Author: J R Reid

Thermochemical Principles

Data book Enthalpies

We can’t measure enthalpies of specific reactants or products directly, but we can measure enthalpy changes.Rather than listing the enthalpy change for every possible reaction, data books list the standard enthalpy of combustion and the standard enthalpy of formation of compounds, and we can use these figures to work out the enthalpy change for other reactions.

Enthalpy change of combustion Hc

Enthalpy change of combustion Hc

the enthalpy change for the complete combustion of 1 mole of a compound in excess pure oxygen under standard conditions of 298k and 1 atmosphere pressure.

C2H5OH(l) + 3O2(g) 2CO2(g) +3H2O(l) Hc (ethanol)

= - 1367 kJ mol-1

Hc are always negative - an exothermic reaction. A fuel contains lots

of energy which is released to the environment when more stable products are formed.

C2H5OH(l) + 3O2(g) 2CO2(g) +3H2O(l) Hc (ethanol)

= - 1367 kJ mol-1

Energy

C2H5OH(l) + 3O2(g)

2CO2(g) +3H2O(l)

Hc (ethanol)

= - 1367 kJ mol-1

A fuel has high chemical potential Energy

stable products

Enthalpy of Formation of a Compound Hf

Enthalpy change of formation Hf :

the enthalpy change for the formation of 1 mole of a compound from its elements under standard conditions of 298k and 100 kPa atmosphere pressure.

2C(s) +3H2(g) + 0.5O2 C2H5OH(l) Hf (ethanol)

= -1367 kJ mol-1

Mg(s) + Cl2(g) MgCl2(s) Hf (MgCl2(s) )

= - 641 kJ mol-1

Hf are almost always negative, since stable compounds have less

energy, are more stable, than their elements

Hf of elements are all of course 0, since to form themselves from

themselves takes no energy !!!

Mg(s) + Cl2(g) MgCl2(s) Hf (MgCl2(s) )

= - 641 kJ mol-1

Energy

Mg(s) + Cl2(g)

MgCl2(s)

Hf (MgCl2(s) )

= - 641 kJ mol-1

Elements have high chemical potential Energy

stable products

If Hf of the reactants and products of a reaction are

known, then Hr of the reaction can be calculated

For the reaction below a = moles of reactant A etc

aA + bB cC + dD Hr

H = c.Hf (C) + d.H

f (D) - a.Hf (A) - b.H

f (B)

Then

Hr = sum H

f of products - sum Hf reactants

(Use correct number of moles as in the equation, because

all Hf are per MOLE )

Hr = sum H

f of products - sum Hf reactants

Hc (ethanol)

C2H5OH(l) + 3O2(g) 2CO2(g) +3H2O(l)

Hf for reactants and products in the equation below are known.

Calculate Hcombustion ethanol. Show that:

We use Hess’s Law, and construct a cycle. If we know the energy by 1 route we can can calculate the energy by another route that ends at the same chemical.

Step 1

Write down the equation for the reaction which we are asked to calculate the unknown H

An example of the equation

Hr = sum H

f of products - sum Hf reactants

Hc (ethanol)

C2H5OH(l) + 3O2(g) 2CO2(g) +3H2O(l)

2C(s) +3H2(g) + 0.5O2

Hf(ethanol)

STEP 2

We are given Hf for the reactants. We can connect

the combustion equation to the elements contained in the reactants.

Make sure that the equation for Hf is

balanced for moles of each element

Hc (ethanol)

C2H5OH(l) + 3O2(g) 2CO2(g) +3H2O(l)

2C(s) +3H2(g) + 0.5O2

Hf(ethanol)

2. Hf (CO2)

3.Hf (H2O)

STEP 3

We are given Hf for the products too; we can

connect the products to the elements contained in the compounds

Hc (ethanol)

C2H5OH(l) + 3O2(g) 2CO2(g) +3H2O(l)

2C(s) +3H2(g) + 0.5O2

Hf(ethanol)

2. Hf (CO2)

3.Hf (H2O)

STEP 4

We can now calculate Hc (ethanol); by applying Hess’s

law.Route 1

Route 2

Enthalpy Change by Route 1 = Enthalpy Change by Route 2

+ 3.Hf (H2O)H

c (ethanol) = + 2. Hf (CO2)- H

f(Ethanol)

Enthalpy change of neutralisation Ho(neutralisation)

Enthalpy change of neutralisation: the enthalpy change when 1 mole of water is produced by the neutralisation of an acid by an alkali under standard conditions of 100 kPa pressure and at 298 k

NaOH(aq) +HCl(aq) = NaCl(aq) + H2O(l) Ho(neut.)= - 57.6 kJmol-1

KOH(aq)+ 0.5H2SO4(aq) =KHSO4(aq)+H2O(l) Ho(n)= -57.6 kJ mol-1These enthalpies of neutralization are the same because the reaction is the same. The acids and alkalis are strong, so they are totally ionised; Na+, K+, Cl-, SO4

2- and HSO4- are spectator ions.

