Chapter 6 Thermochem 110 F11 IP - Palomar College 6 Thermochem 2Up.pdf · The system in this case is the contents of the reaction flask. ... An Auto A Battery A perfect Thermos

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Slide 6-1

Chapter 6

Thermochemistry: Energy Flow and Chemical Change

Chapter 6: First Law

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Slide 6-2

Thermochemistry: Energy Flow and Chemical Change

6.1 Forms of Energy and Their Interconversion

6.2 Enthalpy: Heats of Reaction and Chemical Change

6.3 Calorimetry: Laboratory Measurement of Heats of Reaction

6.4 Stoichiometry of Thermochemical Equations

6.5 Hess’s Law of Heat Summation

6.6 Standard Heats of Reaction (ΔH0rxn)


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Slide 6-3

Thermodynamics: First Law

✔Energy flow is key in science; understanding it will help us to understand WHY things occur. When energy is transferred from one object to another, it appears as work and/or as heat

Thermodynamics was developed independent of theory through observation of bulk matter

Thermodynamics: Study and measure of the nature of heat and energy and its transformations

To deal with large groups of molecules and atoms we need to define THREE parameters: System, Surroundings and Thermodynamic Universe

Thermochemistry is a branch of thermodynamics that deals with the heat involved with chemical and physical changes.


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Slide 6-4 Chapter 6: First Law

Thermodynamics: First Law System:��

The part of the universe of immediate interest; composed of particles with their own internal energies (E or U). The system has an internal energy. When a change occurs, the internal

energy changes. Can be closed, open or isolated;

Surroundings:��The portion of the universe that can exchange

energy and matter with the system;

Thermodynamic Universe:��System + Surroundings

Open System: Exchanges both E and matter with Surroundings;

Closed System: Exchanges only E with Surroundings;

Isolated System: Exchanges NEITHER with Surroundings

Sign is from the point of view of the System

6-4

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Slide 6-5

Figure 6.1

The system in this case is the contents of the reaction flask. The surroundings comprise everything else, including the flask itself.

A chemical system and its surroundings.

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Slide 6-6

An Auto A Battery A perfect Thermos Chapter 6: First Law

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Slide 6-7

ΔE = Efinal - Einitial = Eproducts - Ereactants

Energy diagrams for the transfer of internal energy (E) between a system and its surroundings.

A. E of system decreasses B. E of system increases


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Slide 6-8

A system transferring energy as heat only.

A E lost as heat B E gained as heat


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Slide 6-9

A system losing energy as work only.


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Slide 6-10

Work (w)

Motion against an opposing force; a directed motion thru a distance; w = F•Δx

Capacity of a system to do work is called Internal Energy (U)

Most important: ΔU = Ufinal – Uinitial

If E transfer is only due to work then ΔU = w

Work is either expansion or nonexpansion (ΔV≠ 0 vs. ΔV= 0)

w = –PextΔV (since w = – when system does work)


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Slide 6-11


Work

If Pext does change in infinitesimally small amts, then process is REVERSIBLE: a process in which an infinitesimally small change in the opposite direction will cause the process

to reverse direction

w = – PextΔV (Pext must be constant) Work is in L-atm: 1 L-atm = 101.325 J If is Pext = 0, then free expansion occurs

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Slide 6-12

Figure 6.5 Pressure-volume work.


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Slide 6-13

Figure 6.7 Pressure-volume work.

An expanding gas pushing back the atmosphere does PV work (w = -PΔV).

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Slide 6-14


6-14

Heat (q)

 More familiar way to transfer E  Sys and Surr are at different T’s  If no work is done, then ΔU = q  Temperature is not heat; it is directly

proportional to it (q ∝ ΔT)!  Temperature: a measure of the relative ‘hotness’

or “coldness” of an object  q = –qsurr  Exothermic vs Endothermic

Energy transfer due to a temperature difference

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Slide 6-15

Calorimetry

q = c x m x ΔT

q = heat lost or gained c = specific heat capacity m = mass in g ΔT = Tfinal - Tinitial

The specific heat capacity (c) of a substance is the quantity of heat required to change the temperature of 1 gram of the substance by 1 K.

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Slide 6-16 Chapter 6: First Law 6-16

Measuring Heat Using a Calorimeter

Constant Pressure Constant Volume

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Slide 6-17

Figure 6.7 Coffee-cup calorimeter.


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Slide 6-18

Figure 6.8 A bomb calorimeter.


