Topic 5-energetics SL
5.1 Exothermic and endothermic reactions
5.2 Calculation of Energy/Enthalpy Changes5.3 Hess law5.4 Bond enthalpy
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 reactantsHproducts- Enthalpy of products
DH= Hproducts - Hreactants = - 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
ENDOTHERMIC
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
DHfo: standard enthalpy of formation (cf.
page 8 in Data Booklet)
The enthalpy difference for the reaction when the substance is formed from it’s elements under standard conditions.
5.2 Calculation of Energy/Enthalpy Changes
• Measurements:Open calorimeter
Bomb calorimeter
Calculation of heat of solution
• q = c . M . DT (page 1 in Data Booklet)• q= energy (J)• m = mass (g)• DT = temperature change (K)• 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:
q = 50*4.18*(60-40) = 8364 J = 8.364 kJ
Energy and heat are always positive
Enthalpy changes
Mg + ½ O2 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
q=1202*0.025 = 30 kJ (energy is always positive)
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=-283,0 kJ
CO +½ O2 CO2 DH2=-110,5 kJ
C + O2 CO2 DH3 = DH1+DH2= -393, 5 kJ
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 Þnot so precise data, but normally within 10 %Þcf. page 7 in Data Booklet
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 oxygen 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, DH= -486 kJ/mol
The ”extra” enthalpy needed (1370 kJ) is called ACTIVATION ENERGy
- 1856 kJ
2 H2O + 486 kJ
N2 (g) + 3 H2 (g) 2 NH3 (g)(Enthalpies involved (see data booklet page 7))
N≡N 945kJ/molH-H 436 kJ/molN-H 391 kJ/mol
DH = (bonds broken) – (bonds formed) =(945 + 3*436) – (2*3*391) = -93 kJ/mol
Exothermic
(If using other data DHf = -92kJ/mol)
Enthalpy
N2 + 3 H2
6 H + 2 N
+ 2253 kJ
- 2346 kJ
2 NH3 + 93 kJ
N2 + 3 H2 2 NH3
The enthalpy of the products are 2346-2253 = 93 kJ lower than the reactantsThe excess enthalpy 96 kJ is released to the surroundings. Exothermic reaction, DH= -93 kJ/mol
1/2 N2 (g) + O2 (g) NO2 (g)
ENDOTHERMIC
Enthalpy
½ N2 + O2
2 O + N
+ 970,5 kJ - 812 kJ
NO2
- 93 kJ
The bonds of the reactants are broken, enthalpy is needed1/2 mol N≡N 1/2* 945= 472,5 kJ1 mol O=O = 498 kJSum 970,5 kJ is spent
The free nitrogen and oxygen atoms form bonds to create the products. The bond enthalpy is released. 1 mol N-O = 222kJ1mol N=O = 590 kJ 812 kJ is formed
The sum of the bond enthalpies
The resulting enthalpy, 93 kJ is taken from the surroundings. Endothermic reaction, DH= 93 kJ/mol
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