THERMOCHEMISTRY

CP Unit 9Chapter 17

Thermochemistry 17.1 Thermochemistry is the study of energy

changes (HEAT) that occur during chemical reactions and changes in state.

ENTHALPY 17.2 Enthalpy = a type of chemical energy, sometimes

referred to as “heat content”, ΔH (the heat of reaction for a chemical reaction)

endothermic reactions (feels cold): q = ΔH > 0 (positive values)

exothermic reactions (feels hot): q (heat) = ΔH (enthaply, heat of rxn) < 0 (negative

values)

Endo vs. Exo-Endothermic reactions:

absorbs heat from surroundings (+). If you touch an

endothermic reaction it feels COLD

Exothermic reactions: release heat to the surroundings (-) If you touch an

exothermic reaction it feels HOT

Magnitude of Heat Flow Units of heat energy:

1 kcal = 1,000 cal = 1 Cal (nutritional) 1 kJ = 1,000 J 1 calorie = 4.184 J 1 kcal = 4.184 kJ

Thermochemical Equations A chemical equation that shows the

enthalpy (H) is a thermochemical equation.

Rule #1The magnitude (value) of H is directly proportional to the amount of reactant or product.

H2 + Cl2 2HCl H = - 185 kJ

* meaning there are 185 kJ of energy RELEASED for every:1 mol H2

1 mol Cl22 moles HCl

Rules of ThermochemistryExample 1: H2 + Cl2 2HCl H = - 185 kJ

Calculate H when 2.00 moles of Cl2 reacts.

Rules of ThermochemistryExample 2: Methanol burns to produce carbon dioxide and water:

2CH3OH + 3O2 2CO2 + 4H2O H = - 1454 kJ

What mass of methanol is needed to produce 1820 kJ?

Rule #2 H for a reaction is equal in the magnitude but opposite in sign to H for the reverse reaction.

(If 6.00 kJ of heat absorbed when a mole of ice melts, then 6.00 kJ of heat is given off when 1.00 mol of liquid water freezes)

Rules of ThermochemistryExample 1: Given: H2 + ½O2 H2O H = -285.8 kJ

Reverse: H2O H2 + ½O2 H = +285.8 kJ

Example 2CaCO3 (s) CaO (s) + CO2 (g) H = 178 kJ

What is the H for the REVERSE RXN?

CaO (s) + CO2 (g) CaCO3 (s) H = ?

Alternate form of thermochem. eq. Putting the heat content of a reaction INTO

the actual thermochemical eq. H2 + ½O2 H2O H = -285.8 kJ

Exothermic

(H is negative)

Heat is RELEASED as a PRODUCT

The alternate form is this:

H2 + ½O2 H2O +285.8 kJ

EX:2 NaHCO3 + 129 kJ Na2CO3 + H2O + CO2

Put in the alternate form

The alternate form is this:

2 NaHCO3 Na2CO3 + H2O + CO2 H = + 129 kJ

Put the following in alternate form1. H2 + Cl2 2 HCl H = -185 kJ

2. 2 Mg + O2 2 MgO + 72.3 kJ

3. 2 HgO 2 Hg + O2 H = 181.66 kJ

Enthalpies of Formation

enthalpy

change (delta)

standard condition

s

formation

fH

Enthalpies of Formation usually exothermic see table for Hf

value (Table A3) enthalpy of formation of an element

in its stable state = 0 these can be used to calculate H

for a reaction

Standard Enthalpy Change

Standard enthalpy change, H, for a given thermochemical equation is = to the sum of the standard enthalpies of formation of the product – the standard enthalpies of formation of the reactants.

) H(-) H(H reactantsfproductsfrxn

sum of (sigma)

Standard Enthalpy Change elements in their standard states can be omitted:

2 Al(s) + Fe2O3(s) 2 Fe(s) + Al2O3(s)

ΔHrxn = (ΔHfproducts) - (ΔHf

reactants)

ΔHrxn = ΔHfAl2O3 - ΔHf

Fe2O3

ΔHrxn = (-1676.0 kJ) – (-822.1 kJ)

ΔHrxn = -853.9 kJ

Standard Enthalpy Change the coefficient of the products and reactants in

the thermochemical equation must be taken into account:

O2(g) + 2SO2 (g) 2SO3 (g)

ΔHrxn = (ΔHfproducts) - (ΔHf

reactants)

ΔHrxn = 2ΔHfSO3 - 2ΔHf

SO2

ΔHrxn = 2(-395.7 kJ) – 2(-296.8 kJ)

ΔHrxn = -197.8 kJ

Standard Enthalpy Change - Calculate the standard heat for formation of benzene, C6H6, given the following thermochemical equation:

C6H6(l) + 15/2 O2(g) 6CO2(g) + 3H2O(l) H = -3267.4 kJ

-3267.4kJ = [6(-393.5kJ)+3(-285.8kJ)]–X

X = +49.0 kJ

-3267.4kJ = -3218.4–X

-49.0kJ = – X

X Total-393.5kJ -285.8kJ

Don’t forget the coefficents!