5.00 g of propane is burned in excess oxygen at constant pressure.

5.00 g C3H8 x 1 mol C3H8 /44 g x -2221 kJ/1 mol C3H8 = –252 kJ

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• Science of measuring heat

• Heat capacity – energy required to raise temp. by 1 degree (Joules/ °C) – Specific heat capacity: (s)

• The energy required to raise the temperature of one gram of a substance by one degree Celsius.

– Molar heat capacity: (Cp)• The energy required to raise the temperature of

one mole of substance by one degree Celsius.

Calorimetry

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• If two reactants at the same temperature are mixed and the resulting solution gets warmer, this means the reaction taking place is exothermic.

• An endothermic reaction cools the solution.

Calorimetry

CalorimetryCalorimetryThe amount of heat absorbed or released during a physical or chemical change can be measured, usually by the change in temperature of a known quantity of water in a calorimetercalorimeter.

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A Coffee–Cup Calorimeter Made of Two Styrofoam CupsCoffee-cup calorimetry – in the lab this is how we experiment to find energy of a particular system. We use a Styrofoam® cup, reactants that begin at the same temperature and look for change in temperature. After all data is collected (mass or volume; initial and final temperatures) we can use the specific formula to find the energy released or absorbed. We refer to this process as constant pressure calorimetry. ** q = ΔH @ these conditions**

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• q = s × m × ΔT• q = Energy released (heat) (or

gained)

s = specific heat capacity (J/°C·g)

m = mass (g)

ΔT = change in temperature (°C)

(Q = Cp x m × ΔT) - some use Cp as specific heat

variable.

Calorimetry

Units for Measuring HeatUnits for Measuring HeatThe JouleJoule is the SI system unit for measuring heat:

The caloriecalorie is the heat required to raise the temperature of 1 gram of water by 1 Celsius degree

1 BTU (British Thermal Unit) is the heat required to raise the temperature of 1 pound of water by 1 F

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Section 6.2

Measuring Energy Changes


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• The common energy units for heat are the calorie and the joule. calorie – the amount of energy (heat)

required to raise the temperature of one gram of water 1oC.

Joule – 1 calorie = 4.184 joules

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Convert 60.1 cal to joules and kJ

Exercise 6

1000 J = 1 kiloJoule (kJ)251 J = 0.251 kJ

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• Energy (q) released or gained at constant pressure:

• q = m x s x ΔT • q = quantity of heat (Joules or calories)

• m = mass in grams • ΔT = Tf −Ti (final – initial)

• s = specific heat capacity ( J/g°C) (or Cp with some)

Specific heat of water (liquid state) = 4.184 J/g°C (or 1.00 cal/g°C)

*Water has one of the highest specific heats known! This property makes life on earth possible and regulates earth’s temperature year round!

Heat lost by substance = heat gained by water

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Calculate the amount of heat energy (in joules) needed to raise the temperature of 6.25 g of water from 21.0°C to 39.0°C.

Where are we going?• We want to determine the amount of energy

needed to increase the temperature of 6.25 g of water from 21.0°C to 39.0°C.

What do we know?• The mass of water and the temperature increase.

Exercise 7

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Calculate the amount of heat energy (in joules) needed to raise the temperature of 6.25 g of water from 21.0°C to 39.0°C.

What information do we need?• We need the specific heat capacity of water.

4.184 J/g°C

How do we get there?

Example 7

q = s × m × ΔT

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Exercise 8

A sample of pure iron requires 142 cal of energy to raise its temperature from 23ºC to 92ºC. What is the mass of the sample? (The specific heat capacity of iron is 0.45 J/gºC.)

a) 0.052 g

b) 4.6 g

c) 19 g

d) 590 g

q = s × m × ΔT

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Exercise 8

A sample of pure iron requires 142 cal of energy to raise its temperature from 23ºC to 92ºC. What is the mass of the sample? (The specific heat capacity of iron is 0.45 J/gºC.)

a) 0.052 g

b) 4.6 g

c) 19 g

d) 590 g

q = s × m × ΔT

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Concept Check

A 100.0 g sample of water at 90°C is added to a 100.0 g sample of water at 10°C.

The final temperature of the water is:

a) Between 50°C and 90°C

b) 50°C

c) Between 10°C and 50°C

Example 9

q = s × m × ΔT

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Concept Check

A 100.0 g sample of water at 90°C is added to a 100.0 g sample of water at 10°C.



b) 50°C - work on next screens


Example 9

q = s × m × ΔT

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Problem

• We start by calling the final, ending temperature 'x.' Keep in mind that BOTH water samples will wind up at the temperature we are calling 'x.' Also, make sure you understand that the 'x' we are using IS NOT the Δt, but the FINAL temperature. This is what we are solving for.

