Chemical Reactions: Thermochemistry

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Thermochemistry

University of Lincoln presentation

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Thermochemistry

• Enthalpy changes in chemical reactions (+ video)

• Enthalpy Diagrams• Thermochemical Equations• Calorimetry and measuring enthalpy

changes



Energy and Chemistry

• Petrol bombs

• What does this show?

• How to ensure your bonfire burns!

• Why does this happen?



Energy and Chemical Reactions

• Formation of new substances – Redox reactions– Acid-base reactions– Precipitation reactions

• Energy released/absorbed– Light (chemiluminescence)– Electrical energy (electrochemistry)– Heat (thermochemistry)



Reaction 2Energy products<reactants so energy flows as heat from the system to the surroundings Temperature (surroundings) increases - Exothermice.g. NaOH(aq) and HCl(aq).

Reaction 1 Energy products>reactants so energy is absorbed from the surroundingsHeat is lost from the surroundings soTemperature (surroundings) decreases - Endothermice.g. Ba(OH)2.8H2O(s) and NH4Cl(s) (video)

Endothermic

ExothermicS u r r o u n d I n g s

S u

r r o

u n

d I

n g

s

System

System

Heat Heat

Heat Heat



Enthalpy level diagrams

For You To DoDraw diagrams for the reactions on the previous slideYou will need to write balanced chemical equations first. For reaction1 assume that the products are NH3(g), H2O(l) and BaCl2(s) and that rH = +135 kJ mol-1

Reaction 2 is a straightforward neutralisation with a rH = -55 kJ mol-1

(g)H2NaOH(aq)O(l)2H2Na(s) 22

The reaction between sodium metal and water – metal floats on water – effervescent reaction moves metal around- yellow flame above the metal- no solid residue

2 mol Na(s) + 2 mol H2O(l)

Enth

alpy

, H

(kJ)

2 mol NaOH(aq) + 1 mol H2(g)

ΔH = -367.5 kJ (367.5 kJ of heat is released



Thermochemical Equations

A thermochemical equation is the chemical equation for a reaction (including state symbols) and the enthalpy of reaction for the molar amounts as given by the equation written directly after the equation.

1r22 molkJ65.1HΔ(s)Ca(OH)O(l)HCaO(s)

122 molkJ367.5ΔH(g)H2NaOH(aq)O(l)2H2Na(s)




Why do we need state symbols?

In a thermochemical equation it is important to note state symbols because the enthalpy change depends on the physical state of the substances.

-1r222 mol kJ 483.7- HΔ O(g)2H(g)O(g)2H

-1r222 mol kJ 571.7- HΔ O(l)2H(g)O(g)2H



Two important rules


1. When a thermochemical equation is multiplied by any factor, the value of H for the new equation is obtained by multiplying the DH in the original equation by that same factor.

2. When a chemical equation is reversed, the value of DH is reversed in sign.



• What is the enthalpy change of reaction for the formation of 1 mole and 6 moles of water?

• -285.9 kJ mol-1; -1715.1 kJ mol-1• What is the enthalpy change for the splitting

of 1 mole of water into hydrogen and oxygen gas?

• +285.9 kJ mol-1

-1r222 mol kJ 571.7- HΔ O(l)2H(g)O(g)2H

Using Thermochemical Equations



Using Thermochemical EquationsConsider the reaction of methane, CH4, burning in the presence of oxygen at constant pressure. Given the following equation, how much heat could be obtained by the combustion of 10.0 grams CH4?