H+(aq) + OH-(aq) H2O(l) Ho(n) = - 57.6 kJ mol-1

Calculate Hro for the reaction below, given the

enthalpies of combustion / kJ mol-1; C2H4 -1411, H2 - 286, C2H6 - 1560

C2H4 + H2 C2H6 Hr

o

2 CO2 + 2 H2O

- 2 x 1411

H2O

- 286

Route 1

Route 2

Enthalpy Change by Route 1 = Enthalpy Change by Route 2

Hro = - 2 x 1411 -

286- ( - 1560)

-1560

= -1548 kJ

Enthalpy change of solution Hsolution

Enthalpy change of solution Hsolution:

the enthalpy change when 1 mole of ionic solid dissolves to form a dilute 1 molar solution in water at 298 k and 100 kPa pressure.

Dissolving an ionic solid is a chemical process.

Ionic bonds between the positive and negative ions must be broken, to free the ions from the giant lattice structure; this is a highly endothermic process.

This breaking of the ionic lattice is only possible in water as strong ion - dipole bonds are formed between water and the ions. This is a highly exothermic process, and is called hydration.

Hydration of ionsThe breaking of the lattice as an ionic solid dissolves in water is possible because strong ion - dipole bonds are formed between water molecules and free ions, releasing energy. This is called hydration, and is highly exothermic. Hydration enthalpy offsets the endothermic breaking of the ionic bonds in the lattice.

Water has a dipole, the O atom carries a slight negative charge and the H atom a slight positive charge

+

- H one ion -dipole chemical bond Na+ O in Na+(H2O)6 i.e. Na+(aq)

H ++

- H Cl- one ion -dipole chemical bond O in Cl-(H2O)6 i.e. Cl-(aq)

H +

Enthalpy of solution of sodium chloride is slightly endothermic, because the lattice enthalpy is slightly

bigger than the hydration enthalpy

Energy

NaCl(s) + aq

Na+(aq) + Cl-(aq)

Na+(g) + Cl-(g) + aq

Hsolution(NaCl)

Lattice enthalpy (breaks up ions)

Hydration of ions, forms ion-dipole strong bonds

Na+ + 6H2O Na+(H2O)6 = Na+(aq)

Cl- + 6H2O Cl-(H2O)6 = Cl-(aq)

hydration

hydration

Solid Sodium Chloride in water

Na+ ion

Water molecule

Cl- ion

One Cl- is being removed by water molecules. Water molecules do not yet completely surround the Cl- ion

Sodium chloride crystal dissolving in water

Chloride ion hydrated by water. Note that the positive H atom is orientated to the Cl- ion - it is now a Cl-(aq) ion

A water molecule

If in a reaction in solution, m g of water is warmed through T oC and c is the specific heat capacity of water

Then the heat absorbed by the water is q, and q = m. c. T

If n moles of product are produced then the standard molar enthalpy change, i.e. the heat change per mole is Hr

,

and Hr = - m.c. T kJ per mole

n .1000

Experimental Determination of enthalpy changes

Hr = - q

n= - m.c. T J per mole n

If the temperature falls, the reaction is endothermic and H has a positive sign.

