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Chapter Six

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Prentice Hall © 2005Hall © 2005

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Our Plan:Test Results Notes - Thermochem

Homework (Write in Planner):Problems from the textbook (due 10/30) You can do up through #38

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11% Curve 73.26% Average (5% higher than 1st test) 61.1% of students scored better on this

test than the first one.Letter Grade # of Students

A 2 (95.5 %)

B 7

C 3

D 3

F 4

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The study of energetics of chemical reactions

Thermodynamics is the branch of science which studies the transformation of energy from one form to another

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Because everything is energy!

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The questions that chemical thermodynamics ask are:

How much heat is evolved during a chemical reaction ("thermo chemistry")?

What determines the direction of spontaneous chemical change?

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For example, if you put a flame to a 2:1 mixture of H2 and O2 there is an almighty bang and the reaction proceeds spontaneously to products, (water) without any further supply of external energy.

Why does it not happen that water spontaneously goes to H2 and O2 when we put a lit taper to it under the same conditions?

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Energy is the capacity to do work (to displace or move matter).

Energy literally means “work within”; however, an object does not contain work.

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Kinetic energy is the energy of motion.

Ek = ½ mv2

m = massv = velocity

Energy has the units of joules (J or kg . m2/s2) or calories

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The energy STORED that is released through motion

It is the energy within a system because of its position or arrangement

Examples: Ball in the air Battery Chemical energy

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At what point in each bounce is the potential energy of the ball at a

maximum?

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Potential energy can be converted to kinetic energy

Kinetic energy can be converted to potential energy

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1. Light Energy – light bulb2. Heat Energy – fire3. Electrical Energy – putting your

finger in an electrical socket4. Mechanical Energy – engine or a

water wheel5. Chemical Energy – burning gasoline

(chemical energy is converted to heat and light)

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Chemical energy is stored inside a substance until a substance undergoes a chemical change. Then it can be released or absorbed.

Chemical energy is stored in chemical bonds between atoms.

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Thermodynamics – the study of heat, energy, and the ability to do work.

Energy – the capacity to do work or produce heat.Both heat and work are energy transfers

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The First Law of Thermodynamics – energy is not created or destroyed it is just transferredAll energy in the universe is constant!

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Work – a force acts on something causing it to move (w)

Heat – energy transfer from thermal interaction (q)

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System – the part of the universe we are interested in studying (we choose the system)

Surroundings – the rest of the universe

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• Open: energy and matter can be exchanged with the surroundings.

• Closed: energy can be exchanged with the surroundings, matter cannot.

• Isolated: neither energy nor matter can be exchanged with the surroundings.

A closed system; energy (not matter) can be exchanged.

After the lid of the jar is unscrewed, which kind of system is it?

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We are going to be talking about our chemical system losing and gaining energy, so it is logical to introduce a quantity which represents the total energy of the system. This is known as the internal energy, E. It’s important to note that your textbook

refers to internal energy as U sometimes.

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Internal energy is defined as the total energy of our system - chemical, nuclear, heat, gravitational, any other type of energy you can think of - the sum of all the system's energy (kinetic + potential).

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A little thought will convince ourselves that it would be impossible to actually measure the total internal energy of our system.

So why define a quantity which we cannot measure?

Well, although we cannot measure the absolute value of the internal energy of a system, we can measure changes in the internal energy. And since thermodynamics is all about changes in energy this makes the change in internal energy of a system a very useful experimental quantity.

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The internal energy, E, of a system may change in 3 different ways. The internal energy, E, of a system will change if :

1. heat passes into or out of the system;

2. work is done on or by the system;

3. mass enters or leaves the system.

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The value of a state function is independent of the history of the system. All that matters are the starting and ending values, not how you got there.Elevation change when climbing

a mountain is an example

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The fact that internal energy is a state function is extremely useful because we can measure the energy change in the system by knowing the initial energy and the final energy.

In other words, we don’t need all of the detail of a process to measure the change in the value of a state function.

