Chem Notes

1. Energetics si ygrene neht srucco egnahc yplahtne na fI dna )rH

In an exothermic change energy is transferred from the system (chemicals) to the surroundings. The products have less energy than the reactants In an endothermic change, energy is transferred from the surroundings to the system (chemicals). The products have more energy than the reactants Enthalpy changes are normally quoted at standard conditions. Standard conditions are : • 1 atm pressure • 298 K (room temperature or 25oC) • Solutions at 1mol dm-3 • all substances should have their normal state at 298K Standard enthalpy change of formation

In an exothermic reaction the ∆H is negative

In an endothermic reaction the ∆H is positive

When an enthalpy change is measured at standard conditions the symbol is used Eg ∆H

The enthalpy of formation of an element = 0 kJ mol-1

Standard enthalpy change of combustion The standard enthalpy of combustion of a substance is defined as the enthalpy change that occurs when one mole of a substance is combusted completely under standard conditions. (298K and 1 atm), all reactants and products being in their standard states Symbol ∆Hc CH4 (g) + 2O2 (g) CO2 (g) + 2 H2O (l)

Incomplete combustion will lead to soot (carbon), carbon monoxide and water. It will be less exothermic than complete combustion.

Measuring the enthalpy change for a reaction experimentally Calorimetric method tluciffid eb nac esir erutarepmet tcaxe eht neht wols si noitcaer eht fI

eht fo eno fo elom rep egnahc

ygrene eht otni detrevnoc

yplahtne ehT( .stnatcaer

For a reaction in solution we use the following equation energy change = mass of solution x heat capacity x temperature change Q N = m (g) cp (J g-1K-1) x ∆T ( K) Copyright(J) Goalby Bancroft's School x

seititnauq lautca eht rof ygrene

eht evig ylno lliw noitauqe sihT

Calculating energy change for an experimental reaction

∆

)s(

si eulav siht yllamroN .desu

)s(

3

O eF

2

2lCgM

,noitcaer fo egnahc

)g(

2

)g(

O 5.1 + )s( eF2

2lC

+ )s( gM

The standard enthalpy change of formation of a compound is the energy transferred when 1 mole of the compound is formed from its elements under standard conditions (298K and 1 atm), all reactants and products being in their standard states Symbol ∆Hf

slacimehc eht si metsys ehT

edistuo gnihtyreve si sgnidnuorrus eht dna

metsys metsys metsys metsys

neewteb derrefsnart .slacimehc eht

. sgnidnuorrus . sgnidnuorrus . sgnidnuorrus

ni nekat ygrene taeh fo tnuoma eht si

erusserp erusserp erusserp erusserp

eb ot deen snoitulos h tob fo erutarepmet eht neht snoitulos era stnatcaer

erofeb setunim wef a rof stnatcaer eht fo erutarepmet eht ekat osla eW

owt eht fI .erutarep met egareva retteb a teg ot rehtegot dedda era yeht

noitcaer eht htiw ylsuoenatlumis srucco gnilooc sa niatbo ot

dna slavretni emit raluger ta sgnidaer ekat ew siht tcaretnuoc oT

desu si erutarepmet egareva na dna noitidda erofeb derusaem

eht dedivorp metsys a ni egnahc yna gnirud tuo nevig ro

eht emit eht ot kcab enil/evruc erutarepmet eht etalopartxe

rehtegot dedda erew stnatcaer

egnahc yplahtnE :noitinifeD

na snoc s ttttnattttsnoc siiii na snoc s na snoc s

Calculating the enthalpy change of reaction, ∆Hr from experimental data General method 1. 2. 3. 4. Using q= m x cp x ∆T calculate energy change for quantities used Work out the moles of the reactants used Divide q by the number of moles of the reactant not in excess to give ∆Hr lliw 3

The heat capacity of water is 4.18 J g-1K-1. In any reaction where the reactants are dissolved in water we assume that the heat capacity is the same as pure water. snoitulos eht taht emussa oslA

Example 1. Calculate the enthalpy change of reaction for the reaction where 25cm3 of 0.2 M copper sulphate was reacted with 0.01mol (excess of zinc). The temperature increased 7oC . Step 1: Calculate the energy change for the amount of reactants in the test tube. Q = m x cp x ∆T
7 x 81.4 x 52 = Q