Initial Temperature = 22.35 oC

Enthalpy of Neutralisation of NaOH and HCl

Final temperature = 29.85 oC

Enthalpy of Neutralisation of NaOH and HCl

Calculation of the Enthalpy of Neutralisation

Mass of liquid (water) being warmed = 50 + 20 = 70 g

Temperature rise of liquid & calorimeter = 29.85 - 22.35 = 7.5 oC

Heat capacity of calorimeter = C(cal) = 78.2 J oC-1

Heat to warm calorimeter = C(cal) .T = 78.2 x 7.5 = 586.5 J

Specific Heat capacity of water = 4.184 J g-1oC-1

Heat to warm up water = q = m.c. T

q = 70 x 4.184 x 7.5 = 2196.6 J

Total heat out in reaction = qtot = 586.5 + 2196.6 = 2783.1 J

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

Calculation of the Enthalpy of Neutralisation

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

20 mL of 3M HCl and 50 mL of 1 M NaOH were mixed

Moles of H+ = 0.02 x 3 = 0.06 ; Moles of OH- = 0.05 x 1 = 0.05

Thus H+ is in excess, and 0.05 moles of water will be produced

Hneutralisation = total heat out per mole of water produced

Hneutralisation = - qtot / n = - 2783.1 / 0.05 J mol-1

H is negative, since the temperature has risen

Hneutralisation = - 55662 J mol-1

Hneutralisation = - 55.6 kJ mol-1

An Energy level diagram for the Enthalpy of Neutralisation of hydrochloric acid by sodium hydroxide

NaOH(aq) +HCl(aq)

NaCl(aq) +H2O(l)

Hneutralisation = - 55.6 kJ mol-1

H+(aq) + OH-(aq)

H2O(l)

Energy

The neutralisation equation can be written:

Na+ (aq) + OH- (aq) + H+ (aq) + Cl- (aq) = Na+ (aq) + Cl- (aq) +H2O(l)

Ionic Equation

But Na+ (aq) and Cl- (aq) are spectator ions and contribute nothing to the enthalpy change; they are ignored in the ionic equation.

Enthalpy of Solution of Ammonium Nitrate

The initial temperature =23.15 oC

Crushing device

ammonium nitrate in glass vial

The final temperature is 19.05 oC

- the reaction is endothermic

Enthalpy of Solution of Ammonium Nitrate

Calculation of the Enthalpy of Solution

Mass of liquid (water) being cooled = 60 g

Temperature fall of liquid & calorimeter = 23.15 - 19.05 = 4.1 oC

Heat capacity of calorimeter = C(cal) = 153 J oC-1

Heat to cool calorimeter = C(cal) x T = 153 x 4.1 = 627.3 J

Specific Heat capacity of water = 4.184 J g-1oC-1

Heat to cool water = q = m.c. T

q = 60 x 4.184 x 4.1 = 1029.3 J

Total heat removed in reaction = qtot = 627.3 + 1029.3= 1656.6 J

NH4NO3 (s) +aq = NH4+(aq) + NO3-(aq)

60 g of water was mixed with 5.00 g of pure ammonium nitrate - Mr= 80.04

Moles of NH4NO3 = 5 / 80.04 = 0.0625

Hsolution = total heat absorbed per mole of NH4NO3

Hsolution = + qtot / n = + 1656.6 / 0.0625 J mol-1

H is positive, since the temperature has fallen. Heat has been taken from the water and calorimeter by the chemicals.

Hsolution = + 26505.6 J mol-1

Hsolution = + 26.5 kJ mol-1

Calculation of the Enthalpy of SolutionNH4NO3 (s) +aq = NH4

+(aq) + NO3-(aq)

An Energy level diagram for the Enthalpy of Solution of ammonium nitrate

NH4+(aq) + NO3

-(aq)

NH4NO3 (s) + aq

Hsolution = + 26.5 kJ mol-1Energy

The ionic lattice in the giant structure has to be broken. With ammonium nitrate this endothermic process is larger than the exothermic hydration enthalpies of the ammonium and nitrate ions. So dissolving the solid is overall endothermic.

The Bomb CalorimeterHow accurate enthalpy changes are measured

High pressure oxygen

Steel Bomb

Reactants in sample cup

ThermometerIgnition wires

Stirrer

Insulated container

Water

Ignition heater

The operation of a Bomb Calorimeter• Benzoic acid is used to determine the heat capacity of the

apparatus. It has an accepted value for its enthalpy of combustion.

• A known mass of benzoic acid is burnt and the temperature rise measured. The heat capacity of the ‘Bomb’ can be calculated.

• To determine the enthalpy of combustion of a new substance a known mass is burnt. The conditions are kept as similar as possible to the standardization experiment; i.e. same temperature rise.

• The Enthalpy of Combustion is then calculated

Bond Energies (enthalpies)

H2(g) 2H(g) H = + 435.9 kJ mole-1

The bond energy is the enthalpy change to break 1 mole of specified covalent bonds in a gaseous molecule under standard conditions of 298k and 100 kPa pressure.