In contrast, we do need all of the details to measure the HEAT or the WORK of a system.

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Internal energy (E) is the total energy contained within a system

Part of E is kinetic energy (from molecular motion) Translational motion, rotational

motion, vibrational motion. Collectively, these are sometimes

called thermal energy

• Part of E is potential energy– Intermolecular and intramolecular forces of

attraction, locations of atoms and of bonds.– Collectively these are sometimes called

chemical energy

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There are two ways to classify conversion of potential energy to kinetic energy and vice versa: Heat – q Work – w

Heat and work are NOT state functions (they are pathway dependent) This makes it difficult to distinguish

between the energy from heat and the energy from work. That’s again why ΔE being a state function is so important.

Car going down mountain example from Crash Course.

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Energy entering a system carries a positive sign (gains energy)

Energy leaving a system carries a negative sign (loses energy)Energy, work, and heat are not really positive or negative. The sign just tells you the direction of flow.

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A negative quantity of work signifies that the system loses energy. - w = work done by system (expansion)

A positive quantity of work signifies that the system gains energy. +w = work done on system

(compression)

In gases, work is related to changes in volume.

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For now we will consider only pressure-volume work.

work (w) = –PV

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Technically speaking, heat is not “energy.”

Heat is energy transfer between a system and its surroundings, caused by a temperature difference. Passes from higher to lower areas of heat.

More energetic molecules …

… transfer energy to less energetic molecules.

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+ q = heat gained by system (heat is transferred to the system)Endothermic = the system gets colder

- q = heat lost by system (heat is transferred from the system to the surroundings)Exothermic = the system gets warmer

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Surroundings are at 25 °C

Hypothetical situation: all heat is instantly released to the surroundings. Heat = qrxn

Typical situation: some heat is released to the surroundings,

some heat is absorbed by the solution.

In an isolated system, all heat is absorbed by the solution.

Maximum temperature rise.

25 °C

32.2 °C 35.4 °C

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Sign Convention

Meaning

+ΔE System gains energy (produces E)

-ΔE System loses energy (requires E)

+w Work done on system

-w Work done by system

+q Heat absorbed by system (endothermic)

-q Heat given off by system (exothermic)

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calorie - the quantity of heat required to change the temp of one gram of water one degree Celsius

SI Unit of basic energy is the Joule (J)

1 cal = 4.184 J

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“Energy cannot be created or destroyed.” A change in the internal energy of a

system is due only to heat and work. Inference: the internal energy change of a

system is simply the difference between its final and initial states:

E = Efinal – Einitial

Additional inference: if energy change occurs only as heat (q) and/or work (w), then:

E= q + w

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Example 6.1A gas does 135 J of work while expanding, and at the same time it absorbs 156 J of heat. What is the change in internal energy?

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Enthalpy is the sum of the internal energy and the pressure-volume product of a system:

H = E + PV

Most reactions occur at constant pressure, so for most reactions, the heat

evolved equals the enthalpy change.

The evolved H2 pushes back the atmosphere;

work is done at constant pressure.

For a process carried out at constant pressure,

q = E + PV so q = H

Mg + 2 HCl --> MgCl2 + H2

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Enthalpy is an extensive property. It depends on how much of the substance is

present. This makes sense because heat transfers into and

out of bonds, the more moles you have, the more bonds you have, therefore more energy can be released to or taken from the environment.

• Since E, P, and V are all state functions, enthalpy (H) must be a state function also.

• Enthalpy changes have unique values. H = q

Two logs on a fire give off twice as much heat as does one log.

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Molar enthalpy is the amount of heat lost or gained when one mole of a compound is formed from its constituent elements.

∆Hᵒf = Standard enthalpies of formation (molar enthalpy at 25 degrees C and 1 atm) have been calculated for nearly all compounds and are in the back of your book. (Appendix C)

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H2(g) + ½O2(g) → H2O(l) ΔH = –285.8 kJ mol–1

2C(s) + H2(g) → C2H2(g) ΔH = 227.0 kJ mol–1

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Values of H are measured experimentally. Negative values indicate exothermic reactions. Positive values indicate endothermic reactions.