(l)22(g)2(g)4(g) O2HCO2OCH 1oc kJmol890.3HΔ

kJ556gkJ55.6g10.0 1

Combustion of methane gives 55.6 kJ g-1

1 g of methane would give

11

1gkJ55.6molg16.0

molkJ890.3

10 g of methane would give



•Measuring enthalpy changes is called calorimetry•Carry out the reaction in a calorimeter and measure the temperature change.•Calculate the energy transferred during the reaction from the temperature change.•Also require the mass of the substance and the specific heat capacity

AssumptionsAll the energy change is transferred to the solution (water) No losses of heat to the other surroundings

Measuring enthalpy changes

HCl(aq)

NaOH (aq)

Thermometer

2 polystyrene coffee cups

Coffee-cup calorimeter



Need to knowSpecific Heat Capacity

DefinitionThe amount of energy required to raise the temperature of a specified mass of an object (substance) by 1 degree kelvin (K) units J g -1K-1 or J kg-1 K-1

Important exampleWater 4.184 J g-1 K-1



An Example:25 cm3 of 2.00 mol dm-3 HCl(aq) is mixed with 25 cm3 of 2.00 mol dm-3 NaOH(aq). The temperature rises from 22.5 oC to 34.5 oC. Find the enthalpy change for the reaction

O(l)HNaCl(aq)NaOH(aq)HCl(aq) 2

mcΔcQ

K12.0KgJ4.1850gQ 11 J2508Q

2508 J of heat is transferred from the reaction of 0.05 mol HCl with 0.05 mol NaOH

mol0.05mol2.0100025n



An Example continued

For 1 mol of HCl and NaOH

kJ50.2J501600.052508ΔH

12 molkJ50.2ΔHO(l)HNaCl(aq)NaOH(aq)HCl(aq)



Now Try This One0.327 g of Zinc powder is added to 55 cm3 of aqueous copper sulfate solution at 22.8 oC. The copper sulfate is in excess of that needed to react all the zinc. The temperature rises to 32.3 oC. Calculate H for the following reaction: Cu(s)(aq)ZnSO(aq)CuSOZn(s) 44



Limitations of this method ??

HCl(aq)

NaOH (aq)

Thermometer

2 polystyrene coffee cups

Coffee-cup calorimeter



A bomb calorimeter

How can we accurately measure enthalpy changes of combustion reactions?

NeedleGas inlet

Insulated jacketSteel bomb

+-

O2

ThermometerCurrent for ignition coil

Stirrer

Ignition coil

Graphite sample



Bomb Calorimetry- Bomb Calorimetry- measurementsmeasurements

Some heat from reaction warms “bomb” qbomb = heat capacity x ∆T

Total heat evolved, qtotal = qwater + qbombTotal heat from the reaction =qtotal

NeedleGas inlet

Steel bomb

+-

O2

ThermometerCurrent for ignition coil

Stirrer

Ignition coil

Graphite sample

Some heat from reaction warms waterqwater = mc∆T

Insulated jacket



Calculate enthalpy of combustion of Calculate enthalpy of combustion of octane. octane.

CC88HH18(l)18(l) + 25/2 O + 25/2 O2(g)2(g) 8CO8CO2(g)2(g) + + 9H9H22OO(l)(l)

• Burn 1.00 g of octaneBurn 1.00 g of octane• Temp rises from 25.00 to 33.20 Temp rises from 25.00 to 33.20 ooCC• Calorimeter contains 1200 g waterCalorimeter contains 1200 g water• Heat capacity of bomb = 837 J KHeat capacity of bomb = 837 J K-1-1

Calculating enthalpy changes from calorimetry data



Step 1Step 1: energy transferred from reaction to : energy transferred from reaction to water.water.q = (4.184 J gq = (4.184 J g-1-1KK-1-1)(1200 g)(8.20 K) = 41170 J)(1200 g)(8.20 K) = 41170 J

Step 2Step 2: energy transferred from reaction to : energy transferred from reaction to bomb.bomb.

q = (bomb heat capacity)(∆T)q = (bomb heat capacity)(∆T) = (837 J K= (837 J K-1-1)(8.20 K) = 6860 J)(8.20 K) = 6860 J

Step 3Step 3:Total energy transferred:Total energy transferred 41170 J + 6860 J = 48030 J41170 J + 6860 J = 48030 J

