Calculation of energy changes

If a body changes its temperature then it changes its energy contentEnergy changes can be calculatedUnits of energy = JoulesGeneral relationship: cardΔE = mass x specific heat x temp change (g) (J g-1 0C-1) (0C)

ΔE = mass x specific heat x temp change

Specific heat = amount of energy to change

temperature of 1.00g by 10C Very variable : water 4.2 j 0C-1 g-1

gold 0.13 j 0C-1 g-1

Iron 0.49 j 0C-1 g-1

ΔE = mass x specific heat x temp change

Copper pipes carrying hot water have a mass of 10 kg. Calculate the energy required to heat up the pipes from 250C to 800C (sp. ht Cu = 0.385 j 0C-1 g-1)

Some vehicles have engines which weigh 0.25 tonnes. 1 mole of petrol when burned completely produces 5512 Kj of energy. Estimate how much petrol is needed just to heat up the engine? (sp. ht steel = 0.49 j 0C-1 g-1)

Calculations involving solutions

What to use for “m” can be tricky – it may be an approximation

For solutions of solids in liquids it is the mass of liquid which is taken into account, the mass of solid tends to be ignored (unless the specific heat of the solution is quoted, in which case it is built in)

E.g. 10g of sodium chloride added to 100g of water – 100g used as “m”

But for two liquids: 25g of sulphuric acid added to 25g of water - 50g

used as m

Signs

If temp of mixture goes up sign for ΔE is

Negative Reaction is …….thermic exo Decrease in temp Positive ΔE endothermic

Examples involving solutions

3.00g of CaO was added to 50cm3 of water and the temperature rose by 150C.

Calculate the (i) energy change (ii) the energy change if 1 mole of CaO had been used

3.00g of CaO was added to 50.0g of water and the temperature rose by 150C.

ΔE = mass x specific heat x temp change

= 50.0 x 4.18 x 15 = 3135j Number of moles of CaO = 3/56 =

0.0536 Energy change per mole of CaO =3135/0.0536 = 58489 j mol-1

=58.5 kj mol-1

ΔE vs ΔH

ΔH is an energy change but measured at constant pressure

Consolidation

Sheet “energy change from temperature

change”

Hess’s Law – enthalpy change accompanying reaction is independent of route depends only on initial and final states

2 (direct) = 3+1 (indirect)1=3=

A B

C

x

yz

X= Z= Y=

A B

C

X

YZ

What can you deduce?

A B

C

E1

E2E3

A B

C

H1

H2

H3

Hess’s Law Triangle worksheet

Answers -325 -200 +230 -600 -495

Corner stone definitions -background

STANDARD CONDITIONS: CARD 100kPa (1 atmosphere)

pressure temperature : 298 K . include state symbols.

Corner stone definitions

Standard Enthalpy of Combustion

H c. (Memory card) The enthalpy change when one

mole of a substance is completely burned in oxygen at 298K and 1 atmosphere (102 kPa) pressure.

Enthalpy of combustion of methane CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

Hc = -890.5 kJmol-1

See notes page 4 Q’s 1&2

Corner stone definitions

Standard Enthalpy of Formation Hf.(Memory card)

The enthalpy change when one mole of a compound is formed from its elements in their standard states (i.e. the normal or most stable form) at 298K and 1 atmosphere (102 kPa) pressure.

Enthalpy of formation of ammonia 0.5 N2(g) + 1.5 H2(g) NH3(g) Hf = -46.0.kJmol-1 See page 5 Q1 (Q2 if completed)

Calculate the enthalpy of formation of methane ΔHθ

comb(CH4)= -882 kj mol-1 ΔHθ

f(CO2)= -394 kj mol-1 ΔHθ

f(H2O)= -286 kj mol-1 Ans = -84 kj mol-1

CH4(g) + 2O2(g) C O2(g) + 2H2O(l)ΔHθ

comb(CH4)

C(s) + 2H2 + 2O2

ΔHθf(CO2)+ ΔHθ

f(H2O)x2ΔHθ

f(CH4) ΔHΔH1

ΔH2

ΔH = ΔH1 + ΔH2

ΔH - ΔH2 = ΔH1

See Hess’s Law calculations 1

Calculate the enthalpy of combustion of methane ΔHθ

f(CH4)= -84 kj mol-1 ΔHθ

f(CO2)= -394 kj mol-1 ΔHθ

f(H2O)= -286 kj mol-1 Ans = -882 kj mol-1

CH4(g) + 2O2(g) C O2(g) + 2H2O(l)ΔHθ

comb(CH4)

C(s) + 2H2 + 2O2

ΔHθf(CO2)+ ΔHθ

f(H2O)x2ΔHθ

f(CH4) ΔHΔH1

ΔH2

ΔH = ΔH1 + ΔH2

ΔH - ΔH1 = ΔH2

See Hess’s Law calculations 2

Calculate the enthalpy of reaction

Reactants ProductsΔHθ

reaction

Elements in normal state

Add ΔHθf (all products)

Add ΔHθf(all

reactants) ΔH

ΔH1

ΔH2

ΔH = ΔH1 + ΔH2

ΔH - ΔH1 = ΔH2

Example of enthalpy of reaction calculation

For the reaction SiCl4 + 2H2O SiO2 + 4HCl Calculate the enthalpy of reaction

given: ΔHθ

f(H2O)= -286 kj mol-1

ΔHθf(SiCl4)= -640.2 kj mol-1

ΔHθf(SiO2)= -859.4 kj mol-1

ΔHθf(HCl)= -92.3 kj mol-1

Ans = -16.6 kj mol-1

See Hess’s Law calcs 3 Q’s 1&2

Ans Q1 = -227 kj mol-1

Ans Q2 = +17 kj mol-1

Further questions – answer by means of a triangle

1. Cyclohexene reacts with hydrogen to form cyclohexane as follows

C6H10 + H2 C6H12 Calculate the enthalpy change of the reaction given

that the enthalpies of formation of cyclohexene and cyclohexane are

-36kJ mol-1 and -156kJ mol-1 respectively.

2. The standard enthalpies of combustion of ethane, carbon and hydrogen are -1411, -393 and -286 kJmol-1 respectively. Calculate the standard enthalpy of formation of ethane.

Enthalpy of reaction from enthalpy of combustion (all components must burn)Calculate the enthalpy of hydrogenation (ΔHθ

reaction) of ethane

C2H4(g) + 2H2(g) C2H6

ΔHθreaction

3H2O + 2CO2

ΔHθcomb(C2H6ΔHθ

comb(C2H4) + ΔHθ

comb(H2) x2ΔH2

ΔH

ΔH1

ΔH = ΔH1 + ΔH2

ΔH - ΔH2 = ΔH1

See Hess’s Law calcs 3 Q’s 3&4