Note the mass is the mass of the copper sulphate solution only

Step 2 : calculate the number of moles of the reactant not in excess. ot deen uoy neht ssecxe ni si tahw dlot ton era uoy fI lov x cnoc =
4

Step 3 : calculate the energy change per mole which is often called ∆Hr (the enthalpy change of reaction)
∆Hr = Q/ no of moles = 731.5/0.005 = 146300 J mol-1 = 146 kJ mol-1 to 3 sf sesaercni pmet fi :egn ahc ygrene eht tneserper ot ngis eht ni dda yllaniF

Remember in these questions: sign, unit, 3 sig figs.
1-

Example 2. 25cm3 of 2M HCl was neutralised by 25cm3 of 2M NaOH. The Temperature increased 13.5oC What was the energy change per mole of HCl? Step 1: Calculate the energy change for the amount of reactants in the test tube. Q = m x cp x ∆T
5.31x 81.4 x 05 = Q

Note the mass is the mass of acid + the mass of alkali

Step 2 : calculate the number of moles of the HCl. lov x cnoc = lCH fo selom

Step 3 : calculate the energy change per mole which is often called ∆Hr (the enthalpy change of reaction)
∆Hr = Q/ no of moles = 2821.5/0.05 = 564300 J mol-1 = -56.4 kJ mol-1 to 3 sf

Remember in these questions: sign, unit, 3 sig figs.

Copyright N Goalby Bancroft's School

mc52 gE .

,retaw fo ytisned eht evah
3-

Add a sign and unit (divide by a thousand to convert Jmol-1 to kJmol-1

mc g1 si hcihw g 52 hgiew tuo krow dna stnatcaer htob fo selom eht tuo krow ssecxe ni si tahw noitauqe decnalab eht gnisu ngis sunim a nevig si os dna cimrehtoxE

llllom Jk 641– om Jk 641– om Jk 641– om Jk 641–

ge n gis sunim a nevig si dna cimrehtoxe si noitcaer eht

0001/52 x 2 =

0001/52 x 2.0 =

lom 50 .0 =

lom 500.0 =

J J J 5.1282 = Q J

J J J J J J 5.137 = Q J J
OSuC fo selom

Hess’s Law
Hess’s law states that total enthalpy change for a reaction is independent of the route by which the chemical change takes place
Hess’s law is a version of the first law of thermodynamics, which is that energy is always conserved.

Using Hess’s law to determine enthalpy changes of reaction from enthalpy changes of formation.

Example 3 What is the enthalpy change for this reaction ? Al2O3 + 3 Mg 3 MgO + 2 Al stnatcaer H

∆Hc= ∆Hf products - ∆Hf reactants ∆
) (

Using Hess’s law to determine enthalpy changes of reaction from enthalpy changes of combustion.

Example 5. Using the following combustion data to calculate the heat of reaction CO (g) + 2H2 (g) CH3OH (g) ∆Hc C0(g) = -283 kJ mol-1 ∆Hc H2 (g)= –286 kJ mol-1 ∆Hf CH3OH(g)= –671 kJ mol-1 ∆ ∆ ∆ ∆

∆Hr = -283+ 2x –286 - -671
1-lom
Jk 481- =

Copyright N Goalby Bancroft's School

)HO HC(

s cudorp H - s na caer H = no caer H s cudorp H - s na caer H = no caer H sttttcudorp H - sttttnattttcaer H = noiiiittttcaer H s cudorp H - s na caer H = no caer H

3

CH

- )

2H(CH

x 2 + )OC(

CH

∆

∆

∆

stcudorp

c c c

cH

- stnatcaer

c c c

∆

∆

∆

fH

- ])

02 – ])242– x 3( + )493– x 3([ =

(

fH

x 3 + )

1-lom

cH

(

Jk 8291- =

fH

= noitcaer H

x 3[ =

= H

cH

∆

∆

CO2

∆

H2O

∆

C3H6

1-lom

Example 4. Using the following data to calculate the heat of combustion of propene ∆Hf C3H6(g) = +20 kJ mol-1 ∆Hf CO2(g)= –394 kJ mol-1 ∆Hf H2O(g)= –242 C3H6 + 4.5 O2 3CO2 + 3H2O
Jk