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Selected Heat Capacities (Cs and Cm)

  Amt of heat to raise 10.0 g of water 5°C: 209.2 J   Temp increase if same amount of heat is added to 10.0 g of gold (Cs = 0.106

J/g-C): 197.4 °C!

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Slide 6-22


6-22

Cm,V vs. Cm,P

Since q = nCmΔT then Cm =q

n!T=

q!T

(n = 1)

However, q = ΔU at constant volume or q = ΔH at constant pressure, So:

Cm,V =!Un!T

Cm,P =!Hn!T

OR

and

Cm,P = Cm,V + R

For Ar: Cm,V =12.8 J/mol-K and Cm,P = 21.1 J/mol-K

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Slide 6-23

Sample Problem 6.3 Finding the Quantity of Heat from Specific Heat Capacity

PROBLEM: A layer of copper welded to the bottom of a skillet weighs 125 g. How much heat is needed to raise the temperature of the copper layer from 25oC to 300.oC? The specific heat capacity (c) of Cu is 0.387 J/g•K.

SOLUTION:

PLAN: Given the mass, specific heat capacity and change in temperature, we can use q = c x mass x ΔT to find the answer. ΔT in oC is the same as for K.

q = 125 g (300oC - 25oC) x x = 1.33 x 104 J 0.387 J/g•K


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Slide 6-24

Figure 6.4 Two different paths for the energy change of a system.


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Slide 6-25


6-25

First Law of Thermodynamics

•  ΔU = q + w •  Isolated system, ΔU = 0 •  Container walls can be

either adiabatic (q = 0) or diathermic (w = 0)

•  Both q and w are path dependent

The internal energy of an isolated system is constant

Internal Energy (U) is a State Function

To determine the change in a state function, choose any convenient path (even if the real path is too complicated to calculate)

Slide 6-26

ΔEuniverse = ΔEsystem + ΔEsurroundings

Units of Energy

Joule (J)

calorie (cal)

British thermal unit (Btu)

1 cal = 4.184 J

1 J = 1 kg•m2/s2

1 Btu = 1055 J


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Slide 6-27

Sample Problem 6.1 Determining the Change in Internal Energy of a System

PROBLEM: When gasoline burns in a car engine, the heat released causes the products CO2 and H2O to expand, which pushes the pistons outward. Excess heat is removed by the car’s cooling system. If the expanding gases do 451 J of work on the pistons and the system loses 325 J to the surroundings as heat, calculate the change in energy (ΔE) in J, kJ, and kcal.

SOLUTION:

PLAN: Define system and surroundings, assign signs to q and w and calculate ΔE. The answer should be converted from J to kJ and then to kcal.

q = - 325 J w = - 451 J

ΔE = q + w = -325 J + (-451 J) = -776 J

-776 J x 103J

kJ = -0.776 kJ -0.776 kJ x

4.184 kJ

kcal = -0.185 kcal


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Slide 6-28


6-28

Enthalpy (H)

•  For Constant Volume systems ΔU = q •  Most Processes WE study are Constant Pressure systems •  Enthalpy is a State Function that allows us to keep track of

q at constant P; By definition: H = U + PV •  ΔH = ΔU + PΔV = q – PextΔV + PΔV = q – PextΔV + PextΔV •  So ΔH = q •  Implication 1: At constant volume you measure ΔU; at

constant pressure you measure ΔH •  Implication 2: The molar heat capacity Cm = q/(nΔT)

Therefore it will be different for a constant volume process than for a constant pressure process

The heat released or absorbed by a system at constant pressure

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Slide 6-29

The Meaning of Enthalpy

w = - PΔV

ΔH = ΔE + PΔV

qp = ΔE + PΔV = ΔH

ΔH ≈ ΔE in

1. Reactions that do not involve gases.

2. Reactions in which the number of moles of gas does not change.

3. Reactions in which the number of moles of gas does change but q is >>> PΔV.

H = E + PV where H is enthalpy


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Slide 6-30


6-30

Mathematical Aspects of State Functions Given X is a state function, then: •  The change in X is equal to the sum of the X values for

the products minus the reactants: ΔX = ΣniXpdts – ΣnjXrxtnts

•  The value for the reverse process is equal in magnitude BUT opposite in sign: ΔXrev = –ΔXfwd

•  Multiplication of a process by number multiplies the X value by that number: If A → 2B ΔX1 = 7 then 3A → 6B ΔX2 = 3ΔX1 = 21

•  The sum of all of the X values making up a process is equal to the X overall for that process: ΔXtotal = ΔXstep 1 + ΔXstep 2 + ΔXstep 3 + …

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Slide 6-31

Figure 6.6

Enthalpy diagrams for exothermic and endothermic processes.