A decrease in enthalpy during the reaction; H

is negative.

An increase in enthalpy during the reaction; H

is positive.

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H changes sign when a process is reversed. Therefore, a cyclic process has the value H = 0.

Same magnitude; different signs.

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Example 6.3Given the equation

(a) H2(g) + I2(s) --> 2 HI(g) H = +52.96 kJ

calculate H for the reaction

(b) HI(g) --> ½ H2(g) + ½ I2(s).

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The complete combustion of liquid octane, C8H18, to produce gaseous carbon dioxide and liquid water at 25 °C and at a constant pressure gives off 47.9 kJ of heat per gram of octane. Write a chemical equation to represent this information.

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What quantity of heat is associated with the complete combustion of 1.00 kg of sucrose (C12H22O11)?

KNOWN: 342.3 g/mol sucrose

-5.65 x103 kJ/mole

1.00 kg x 1000 g

1 kg

x 1 mol

342.3 g

x -5.65 x 103 kJ

1 mol

=

-1.65 x 104 kJ

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For problem-solving, heat evolved (exothermic reaction) can be thought of as a product. Heat absorbed (endothermic reaction) can be thought of as a reactant.

We can generate conversion factors involving H. For example, the reaction:

H2(g) + Cl2(g) --> 2 HCl(g) H = –184.6 kJ

can be used to write:

–184.6 kJ ———— 1 mol H2

–184.6 kJ ———— 1 mol Cl2

–184.6 kJ ———— 2 mol HCl

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Example 6.5

What is the enthalpy change associated with the formation of 5.67 mol HCl(g) in this reaction?

H2(g) + Cl2(g) --> 2 HCl(g) H = –184.6 kJ

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6.5 A – What is the enthalpy change when 12.8 g H2 (g) reacts with excess Cl2 (g) to form HCl (g)

H2 (g) + Cl2 (g) → 2 HCl (g) ∆H = -184.6 kJ

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Turn in: Get out Notes

Our Plan: Clicker Review Notes Calorimetry POGIL (as practice) Find Someone Who

Homework (Write in Planner): Problems from the textbook (due 10/30)

You can complete up through #60

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Heat capacity – quantity of heat required to change the temperature of a system by one degree

Molar heat capacity – quantity of heat required to change the temperature of a mole of a substance one degree

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C = q/∆T Example 6.6 – Calculate the heat

capacity of an aluminum block that must absorb 629 J of heat from its surroundings in order for its temperature to rise from 22ᵒC to 145ᵒC.

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The science of measuring changes of heat.

A calorimeter is a device used to make this measurement.

General principle of calorimetery:Heat Lost = - Heat Gained

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For a reaction carried out in a calorimeter, the heat evolved by a reaction is absorbed by the calorimeter and its contents.

qrxn = – qcalorimeter

qcalorimeter = mass x specific heat x T

By measuring the temperature change that occurs in a calorimeter, and using the specific heat and mass of the contents, the heat evolved (or absorbed) by a reaction can be determined and the enthalpy change calculated.

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Heat evolved in a reaction is absorbed by the calorimeter and its contents.

The specific heat (c) is the heat capacity of one gram of a pure substance (or homogeneous mixture).

q = c m T

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High specific heat means that the substance is resistant to changes in temperature.

Many metals have low specific heat making them easy to heat up or cool down.

Water has a high specific heat. If it did not, life on Earth would hardly be possible.

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q = mass x specific heat x T If T is positive (temperature increases),

q is positive and heat is gained by the system.

If T is negative (temperature decreases), q is negative and heat is lost by the system.

The calorie, while not an SI unit, is still used to some extent.

Water has a specific heat of 1 cal/(g oC). 4.184 J = 1 cal One food calorie (Cal or kcal) is actually

equal to 1000 cal.

