thermochemistry cp unit 9 chapter 17 thermochemistry 17.1 thermochemistry is the study of energy...

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!