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