Calculation of energy changes  If a body changes its temperature then it changes its energy content  Energy changes can be calculated  Units of energy =  Joules  General relationship: card ΔE = mass x specific heat x temp change (g) (J g -1 0 C -1 ) ( 0 C)

ΔE = mass x specific heat x temp change  Specific heat =  amount of energy to change temperature of 1.00g by 1 0 C  Very variable :  water 4.2 j 0 C -1 g -1  gold 0.13 j 0 C -1 g -1  Iron 0.49 j 0 C -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 25 0 C to 80 0 C (sp. ht Cu = j 0 C -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 0 C -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 50cm 3 of water and the temperature rose by 15 0 C.  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 15 0 C.  ΔE = mass x specific heat x temp change  = 50.0 x 4.18 x 15 = 3135j  Number of moles of CaO = 3/56 =  Energy change per mole of CaO  =3135/ = 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= AB C x y z

 X=  Z=  Y= AB C X YZ

What can you deduce? AB C E1 E2 E3

AB 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  CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) H  c = kJmol-1  See notes page 4 Q’s 1&2

Corner stone definitions  Standard Enthalpy of Formation H  f.(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 N 2 (g) H 2 (g)  NH 3 (g)  H  f = kJmol-1  See page 5 Q1 (Q2 if completed)

Calculate the enthalpy of formation of methane ΔH θ comb (CH 4 )= -882 kj mol -1 ΔH θ f (CO 2 )= -394 kj mol -1 ΔH θ f (H 2 O)= -286 kj mol -1 Ans = -84 kj mol -1 CH 4 (g) + 2O 2 (g)C O 2 (g) + 2H 2 O(l) ΔH θ comb (CH 4 ) C(s) + 2H 2 + 2O 2 ΔH θ f (CO 2 )+ ΔH θ f (H 2 O)x2 ΔH θ f (CH 4 )ΔHΔH ΔH1ΔH1 ΔH2ΔH2 ΔH = ΔH 1 + ΔH 2 ΔH - ΔH 2 = ΔH 1

See Hess’s Law calculations 1

Calculate the enthalpy of combustion of methane ΔH θ f (CH 4 )= -84 kj mol -1 ΔH θ f (CO 2 )= -394 kj mol -1 ΔH θ f (H 2 O)= -286 kj mol -1 Ans = -882 kj mol -1 CH 4 (g) + 2O 2 (g)C O 2 (g) + 2H 2 O(l) ΔH θ comb (CH 4 ) C(s) + 2H 2 + 2O 2 ΔH θ f (CO 2 )+ ΔH θ f (H 2 O)x2 ΔH θ f (CH 4 )ΔHΔH ΔH1ΔH1 ΔH2ΔH2 ΔH = ΔH 1 + ΔH 2 ΔH - ΔH 1 = ΔH 2

See Hess’s Law calculations 2

Calculate the enthalpy of reaction ReactantsProducts ΔH θ reaction Elements in normal state Add ΔH θ f (all products) Add ΔH θ f (all reactants) ΔHΔH ΔH1ΔH1 ΔH2ΔH2 ΔH = ΔH 1 + ΔH 2 ΔH - ΔH 1 = ΔH 2

Example of enthalpy of reaction calculation  For the reaction  SiCl 4 + 2H 2 O  SiO 2 + 4HCl  Calculate the enthalpy of reaction given:  ΔH θ f (H 2 O)= -286 kj mol -1  ΔH θ f (SiCl 4 )= kj mol -1  ΔH θ f (SiO 2 )= kj mol -1  ΔH θ f (HCl)= kj mol -1  Ans = 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 C 6 H 10 + H 2  C 6 H 12 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 (ΔH θ reaction ) of ethane C 2 H 4 (g) + 2H 2 (g)C2H6C2H6 ΔH θ reaction 3H 2 O + 2CO 2 ΔH θ comb (C 2 H 6 ΔH θ comb (C 2 H 4 ) + ΔH θ comb (H 2 ) x2 ΔH2ΔH2 ΔHΔH ΔH1ΔH1 ΔH = ΔH 1 + ΔH 2 ΔH - ΔH 2 = ΔH 1

See Hess’s Law calcs 3 Q’s 3&4