The warmer water goes down from to 90 to x, so this means its Δt equals 90 minus x. The colder water goes up in temperature, so its Δt equals x minus 10.

That last paragraph may be a bit confusing, so let's compare it to a number line:

---10-------------------------x-----------------------------------90----

To compute the absolute distance, it's the larger value minus the smaller value, so 90 to x is 90 minus x and the distance from x to 10 is x - 10.

These two distances on the number line represent our two Δt values:

a) the Δt of the warmer water is 90 minus x

b) the Δt of the cooler water is x minus 10

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Problem

• q lost = q gain

• q = (mass) (Δt) (s)

So:

(mass) (Δt) (s) = (mass) (Δt) (s)

With qlost on the left side and qgain on the right side. Since these are both liquid water the specific heat capacity (s) are equal and can be eliminated from the equation. Also, since the amount of water is the same, it too could be eliminated. Then we let x be final temperature

Substituting values into the above, we then have

90 - x = x - 10 Then solve for x (the final temperature)100 = 2x so x = 50 The final temperature is at 50 degrees C.

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Concept Check

A 100.0 g sample of water at 90.°C is added

to a 500.0 g sample of water at 10.°C.



b) 50°C


Calculate the final temperature of the water.

100 x 4.19 x (90-t) = 500 x 4.19 x (t - 10) =

9000 -100t = 500t - 5000 so 14000 = 600t

t = 23.333 = 23°C

----- -----

Exercise 10

q = s × m × ΔT

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Concept Check

You have a Styrofoam cup with 50.0 g of water at 10.C. You add a 50.0 g iron ball at 90.C to the water. (sH2O = 4.18 J/°C·g and sFe = 0.45 J/°C·g)



b) 50°C


Calculate the final temperature of the water.

(4.18) (t-10) = (.45) (90-t) t = 17.77 = 18°C

Exercise 11q = s × m × ΔT


Table 6.1 The Specific Heat Capacities of Some Common Substances

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Type 2 Problem Exercise 12

A 110.0 g sample of Cu (specific heat capacity = .20 J/°C·g is heated to 82.4 °C and then placed in a container of water at 22.3°C. The final temperature of the water and the copper is 24.9°C. What was the mass of the water in the original container, assuming that all of the heat lost by the copper is gained by the water?

The heat lost by copper is equal to:110.0g Cu x 0.20 J/°C·g x (82.4°C - 24.9°C)= 1265 J

The heat lost (-value) by Cu is equal to the heat gained by water.mH2O = q/(s x Δt) so 1265 J / (4.184 J x (24.9-22.3)) = 116.3 g water.

7878

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The Heat Curve for Water, going from -20 to 120 oC,

7979

The solid temperature is rising from -20 to 0 oC (use q = s x m x ΔT)

The solid is melting at 0o C; no temperature change (use q = mol x ΔHfusion)

The liquid temperature is rising from 0 to 100 oC (q = s x m x ΔT)

The liquid is boiling at 100o C; no temperature change (use q = mol x ΔHvap.)

The gas temperature is rising from 100 to 120 oC (use (use q = s x m x ΔT)

120The liquid is boiling at 100o C; no temperature change (use q = mol x ΔHvaporization)

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Exercise 13

• 20.0 mL of pure water at 285 K is mixed with 48.0 mL of water at 315 K. What is the final temperature of the mix? Requires 5 steps. Assume all heat from warm water transferred to cold water. q gain = q lost (Density pure water is 1.0 g/mL so 1 gm/L = m/20 mL

• a) m × ΔT x s = m × ΔT x s • b) (20 g) (tf-285) (s) = (48 g) (315 - tf) (s) • c) 20tf - 5700 = 15120 - 48tf

• d) 68tf = 20820• e) tf = 306.17 = 306 K

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Exercise 14

• 20.0 g of water at 28.5 C is mixed with 32.5 g of water at 77.0 C. What is the final temperature after complete mixing? (Assume no energy is lost to the surroundings.)

q cold = q warm

• a) m × ΔT x s = m × ΔT x s • b) (20 g) (tf-28.5) (s) = (32.5 g) (77 - tf) (s) (liquids - s cancels)• c) 20tf - 570 = 2502.5 - 32.5tf

• d) 52.5tf = 3072.5• e) tf = 58.5 C

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Exercise 15

• Exercise 15 In a coffee cup calorimeter, 100.0 mL of 1.0 M NaOH and 100.0 mL of 1.0 M HCl are mixed. Both solutions were originally at 24.6°C. After the reaction, the final temperature is 31.3°C. Assuming that all solutions have a density of 1.0 g/cm3 and a specific heat capacity of 4.184 J/g°C, calculate the enthalpy change for the neutralization of HCl by NaOH. Assume that no heat is lost to the surroundings or the calorimeter.