Heat of combustion of 1.00 g of octane = - 48.0 Heat of combustion of 1.00 g of octane = - 48.0 kJkJFor 1 kg = -48 MJ kgFor 1 kg = -48 MJ kg-1-1

H=-48 kJ x 114 g mol-1=-5472 kJ mol-1

Calculating enthalpy changes from calorimetry data



Video

Click to link to “Thermochemistry

” video



A case study- Self-heating cans

Water

Can

InsertQuicklimeFoil separator

Button



The ChemistryCaO(s) + H2O(l) Ca(OH)2(s)

quicklime slaked lime

• Water and quicklime packaged separately• When mixed, exothermic reaction takes

place and the temperature of the water increases

• Heat transferred to the drink rH = -65.1 kJ mol-1



How much quicklime is needed to heat up a coffee can?

• Think about what information you need to know for the calculation before doing the calculation – do some research and find approximate values

Homework• Draw an enthalpy level diagram for the

reaction• Do the calculation



FRS1027 Introductory Chemistry

• Hess’s law• Standard enthalpy of formation• Calculating enthalpy changes



Hess’s Law• The enthalpy change on going from

reactants to products is independent of the reaction path taken

• Can be used to calculate enthalpy changes



Hess’s Law & Energy Level DiagramsHess’s Law & Energy Level DiagramsReaction can be shown as a single step or in a two steps. ∆Htotal is the same no matter which path is followed.

∆∆H reaction path H reaction path 1=1=∆H reaction ∆H reaction path 2path 2

C(s) + O2(g)En

ergy

CO2(g)

CO(g) + ½ O2(g)

ΔH3 =

ΔH

1 + Δ

H2 =

-393

.5 k

J

ΔH2 =

-283

.0 k

JΔH

1 = -1

10.5

kJ



Standard enthalpy values (Ho)

1.Initial and and finalfinal species are in their species are in their standard statesstandard states

2.2.The The standard state of a substancestandard state of a substance at at a specified temperature is its pure form a specified temperature is its pure form at 1 bar (100 kPa)at 1 bar (100 kPa). . T T isis usually 298 K (25 usually 298 K (25 ooC) but not alwaysC) but not always

3.3. rrHHoo(298 K)(298 K) is the is the standard enthalpy standard enthalpy of reactionof reaction at 298 K at 298 K

Some Physical States at 298 KSome Physical States at 298 KC = graphite; OC = graphite; O22 = gas; CH = gas; CH4 4 = gas; H= gas; H22O = O =

liquidliquid



Standard Enthalpy of FormationStandard Enthalpy of Formation∆∆ffHHoo (298 K) = standard (298 K) = standard molar enthalpy molar enthalpy of formationof formation at 298 K at 298 Kenthalpy change when 1 mole of enthalpy change when 1 mole of compound is formed from elements in compound is formed from elements in their standard states at 298 K their standard states at 298 K (These are available from data books)(These are available from data books)

HH22(g) + 1/2O(g) + 1/2O22(g)(g) HH22O(g)O(g)∆∆ffHHoo (H (H22O, g) = -241.8 kJ molO, g) = -241.8 kJ mol-1-1

∆∆ffHHoo is zero for elements in their is zero for elements in their standard states.standard states.



Standard enthalpies of formation Standard enthalpies of formation and Hess’s Law can be used to and Hess’s Law can be used to calculate unknown ∆calculate unknown ∆rrHHoo ∆∆rrHHoo = ∆H = ∆Hff

oo (products) - ∆H(products) - ∆Hff

oo

(reactants)(reactants)Enthalpy of reaction = sum of the enthalpies of formation of the products (correct molar amounts) – sum of the enthalpies of formation of the reactants (correct molar amounts)

Why is this an application of Hess’s law?