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Many metals have low specific heats.

The specific heat of water is higher than that of almost any other substance.

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q = m x c x T The mass and specific heat

of an object are often combined into a single quantity (heat capacity – C)

Thus, q = C x T

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Constant Pressure Calorimeter “coffee cup” calorimeter Open to atmosphere Appropriate for solution chemistry

Constant Volume Calorimeter Bomb calorimeter Sealed and isolated Appropriate for gas phase

chemistry

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Example 6.7

How much heat, in joules and in kilojoules, does it take to raise the temperature of 225 g of water from 25.0 to 100.0 °C?

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Example 6.8

What will be the final temperature if a 5.00-g silver ring at 37.0 °C gives off 25.0 J of heat to its surroundings? Use the specific heat of

silver listed in Table 6.1.

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Example 6.9

A 15.5-g sample of a metal alloy is heated to 98.9 °C and then dropped into 25.0 g of water in a calorimeter. The temperature of the water rises from 22.5 to 25.7 °C. Calculate the specific heat of the alloy.

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Example 6.11

A 50.0-mL sample of 0.250 M HCl at 19.50 °C is added to 50.0 mL of 0.250 M NaOH, also at 19.50 °C, in a calorimeter. After mixing, the solution temperature rises to 21.21 °C. Calculate the heat of this reaction.

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A copper bottomed saucepan has a heat capacity of 227.3 J/ºC. Calculate how much heat the saucepan gains when its temperature changes from 24.7 ºC to 110.2 ºC.

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6.8A – A 454 gram block of lead is at an initial temperature of 22.5 degrees C. What will be the temperature of the lead after it absorbs 4.22 kJ of heat from its surroundings?

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I have finished notes up through homework problem #60. DO AT LEAST THOSE PROBLEMS BY MONDAY!

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Turn in:Get out notes & textbook

Our Plan:Review RaceFinish the Unit Notes

Homework (Write in Planner):Problems from the textbook (due 10/30)

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Some reactions cannot be carried out “as written.”

Consider the reaction:C(graphite) + ½ O2(g) --> CO(g).

If we burned 1 mol C in ½ mol O2, both CO and CO2 would probably form. Some C might be left over. However …

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… enthalpy change is a state function. The enthalpy change of a reaction is the

same whether the reaction is carried out in one step or through a number of steps.

Hess’s Law: If an equation can be expressed as the sum of two or more other equations, the enthalpy change for the desired equation is the sum of the enthalpy changes of the other equations.

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Example 6.14Calculate the enthalpy change for reaction (a) given the data in equations (b), (c), and (d).

(a) 2 C(graphite) + 2 H2(g) → C2H4(g) H = ?

(b) C(graphite) + O2(g) → CO2(g) H = –393.5 kJ

(c) C2H4(g) + 3 O2 → 2 CO2(g) + 2 H2O(l) H = –1410.9 kJ

(d) H2(g) + ½ O2 → H2O(l) H = –285.8 kJ

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Example: Find the Hrxn of 2 NO (g) + O2 (g) N2O4 (g), given the Hrxn for the following reactions:

N2O4 (g) 2 NO2 (g) Hrxn = 57.93 kJ/mol

2 NO (g) + O2 (g) 2 NO2 (g) Hrxn = -113.14 kJ/mol

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Calculate the enthalpy of reaction for the following reaction: 2 Al(s) + 3 Cl2(g) 2 AlCl3(s), given the reactions below.

2 Al(s) + 6 HCl(aq) 2 AlCl3(aq) + 3 H2(g) H = -1049 kJ/molHCl(g) HCl(aq) H = -74.8 kJ/molH2(g) + Cl2(g) 2 HCl(g) H = -185 kJ/molAlCl3(s) AlCl3(aq) H = -323 kJ/mol

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H2 (g) + ½ O2 (g) H2O (l) H = -286 kJ

2 Na (s) + ½ O2 (g) Na2O (s) H = -414 kJ

Na (s) + ½ O2 (g) + ½ H2 (g) NaOH (s) H = -425 kJ

Based on the information above, what is the standard enthalpy change for the following reaction? Na2O (s) + H2O 2 NaOH (s)

(A) -1,125 kJ(B) -978 kJ(C) -722 kJ(D) -150 kJ(E) +275 kJ

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• It would be convenient to be able to use the simple relationship

ΔH = Hproducts – Hreactants

to determine enthalpy changes.