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Slide 6-32


6-32

ΔH of Physical Processes

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Slide 6-33


6-33

Heating Curves

q q q

ΔHfus ΔHvap

T1 (–5 C)) T2 (+120 C)

o

o

Slide 6-34


6-34

Heating Curve Calculations To take 0.5 mole (9.0 g) of ice from –5ºC to vapor at 120ºC:

Total heat = (heat to take 0.5 mol of ice from –5ºC to 0ºC) + (heat to melt 0.5 mol of ice) + (heat to take 0.5 mol of water from 0ºC to 100ºC) + (heat to vaporize 0.5 mol of ice) + (heat to take 0.5 mol of vapor from 100ºC to 120ºC)

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Slide 6-35


6-35

ΔH of Chemical Change • Enthalpy change during chemical reactions important to

understanding chemistry. • Reaction Enthalpies: ΔH or ΔHrxn • ΔH associated with Thermochemical Equations:

2Al (s) + 6HCl (aq) → 2AlCl3 (aq) + H2 (g) ΔH = –1049 kJ • ΔHrxn associated with certain types of reactions; corrected

to a per mole basis; used in tabular collections of data: ΔHrxn = –524.5 kJ/mol for the HCl oxidation of Al ΔHcombustion = –1560 kJ/mol for ethane (C2H6)

• Relation between ΔH and ΔU: For reactions in which NO gas is generated: ΔH = ΔU For reactions in which gas is generated: ΔH = ΔU + ΔngasRT

Slide 6-36


6-36

Standard Reaction Enthalpies ΔH°

ΔH depends on conditions and phases; to allow standardization for reporting the idea of Standard State was developed.

The standard state of a property X (designated as X°) is defined as such:

  Substances and elements: most stable form of that material at the T of interest and 1 bar of pressure;

  Solutions: 1.0 M concentration; Allows the tabulation and reporting of data from

various sources to be accomplished!

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Slide 6-37


6-37

Common ΔHx°’s Enthalpy of Formation (ΔHf°)   Defined as: the standard reaction enthalpy for the formation of one mole

of a substance in its most stable form from its elements in their most stable form at the temperature of interest:

2 Na(s) + C(s,gr) + 3/2 O2(g) Na2CO3 (s)   By Definition ΔHf°≡ 0 of an element in its most stable state;   Allows ΔHrxn° to be calculated using ΔHf° of pdts - rxtnts:

aA + bB dD

ΔHf° = [dΔH°f,D] – [aΔH°f,A – bΔH°f,B]

Enthalpy of Combustion (ΔHc° or ΔH°comb)   Defined as: the standard reaction enthalpy for the combustion of one mole

of a substance in its most stable form at T of interest: C2H2(g) + 5/2 O2(g) 2 CO2(g) + H2O (l)

Slide 6-38

Sample Problem 6.2 Drawing Enthalpy Diagrams and Determining the Sign of ΔH

PROBLEM: In each of the following cases, determine the sign of ΔH, state whether the reaction is exothermic or endothermic, and draw an enthalpy diagram.

SOLUTION: PLAN: Determine whether heat is a reactant or a product. As a reactant, the products are at a higher energy and the reaction is endothermic. The opposite is true for an exothermic reaction.

(a) H2(g) + O2(g) H2O(l) + 285.8 kJ

(b) 40.7 kJ + H2O(l) H2O(g)

1 2


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Slide 6-39

Sample Problem 6.4 Determining the Heat of a Reaction

PROBLEM: You place 50.0 mL of 0.500 M NaOH in a coffee-cup calorimeter at 25.00oC and carefully add 25.0 mL of 0.500 M HCl, also at 25.00oC. After stirring, the final temperature is 27.21oC. Calculate qsoln (in J) and ΔHrxn (in kJ/mol). (Assume the total volume is the sum of the individual volumes and that the final solution has the same density and specific heat capacity as water: d = 1.00 g/mL and c = 4.184 J/g•K)

PLAN: We need to determine the limiting reactant from the net ionic equation. The moles of NaOH and HCl as well as the total volume can be calculated. From the volume we use density to find the mass of the water formed. At this point qsoln can be calculated using the mass, c, and ΔT. The heat divided by the M of water will give us the heat per mole of water formed.