Calculating Enthalpy Changes using standard enthalpies of formation



Calculate ∆Calculate ∆ccHHoo for methanol for methanolStandard state of methanol at 298 K is liquidCHCH33OH(l) + 3/2OOH(l) + 3/2O22(g)(g) COCO22(g) + 2H(g) + 2H22O(l)O(l) ∆ ∆ccHHoo = ∆H = ∆Hff

oo (products) - ∆H(products) - ∆Hff

oo (reactants)(reactants)

= {∆H= {∆Hffoo (CO(CO22) + 2 ∆H) + 2 ∆Hff

oo (H(H22O)} - {3/2 ∆HO)} - {3/2 ∆Hff

oo (O(O22) + ) +

∆H∆Hffoo(CH(CH33OH)} OH)}

= {(-393.5 kJ) + 2 (-285.8 kJ)} - {0 + (-238.9 kJ)}= {(-393.5 kJ) + 2 (-285.8 kJ)} - {0 + (-238.9 kJ)}

∆∆ccHHoo(298K)= -726.2 kJ mol(298K)= -726.2 kJ mol-1-1

Now try the problems on the separate sheet

Calculating Enthalpy Changes



Video

Link to “Hess’s Law” video



Some Other Problems to do



Describe the reaction



The enthalpy of reaction for black powder

• Black powder is a mixture of potassium nitrate (75%), charcoal (13%) and sulfur (12%).

• A simplified equation is:

)(2)(2)(2)()()(3 332 ggssss CONSKCSKNO

fHo values/kJ mol-1

KNO3(s) -494.6 K2S(s) -380.70CO2(g) -393.51Calculate the enthalpy of reaction in kJ mol-1 and kJ kg-1 of black powder.



The reaction of barium hydroxide with ammonium chloride

Equation

fHo values/ kJ mol-1Ba(OH)2.8H2O(s) -3345NH4Cl(s) -314NH3(g) -46H2O(l) -286BaCl2(s) -859Calculate rHo

)(2)(2)(3)(4)(22 10228.)( slgss BaClOHNHClNHOHOHBa



Calculate the standard enthalpy of combustion of octane at 298 K

fHo (octane)= -249.9 kJ mol-1



FRS1027 Introductory Chemistry

Bond Dissociation Enthalpies



DefinitionThe bond dissociation enthalpy (dissH) for an X-X diatomic molecule refers to the process:X2(g)2X(g)

at a given temperature (often 298 K)

dissH = D(X-X) is the bond enthalpy for a specific process (a particular bond in a molecule)

Breaking bonds is an endothermic process



D and DCH4(g) CH3(g) + H(g) H=436 kJ mol-1

CH3(g) CH2(g) + H(g) H=461 kJ mol-1

CH2(g) CH(g) + H(g) H=428 kJ mol-1

CH(g) C(g) + H(g) H=339 kJ mol-1

• D depends on the bond• D is an average value and is obtained fromCH4(g) C(g) + H(g) H=1664 kJ mol-1D (C-H) = 416 kJ mol-1



Enthalpy Changes in Chemical ReactionsEnthalpy difference between products and

reactants because different chemical bonds are formed.Enthalpy change can be estimated from the chemical bonds that are broken and made.Breaking bonds is an endothermic process and making bonds is an exothermic processOnly an estimate because

•Bond enthalpies are mean values•All species are in the gaseous state

A Hess’s law problem



Now have a go at the bond enthalpy problems and set up as an Excel spreadsheet. Can you set it up so that you only need to enter the number of C and H atoms to calculate the enthalpy change ???

Enthalpy changes of combustion for hydrocarbons Compound CH4

Bond typeBond energy

Number broken

Energy in KJ mol-1

Number made

Energy out KJ mol-1

C-C 347 0 0 0 0C-H 413 4 1652 0 0C=O 805 0 0 2 -1610O=O 498 2 996 0 0O-H 464 0 0 4 -1856

Total energy in/kJ mol-1 2648

Total energy out/kJ mol-1 -3466

Tota enthalpy change of combustion kJ mol-1 -818


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http://www.lincoln.ac.uk/cerd/

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