• Although we don’t know absolute values of enthalpy, we don’t need them; we can use a relative scale.

• We define the standard state of a substance as the state of the pure substance at 1 atm pressure and the temperature of interest (usually 25 °C).

• The standard enthalpy change (ΔH°) for a reaction is the enthalpy change in which reactants and products are in their standard states.

• The standard enthalpy of formation (ΔHf°) for a reaction is the enthalpy change that occurs when 1 mol of a substance is formed from its component elements in their standard states.

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When we say “The standard enthalpy of formation of CH3OH(l) is –238.7 kJ”, we are saying that the reaction:

C(graphite) + 2 H2(g) + ½ O2(g) → CH3OH(l)

has a value of ΔH of –238.7 kJ.

We can treat ΔHf° values as though they were absolute enthalpies, to determine enthalpy changes for reactions.Question: What is ΔHf° for an element in its standard state [such as O2(g)]? Hint: since the reactants are the same as the products …

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H°rxn = p x Hf°(products) – r x Hf°(reactants) The symbol signifies the summation of several

terms. The symbol signifies the stoichiometric coefficient

used in front of a chemical symbol or formula. In other words …

1. Add all of the values for Hf° of the products.

2. Add all of the values for Hf° of the reactants.

3. Subtract #2 from #1(This is usually much easier than using Hess’s Law!)

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Example 6.15Synthesis gas is a mixture of carbon monoxide and hydrogen that is used to synthesize a variety of organic compounds. One reaction for producing synthesis gas is 3 CH4(g) + 2 H2O(l) + CO2(g) --> 4 CO(g) + 8 H2(g) ΔH° = ?Use standard enthalpies of formation from Table 6.2 to calculate the standard enthalpy change for this reaction.

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Try it Out - Example 6.16The combustion of isopropyl alcohol, common rubbing alcohol, is represented by the equation 2 (CH3)2CHOH(l) + 9 O2(g) --> 6 CO2(g) + 8 H2O(l) ΔH° = –

4011 kJUse this equation and data from Table 6.2 to establish the standard enthalpy of formation for isopropyl alcohol.

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C2H4 (g) + 3 O2 (g) 2 CO2 (g) + 2 H2O (g)For the reaction of ethylene represented above, H is -1,323 kJ. What is the value of H if combustion produced liquid water H2O (l), rather than water vapor H2O (g)? (H for the phase change H2O (g) H2O (l) is -44 kJ mol-1.)

(A) -1,235 kJ(B) -1,278 kJ(C) -1,323kJ(D) -1,367 kJ(E) -1,411 kJ

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We can apply thermochemical concepts to reactions in ionic solution by arbitrarily assigning an enthalpy of formation of zero to H+(aq).

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Example 6.18

H+(aq) + OH–(aq) --> H2O(l) ΔH° = –55.8 kJ

Use the net ionic equation just given, together with ΔHf° = 0 for H+(aq), to obtain ΔHf° for OH–

(aq).

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Turn in:Homework

Our Plan:Homework QuestionsSpecific Heat Review ProblemHess’s Law ActivityBegin Hand Warmer Lab

Homework (Write in Planner):Work on Test ReviewHave Pre-Lab and Procedure ready

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http://www.youtube.com/watch?v=JuWtBR-rDQk

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Turn in:Nothing

Our Plan:Hand Warmer LabTest Review

Homework (Write in Planner):TEST MONDAY

Breakfast Club Monday morningLab Report Due Monday

Chapter Six

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