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Exercise 15

• Exercise 15 In a coffee cup calorimeter, 100.0 mL of 1.0 M NaOH and 100.0 mL of 1.0 M HCl are mixed. Both solutions were originally at 24.6°C. After the reaction, the final temperature is 31.3°C. Assuming that all solutions have a density of 1.0 g/cm3 and a specific heat capacity of 4.184 J/g°C, calculate the enthalpy change for the neutralization of HCl by NaOH. Assume that no heat is lost to the surroundings or the calorimeter.

• ANSWER: ‒5.6 kJ/mol (m = mass = 200 g (because we have 100 mL NaOH plus 100 mL HCl

each with a density of 1 g / mL) Then using the equation q = m * s * ΔT = (200 g) (4.184 J/g*C) * 6.7 C) = 5606.56 J = 5.6 kJ

• Since this is an exothermic reaction, q is negative.

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Exercise 16

• When 1 mole of methane (CH4) is burned at constant pressure, 890 kJ/mol of energy is released as heat. Calculate ΔH for a process in which a 5.8 gram sample of methane is burned at constant pressure.

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Exercise 16

• When 1 mole of methane (CH4) is burned at constant pressure, 890 kJ/mol of energy is released as heat. Calculate ΔH for a process in which a 5.8 gram sample of methane is burned at constant pressure. • ANSWER: ΔH = heat flow = ‒320 kJ/mol

5.8 g / 16.0 g = .3625 moles so set up proportion1 mole/890 kJ = 0.36 mol/x so x = (890)(.36)/(1) = 320.4 = 320 kJ/mol and is -320

kJ/mol since exothermic reaction

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Exercise 17

• Exercise 17 Constant-Pressure Calorimetry When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0°C is mixed

with 1.00 L of 1.00 M Na2SO4 solution at 25°C in a calorimeter, the white solid BaSO4 forms and the temperature of the mixture increases to 28.1°C. Assuming that the calorimeter absorbs only a negligible quantity of heat, and that the specific heat capacity of the solution is 4.18 J/°C g, and ⋅that the density of the final solution is 1.0 g/mL, calculate the enthalpy change per mole of BaSO4 formed.

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Exercise 17

• Exercise 17 Constant-Pressure Calorimetry When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0°C is mixed

with 1.00 L of 1.00 M Na2SO4 solution at 25°C in a calorimeter, the white solid BaSO4 forms and the temperature of the mixture increases to 28.1°C. Assuming that the calorimeter absorbs only a negligible quantity of heat, and that the specific heat capacity of the solution is 4.18 J/°C g, and ⋅that the density of the final solution is 1.0 g/mL, calculate the enthalpy change per mole of BaSO4 formed.

• ANSWER = ‒26 kJ/mol• 2 L total volume = 2000 mL x (1.0 g/mL) = 2000 grams of sol.• ΔH = q at constant pressure = q = s × m × ΔT

• = (4.18 J/C*g) (2000 g) (28.1-25.0 C) = 25,916 Joules or 25.9 = 26 kJ per mole

Specific HeatSpecific HeatThe amount of heat required to raise the temperature of one gram of substance by one degree Celsius.

SubstanceSubstance Specific Heat (J/g·C)Specific Heat (J/g·C)Water (liquid) 4.184.18Ethanol (liquid) Ethanol (liquid) 2.442.44Water (solid) Water (solid) 2.062.06Water (vapor) Water (vapor) 1.871.87Aluminum (solid) Aluminum (solid) 0.8970.897Carbon (graphite,solid) Carbon (graphite,solid) 0.7090.709Iron (solid) Iron (solid) 0.4490.449Copper (solid) Copper (solid) 0.3850.385Mercury (liquid) Mercury (liquid) 0.1400.140Lead (solid)Lead (solid) 0.1290.129Gold (solid) Gold (solid) 0.1290.129

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END OF NOTES 6.1-6.2

• HW: Read Section 6.1-6.3 by next Tuesday.• HW: Pre-lab Calorimetry due on Friday for lab• Turn in assignments including take home test ch. 5 if not

done already no later than Friday if not turned in today.

Chapter 6.1-6.2 Thermochemistry PHALL not integrated into presentation.

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