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Slide 6-40

total volume after mixing = 0.0750 L

0.0750 L x 103 mL/L x 1.00 g/mL = 75.0 g of water

q = mass x specific heat x ΔT

= 75.0 g x 4.184 J/g•oC x (27.21oC - 25.00oC)

= 693 J

(693 J/0.0125 mol H2O)(kJ/103 J) = -55.4 kJ/ mol H2O formed

SOLUTION:

For NaOH 0.500 M x 0.0500 L = 0.0250 mol OH-

For HCl 0.500 M x 0.0250 L = 0.0125 mol H+

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

H+(aq) + OH-(aq) H2O(l)

HCl is the limiting reactant. 0.0125 mol of H2O will form during the reaction.

Sample Problem 6.4 Determining the Heat of a Reaction continued


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Slide 6-41

Sample Problem 6.5 Calculating the Heat of a Combustion Reaction

PROBLEM: A manufacturer claims that its new dietetic dessert has “fewer than 10 Calories per serving.” To test the claim, a chemist at the Department of Consumer Affairs places one serving in a bomb calorimeter and burns it in O2 (the heat capacity of the calorimeter = 8.151 kJ/K). The temperature increases 4.937oC. Is the manufacturer’s claim correct?

SOLUTION:

PLAN: - q sample = qcalorimeter (First Law of Thermodynamics)

qcalorimeter = heat capacity x ΔT

= 8.151 kJ/K x 4.937 K = 40.24 kJ

1 Calorie = 1 kcal = 4.184 kJ

The manufacturer’s claim is true.

10 Calorie = 41.84 kJ


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Slide 6-42

Figure 6.9 Summary of the relationship between amount (mol) of substance and the heat (kJ) transferred during a reaction.


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Slide 6-43

Sample Problem 6.6 Using the Heat of Reaction (ΔHrxn) to Find Amounts

SOLUTION: PLAN:

PROBLEM: The major source of aluminum in the world is bauxite (mostly aluminum oxide). Its thermal decomposition can be represented by

If aluminum is produced this way, how many grams of aluminum can form when 1.000 x 103 kJ of heat is transferred?

Al2O3(s) 2Al(s) + O2(g) ΔHrxn = 1676 kJ

1.000 x 103 kJ x 2 mol Al

1676 kJ

26.98 g Al

1 mol Al

= 32.20 g Al

x

3 2


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Slide 6-44

Sample Problem 6.8 Using Hess’s Law to Calculate an Unknown ΔH

PROBLEM: Two gaseous pollutants that form in auto exhausts are CO and NO. An environmental chemist is studying ways to convert them to less harmful gases through the following reaction:

CO(g) + NO(g) → CO2(g) + ½N2(g) ΔH = ?

Given the following information, calculate the unknown ΔH:

Equation A: CO(g) + ½ O2(g) → CO2(g) ΔHA = -283.0 kJ

Equation B: N2(g) + O2(g) → 2NO(g) ΔHB = 180.6 kJ

PLAN: Manipulate Equations A and/or B and their ΔH values to get to the target equation and its ΔH. All substances except those in the target equation must cancel.

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Slide 6-45

Sample Problem 6.8

SOLUTION:

Multiply Equation B by ½ and reverse it:

ΔHrxn = -373.3 kJ

CO(g) + NO(g) → CO2(g) + ½ N2(g)

Equation A: CO(g) + ½ O2(g) → CO2(g) ΔH = -283.0 kJ

NO(g) → ½ N2(g) + ½ O2(g); ΔH = - 90.3 kJ

Add the manipulated equations together:

½ Equation B: NO(g) → ½ N2(g) + ½ O2(g) ΔH = - 90.3 kJ (reversed)

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Slide 6-47

Table 6.3 Selected Standard Enthalpies of Formation at 25°C (298K)

Formula ΔH°f (kJ/mol)

Calcium Ca(s) CaO(s) CaCO3(s)

Carbon C(graphite) C(diamond) CO(g) CO2(g) CH4(g) CH3OH(l) HCN(g) CSs(l)

Chlorine Cl(g)

0-635.1

-1206.9

0 1.9

-110.5 -393.5 -74.9

-238.6 135

87.9

121.0

Hydrogen

Nitrogen

Oxygen


H(g) H2(g)

N2(g) NH3(g) NO(g)

O2(g) O3(g) H2O(g)

H2O(l)

Cl2(g)

HCl(g)

0

0

0

-92.3 0

218

-45.9 90.3

143 -241.8

-285.8

107.8


Silver Ag(s) AgCl(s)

Sodium

Na(s) Na(g) NaCl(s)

Sulfur S8(rhombic) S8(monoclinic) SO2(g)

SO3(g)

0

0

0

-127.0

-411.1

0.3 -296.8

-396.0

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Slide 6-48

Sample Problem 6.9 Writing Formation Equations

PROBLEM: Write balanced equations for the formation of 1 mol of the following compounds from their elements in their standard states and include ΔH°f.

