topic 5 energetics thermochemistry
DESCRIPTION
Topic 5 Energetics Thermochemistry. SL. 5.1 Exothermic reactions. Heat is produced and transferred to surroundings NaOH (s ) + H 2 O NaOH ( aq ) + heat HCl + NaOH NaCl + H 2 O + heat Neutralisation Wood + O 2 CO 2 + H 2 O + heat Combustion. - PowerPoint PPT PresentationTRANSCRIPT
Topic 5 Energetics
Thermochemistry
SL
5.1 Exothermic reactions
• Heat is produced and transferred to surroundings
• NaOH(s) + H2O NaOH(aq) + heat
• HCl + NaOH NaCl + H2O + heat Neutralisation
• Wood + O2 CO2 + H2O + heat Combustion
Exothermic- compare explosion
Endothermic reactions
• Heat is consumed from surroundings- it gets colder or you need to heat
• Ba(OH)2(s) + 2 NH4SCN(s) + heat
Ba2+(aq) + 2 SCN-
(aq) + 2 H2O(l) + NH3(aq)
Enthalpy, H
• H = internal energy. The total chemical energy of a system. Some of the energy is stored in chemical bonds.
DH = enthalpy change
• There is no “absolute zero” for enthalpy => enthalpy for a particular state cannot be measured but changes in enthalpy during reactions can be measured.
•DH = Hproducts – Hreactants
In the reaction2 H2 (g) + O2 (g) 2 H2O (g)
Enthalpy, H
2 H2 + O2
2 H2O
- 486 kJ
Hreactants- Enthalpy of reactants- 1856 kJHproducts- Enthalpy of products- 1370 kJ
DH= Hproducts - Hreactants=1370-1856 = - 486 kJ
EXOTHERMIC
In the reaction1/2 N2 (g) + O2 (g) NO2 (g)
Enthalpy, H
1/2 N2 + O2
NO2
33,9 kJ
DH= Hproducts - Hreactants = 33,9 kJ/mol
ENDTHERMIC
Exothermic reactionCH4 + 2 O2 CO2 + 2H2O + heat
Energy rich Energy poor
• DH = (Energy poor) – (Energy rich) => negative value
=> In Exothermic reactions: DH < 0 => Gives more stable products
=> In Endothermic reactions DH > 0 => Gives more reactive products.
Type of reaction
Heat energy change
Temperature change
Sign of DH
Exothermic Heat energy evolved
Becomes hotter
Negative (-)
Endothermic Heat energy absorbed
Becomes colder orenergy must be added
Positive (+)
DHo: standard enthalpy change of reaction
Standard conditions: p =101.3 kPa, T =298 KFactors affecting DHo
• The nature of the reactants and products• The amount• Changing state involves the enthalpy change• The temperature and pressure of the reaction
surroundings
5.2 Calculation of Energy/Enthalpy Changes
• Measurements:Open calorimeter
Bomb calorimeter
E = c. m. DT
• E= energy• m = mass• DT = temperature change• c = specific heat capacity, different for all
substances– E.g. 4.18 J/g*K for Water
Example
• The heat energy required to heat 50 g of water from 20oC to 60oC is:
• E = 50*4.18*40 = 8364 J = 8.364 kJ
Enthalpy changes
• 2 Mg + O2 2 MgO DH = -1202 kJ/mol Exothermic
• The amount of energy released when 0.6 g of Mg is burnt?
Mgm 0.6 gM 24.3 g/moln 0.025 mol
• 1202*0.025 = 30 kJ
5.3 Hess’s law
• The principle of conservation of energy states that energy cannot be created or destroyed.
• The total change in chemical potential (enthalpy change) must be equal to the energy gained or lost.
• The total enthalpy change on converting a given set
of reactants to a particular set of products is constant, irrespective of the way in which the change is carried out.
C + ½ O2 CO DH1
CO +½ O2 CO2 DH2
C + O2 CO2
DH3 = DH1+DH2
http://www.ausetute.com.au/hesslaw.html
http://www.mikeblaber.org/oldwine/chm1045/notes/Energy/HessLaw/Energy04.htm
5.4 Bond Enthalpies
• Break chemical bonds requires energy => Endothermic process
• Form chemical bonds => Exothermic process
• Approximate enthalpy change, DH, can be calculated by looking at bonds being broken and formed in the reaction.
Average bond enthalpies
• gaseous molecule into gaseous atoms (not necessary the normal state)
• approx in different molecules => not so precise data, but normally within 10 %
Calculate DH for the reaction:2 H2 (g) + O2 (g) 2 H2O (g)
H-H 436 kJ/molH-O 464 kJ/molO=O 498 kJ/mol
Enthalpy
2 H2 + O2
4 H + 2 O
+ 1370 kJ
1. The bonds of the reactants are broken, enthalpy is needed2 mol H-H = 2* 436= 872 kJ1 mol O=O = 498 kJSum 1370 kJ is spent
Enthalpy
2 H2 + O2
4 H + 2 O
+ 1370 kJ
2.The free hydrogen and oxyden atoms form bonds to create the products. The bond enthalpy is released
2*2 mol H-O = 4* 464= 1856 kJ is formed
- 1856 kJ
2 H2O
Enthalpy
2 H2 + O2
4 H + 2 O
+ 1370 kJ
3. The enthalpy of the products are 1856-1370 = 486 kJ lower than the reactants
The excess enthalpy 486 kJ is released to the surroundings. Exothermic reaction
The ”extra” enthalpy needed (1370 kJ) is called ACTIVATION ENERGy
- 1856 kJ
2 H2O + 486 kJ
N2 + 3 H2 2 NH3
(Energies involved (see data booklet))
• N≡N 944kJ/mol• H-H 436 kJ/mol• N-H 388 kJ/mol
DH = (bonds broken) – (bonds formed) =(944 + 3*436) – (2*3*388) = -76 kJ/mol Exothermic
• (If using other data DHf = -92kJ/mol)