1-

lom Jk 7.5761- = ) O lA( H

1-

lom Jk 7.106- =)OgM( H

3

2

f

f

7.5761- - )7.106– x 3( =

1-lom

Jk 4.921- =

cH

r

rH

∆

) O lA(

3

2

fH

- )OgM(

fH

x 3= H

r

∆

∆

∆

∆ ∆

0 = H

f

f

- stcudorp H

f

= = = =

rH

∆

∆

s na caer H - s cudorp H = no caer H s na caer H - s cudorp H = no caer H sttttnattttcaer H - sttttcudorp H = noiiiittttcaer H s na caer H - s cudorp H = no caer H
∆
f f f f f f

∆

∆

∆

Remember elements have

∆

Example 6. Using the following combustion data to calculate the heat of formation of propene 3C (s) + 3H2 (g) C3H6 (g) ∆Hc H2 (g)= –286 kJ mol-1 ∆Hf C3H6(g)= –-2058 kJ mol-1 ∆Hc C (s) = -393kJ mol-1 ∆ ∆ ∆ ∆

∆Hf = 3x -393+ 3x –286 - -2058
1-lom
Jk 12+ =

Mean Bond energies The Mean bond energy is the enthalpy needed to break the bond into gaseous atoms, averaged over different molecules
We use values of mean bond energies because in reality every single bond in a compound has a slightly different value. E.g. In CH4 there are 4 C-H bonds. Breaking each one will require a different amount of energy. However, we use an average value for the C-H bond for all hydrocarbons.

In general (if all substances are gases) ∆Hr = Σ bond energies broken - Σ bond energies made ∆H values calculated using this method will be less accuate than using formation or combustion data because the mean bond energies are not exact

Example 7. Using the following mean bond enthalpy data to calculate the heat of combustion of propene

∆H = Σ bond energies broken - Σ bond energies made
= [E(C=C) + E(C-C) + 6 x E(C-H) + 4.5 x E(O=O)] – [ 6 xE(C=O) + 6 E(O-H)]

= [ 612 + 348 + (6 x 412) + (4.5 x 496) ] – [ (6 x 743) + (6 X 463)] = - 1752 kJmol-1

Example 8. Using the following mean bond enthalpy data to calculate the heat of formation of NH3 ½ N2 + 1.5 H2 NH3 (note the balancing is to agree with the definition of heat of formation (i.e. one mole of product)
E(H-H) = 436 kJ mol-1 E(N-H) = 388 kJ mol-1

E(N≡N) = 944 kJ mol-1

∆H = Σ bond energies broken - Σ bond energies made
= [0.5 x E(N≡N) + 1.5 x E(H-H)] – [ 3 xE(N-H)]

= [ (0.5 x 944) + (1.5 x 436) ] – [ 3 x 388)] = - 38 kJmol-1

Copyright N Goalby Bancroft's School

) H C(

6 3

CH

- )

2H(CH

x 3 + )C(

CH

∆

∆

stcudorp

cH

- stnatcaer

cH

x 3 = H

=

f

H

∆

These values are positive because energy is required to break a bond. The definition only applies when the substances start and end in the gaseous state.

Bond

Mean enthalpy (kJ mol-1)

C=C C-C O=O O=C O-H C-H

612 348 496 743 463 412

More complicated examples that may occur at AS Working out ∆Hf of a compound using bond energies and other data
This is a more complicated example of the type in example 8

elements

Compound in standard state ∆H to turn to compound into gaseous atoms

∆H to turn to elements into gaseous atoms

The ∆H’s can be combinations of different data

Gaseous atoms

Can be bond energies E(Cl-Cl) Cl2 2Cl Σ bond energies of compound + (∆H to turn to gas if compound is not gaseous)

Or atomisation energies (if the substance is not diatomic C(s) C(g)

∆Hf =

∆H to turn to elements into gaseous atoms

∆ Hf = (3x∆Hat [C] + 4 x E[H-H] ) – (2 x E[C-C]+ 8 x E[C-H] ) ∆ = (3x715 + 4 x 436 ) – (2 x 348+ 8 x 412 ) =-103 kJ mol-1

Copyright N Goalby Bancroft's School

.atad gniwollof eht nevig ,)g( H C ,enaporp rof ,)g( H C

8 3

)g(

517 = H

∆

1-lom Jk 8 3

2H4

+ )s( C3

f

HΔ etaluclaC
)g(C )s(C

9 elpmaxE

Bond kJ mol-1

C–C 348

C–H 412

H–H 436

- ∆H to turn to compound into gaseous atoms