(a) Silver chloride, AgCl, a solid at standard conditions.

(b) Calcium carbonate, CaCO3, a solid at standard conditions.

PLAN: Write the elements as reactants and 1 mol of the compound as the product formed. Make sure all substances are in their standard states. Balance the equations and find the value of ΔH°f in Table 6.3 or Appendix B.

(c) Hydrogen cyanide, HCN, a gas at standard conditions.

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Slide 6-49

Sample Problem 6.9

SOLUTION:

(a) Silver chloride, AgCl, a solid at standard conditions.

(b) Calcium carbonate, CaCO3, a solid at standard conditions.

Ag(s) + ½Cl2(g) → AgCl(s) ΔH°f = -127.0 kJ

Ca(s) + C(graphite) + O2(g) → CaCO3(s) ΔH°f = -1206.9 kJ 3 2

(c) Hydrogen cyanide, HCN, a gas at standard conditions.

½H2(g) + C(graphite) + ½N2(g) → HCN(g) ΔH°f = 135 kJ

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Slide 6-52


6-52

Hess’ Law of Heat Summation Let's say we want to determine the reaction enthalpy for the following reaction:

H2 (g) + Cl2 (g) ! 2HCl (g) H°rxn = ?? given the following individual reactions:

NH3 (g) + HCl (g) ! NH4Cl (s) H1° = -176.0 kJ N2 (g) + 3H2 (g) ! 2NH3 (g) H2° = -92.22 kJ N2 (g) + 4H2 (g) + Cl2 (g) ! 2 NH4Cl (s) H3° = -628.66 kJ

Start by adding two reactions that give us the reactants we want on the correct side: Step 1

N2 (g) + 4H2 (g) + Cl2 (g) ! 2 NH4Cl (s) H3° 2NH3 (g) ! N2 (g) + 3H2 (g) -( H2°)

2NH3 (g) + H2 (g) + Cl2 (g) ! 2 NH4Cl (s) H°(step 1) = H3° + -( H2°) Step 2 2NH3 (g) + H2 (g) + Cl2 (g) ! 2 NH4Cl (s) H°(step 1)

2[NH4Cl (s) ! NH3 (g) + HCl (g)] 2(- H1°) H2 (g) + Cl2 (g) ! 2HCl (g)

So H°rxn = H°(step 1) + 2(- H1°) = H3° + -( H2°)+ 2(- H1°) H°rxn = -184.64 kJ

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Slide 6-53

Figure 6.10

The general process for determining ΔHorxn from ΔHo

f values.


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Slide 6-54

Sample Problem 6.9 Calculating the Heat of Reaction from Heats of Formation

SOLUTION:

PLAN:

PROBLEM: Nitric acid, whose worldwide annual production is about 10 billion kilograms, is used to make many products, including fertilizer, dyes, and explosives. The first step in the industrial production process is the oxidation of ammonia:

Calculate ΔHorxn from ΔHo

f values.

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

Look up the ΔHof values and use Hess’s law to find ΔHrxn.

ΔHrxn = Σ mΔHof (products) - Σ nΔHo

f (reactants)

ΔHrxn = {4[ΔHof NO(g)] + 6[ΔHo

f H2O(g)]} - {4[ΔHof NH3(g)] + 5[ΔHo

f O2(g)]}

= (4 mol)(90.3 kJ/mol) + (6 mol)(-241.8 kJ/mol) -

[(4 mol)(-45.9 kJ/mol) + (5 mol)(0 kJ/mol)]

ΔHrxn = -906 kJ


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Slide 6-55

Figure 6.11

The trapping of heat by the atmosphere. Chapter 6: First Law

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Slide 6-56

End of Chapter 6

Silberberg HW: 10, 14, 15, 19, 35, 37, 39, 41, 48, 52, 54, 57, 59, 64, 65, 67, 71, 77, 79, 81, 82, 92,

97, 98, 100


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Chapter 6 Thermochem 110 F11 IP - Palomar College 6 Thermochem 2Up.pdf · The system in this case is the contents of the reaction flask. ... An Auto A Battery A perfect Thermos