H-H 2H. H= + 435.9 kJ mole-1

Bond energies are always positive, since they are the energies to break covalent bonds. I.e to break the electrostatic attraction between a shared pair of electrons and 2 adjacent positive nuclei.

The H-H Bond Energy is + 435.9 kJ mole-1

Average bond energiesThese are averages of bond energies for a particular bond, taken from a wide range of compounds containing that bond.The Average Bond Energy is the average energy to break 1 mole of specified bonds from a wide number of compounds at 298 k and 1 atmosphere pressure. • They are averages, and the actual bond energy in a molecule

may not equal the average bond energy.

• Factors that change actual bond energies from the average are the electro-negativities of adjacent atoms.

• When one breaks up methane into C(g) + 4 H(g) atoms 4 C-H bonds are broken; the average bond energy is the total enthalpy for breaking up methane /4

• The average bond energy for a C-H bond is quoted as 413 kJ mole-1, which is the average for methane, ethane and a host of other hydrocarbons etc.

If all the Bond energies (B.E.) are known for the reactants and products of a reaction then Hr

can be calculated. If B.E.(A-B) is the bond energy in the molecule A-B etc.

Hr

A-B + C-D A-C + B-D

A B C DB.E.(A-B)

B.E.(C-D)B.E.(A-C)

B.E.(B-D)

Gaseous elements

By Hess’s law; energy change route 1 = energy change route 2

Hr = B.E.(A-B) + B.E.(C-D) - B.E.(A-C) - B.E.(B-D)

Calculation of Hr from the bond energies

of reactants and productsEstimate the H for the reaction, given the Average Bond Enthalpies / kJ mol-1 C=C 699, C-C 368, C-H 435, C-F 484, F-F 155

H H

C=C

H HF F

H H

F C C F

H H

Hr

Break ALL bonds

4 C-H = + 4 x 435

1 C=C = + 699

2 C + 4 H + 2F

1 F-F = +155

4 C-H = + 4 x 435

1 C-C = + 369

2 C-F = + 2 x 484

H H

C=C

H HF F

H H

F C C F

H H

Hr

Break ALL bonds

4 C-H = + 4 x 435

1 C=C = + 699

2 C + 4 H + 2F

1 F-F = +1554 C-H = + 4 x 435

1 C-C = + 369

2 C-F = + 2 x 484

Hr = + 4 x 435 + 699 + 155

Enthalpy change by route 1 = change by route 2

The - sign arise as bonds are formed in 1,2-difluoroethane

- (4 x 435) - 369 - (2 x 484)

= - 478 kJ

Route 1

Route 2

An estimate, as average bond energies used

Using Bond Enthalpies (B.E.) to calculate the

enthalpy of combustion of ethanol

H H

H-C-C-O-H

H H

3 O=O

2 C + 6H + 7O

2 O=C=O + 3 H-O-H

BREAK 1 C-C, 5 C-H, 1 C-O and 1 O-H BONDS

Endothermic(+)

MAKE 4 C=O, and 6O-H BONDS

Exothermic (-)H

c(ethanol)

B.E./ kJ: C-H +413, C-C 348, O=O 498, C-O 360, C=O 743,O-H 463H

c(ethanol) = + 348 + 5 x 413 +360 + 463

- 4 x 743 - 6 x 463

= -1368 kJ mol-1

Enthalpy and Changes of State

SOLID LIQUID

GAS

H FUSION (+ve)

- H FUSION (+ve)

H VAPORISATION (+ve)

- H VAP

H SUBLIMATION

Liquids have higher kinetic energies than solids. Heat energy must be supplied to change a solid into a liquid, H

FUSION is thus +ve. Similarly gases have more KE than

liquids, and so H VAPORISATION is also +ve.

Enthalpy and Changes of State

H2(g) + 1/2 O2(g) H2O(g)

H = -- 286 kJ mol-1H

Vaporisation (Water)

H2O(l)

H = -- 242 kJ mol-1

By Hess’s law

- 242 = - 286 + H Vaporisation (Water)

H Vaporisation (Water) = - 242 +286

= + 44 kJ mol-1

4th April 2012Enthalpy of combustion

• AIM – to calculate enthalpy of combustion from experimental data

Complete investigation 4.2 from p.168