Standard Enthalpy Changes of Reaction

Section 15.1

Introduction

Enthalpy change of formation: specified by ΔHθ

f is the heat change (at constant

pressure) on production of one mole of the pure substance from its elements in their standard state under standard thermodynamic conditions (298 K and 1 atm pressure)

More

Standard state: generally the most thermodynamically stable form of the pure element that exists under standard thermodynamic conditions

For example, for carbon it is graphite

Example

ΔHθf of silver bromide is the enthalpy

change for the reaction: Ag(s) + ½Br2(l) → AgBr(s) ΔHθ

f = -99.5kJ mol-1

Notice the formation of one mole and the standard states

2Ag(s) + Br2(l) → 2AgBr(s) makes 2 mol

Ag(s) + ½Br2(g) → AgBr(s) Br not in standard state

Calculations

Enthalpy change of a reaction = the sum of enthalpies of formation of products – sum of enthalpies of formation of reactants

ΔH = ΣΔHθf [products] – ΣΔHθ

f [reactants]

reactants products

elements

Example Problem #1

Calculate the enthalpy change of the following reaction:

3CuO(s)

+ 2Al(s)

→ 3Cu(s)

+ Al2O

3(s)

ΔHθf [CuO] = -155 kJmol-1

ΔHθf [Al

2O

3] = -1669 kJmol-1

See the board for the working out

Example Problem #2

Calculate the enthalpy change of the following reaction:

NH4NO

3(s) → N

2O

(g) + 2H

2O

(l)

ΔHθf [NH

4NO

3] = -366 kJ mol-1

ΔHθf [N

2O] = +82 kJ mol-1

ΔHθf [H

2O] = -285 kJ mol-1

Continued

ΔHθreaction

= ΔHθf,products

– ΔHθf, reactants

Have to consider 2 mol of water ΔHθ = (82 + (2) (-285)) - (-366) ΔHθ = (82 + (-570)) - (-366) ΔHθ = -488 + 366 ΔHθ = -122 kJ mol-1

Enthalpy Change of Combustion ΔHθ

c is the enthalpy change (at constant

pressure) when one mole of a pure substance undergoes complete combustion under standard thermodynamic conditions

Products of complete combustion are CO2

and H2O

Can be used to calculate the total enthalpy change only if both sides can be burnt in oxygen

Using Hess's Law elements compounds

+ O2

+ O2

combustion products

ΔHθ =ΣΔHθc [reactants] – ΣΔHθ

c [products]

Combustion is from the compound while formation is to the compound

Example Problem #3

Calculate the enthalpy change for the hydration of ethene according to the following equation:

C2H

4(l) + H

2O

(l) → C

2H

5OH

(l)

ΔHθc [C

2H

4] = -1409 kJ mol-1

ΔHθc [C

2H

5OH] = -1371kJ mol-1

ΔHθ = -1409 - (-1371) = - 38 kJ mol-1