Enthalpy

A Quick Guide

Introduction

• What is enthalpy?

– The energy of a system

– For example, the energy contained within a compound

– Given the symbol H

Enthalpy

• Chemical reactions that give out energy in the form of heat are known as …

• Chemical reactions that take in energy in the form of heat are known as …

exothermic

endothermic

Enthalpy

• In exothermic reactions the products have less energy than the reactants

• Since you can’t destroy energy the energy is given out to the surroundings as heat

• This is known as an …

Reactants Products

enthalpy change

Measuring Enthalpy

• We can’t measure how much energy a compound has in it …

… instead we measure the change in energy (ΔH) when it reacts instead

ΔH = Hproducts - Hreactants

If ΔH is negative then the reactants give out energy (exothermic)

If ΔH is positive then the reactants take in energy (endothermic)

Measuring Enthalpy

• Enthalpy level diagrams chart this change

reactants

products

en

thalp

y

Energy given out therefore

change in enthalpy is negative

products

reactants

Energy taken in therefore change in enthalpy is positive

Units of Enthalpy

• Enthalpy is measured in kilojoules per mole (kJ mol-1)

• For example:

CH4 + 2O2 → 2H2O + CO2 (ΔH = -890kJ mol-1)

For every one mole of CH4

…

… 890kJ of energy is given

out

Measuring Enthalpy

• The ΔH can vary with conditions

• We assume our reactions to have taken place at standard conditions:

– A specified temperature (25oC / 298K)– A standard pressure of 1 atmosphere– A standard concentration of 1 mol dm-3

Measuring Enthalpy

• Enthalpy can be measured by measuring the transfer of energy to or from water in a bomb calorimeter

Energy transferred = cmΔT

• Where:– c = specific heat of water (4.17J g-1 K-

1)– m = mass of water in grams– ΔT = change in temperature of water

Showing Enthalpy Changes

• Enthalpy changes must always be accompanied by a chemical equation

• For example:

H2 + ½O2 → H2O (ΔH = -286kJ mol-1)

• But

2H2 + O2 → 2H2O (ΔH = -572kJ mol-1)

Enthalpy

• When 1 mole of fuel is burned under standard conditions we get …

• Remember the bigger the negative number …

… the more exothermic …

… the more energy given out

standard enthalpy change of combustion (ΔHc)

Enthalpy

• When 1 mole of compound is formed under standard conditions we get …

• Just because ΔHc is exothermic doesn’t mean that ΔHf is endothermic

2H2 + O2 → 2H2O (ΔH = -572kJ mol-1)

standard enthalpy change of formation (ΔHf)

Enthalpy Cycles

• Since we can’t measure the energy given out or taken in when all compound are created we have to use an enthalpy cycle

• For example:

C + 2H2 → CH4 (ΔHf = -75kJ mol-1)

… doesn’t happen under normal conditions

Enthalpy Cycles

C + 2H2 CH4

CO2 + 2H2O

ΔH can’t be measured this way

ΔH can be measured this way

• Theoretically the energy changes that occur in either route are the same

• Therefore, if we can measure one – we will know the other

Enthalpy Cycles

• The Universe states that energy cannot be created or destroyed …

• So, as long as your starting and finishing points are the same …… the enthalpy change will be the same

no matter how you get there

Law of Conservation of Energy

Hess’s Law

• Hess stated that the sum of enthalpy changes for a series of reaction …

… is the same as the overall enthalpy change

• So, if you know the enthalpy change in two parts of the cycle, then you can work out the third part of the cycle

ΔH1 = ΔH2 – ΔH3

Hess’s Law

C + 2H2 CH4

CO2 + 2H2O

ΔH1

ΔH2ΔH3

• ΔH1 = ΔHf of 1 mole of methane• ΔH2 = ΔHc of 1 mole of carbon with 2 moles of H2

• ΔH3 = ΔHc of 1 mole of methane

Hess’s Law

ΔH1 = ΔH2 – ΔH3

ΔHf (CH4) = (ΔHc (C) + 2ΔHc (H2)) – ΔHc (CH4)

• Where:– ΔHc (C) = -393kJ mol-1

– ΔHc (H2) = -286kJ mol-1

– ΔHc (CH4) = -890kJ mol-1

C + 2H2 CH4

CO2 + 2H2O

ΔH1

ΔH2ΔH3

Hess’s Law

ΔHf (CH4) = (ΔHc (C) + 2ΔHc (H2)) – ΔHc (CH4)

ΔHf (CH4) = (-393 + (2 x -286)) – 890

ΔHf (CH4) = -75kJ mol-1

• Therefore, even though we can’t directly measure it, the energy given out when creating methane from carbon and hydrogen is 75kJ mol-1

Hess’s Law

• The direction of the arrows in the cycle make a difference

NH3 + HCl NH4Cl

½N2 + 2H2 + ½Cl2

ΔH1

ΔH2ΔH3

ΔH1 = -ΔH2 + ΔH3

Hess’s Law

• Where:– ΔHf (NH3) = -46.1kJ mol-1

– ΔHf (HCl) = -92.3kJ mol-1

– ΔHf (NH4Cl) = -315kJ mol-1

NH3 + HCl NH4Cl

½N2 + 2H2 + ½Cl2

ΔH1

ΔH2ΔH3

ΔH1 = -ΔH2 + ΔH3

ΔH2 = ΔHf (NH3) - ΔHf (HCl)ΔH3 = ΔHf (NH4Cl)

Hess’s Law

ΔH1 = -(ΔHf (NH3) - ΔHf (HCl)) + ΔHf (NH4Cl)

ΔH1 = -(-46.1) – (-92.3) + (-315)

ΔH1 = -176.6kJ mol-1

• Therefore, 176.6 kJ of energy are given out every time one mole of NH4Cl is created

Bond Enthalpy

• The energy needed to break a particular bond is called …

H2 → 2H ΔH = +436kJ mol-1

• Notice that bond enthalpy is always positive (endothermic) because it requires energy to be put in to break the bond

bond enthalpy

Bond Enthalpy

• These can be worked out using enthalpy cycles

CH4 + 2O2 CO2 + H2O

C + 4H + 4O

ΔH1

ΔH2ΔH3

• Using this cycle we can work out the enthalpy change for the combustion of methane (ΔHc (CH4))

Bond Enthalpy

• ΔH2 = enthalpy change when bonds are broken

(4 x C-H bonds) + (2 x O=O bonds)

• ΔH3 = enthalpy change when bonds are formed

(2 x C=O bonds) + (4 x H-O bonds)

CH4 + 2O2 CO2 + H2O

C + 4H + 4O

ΔH1

ΔH2ΔH3

Bond Enthalpy• Bond enthalpies:

– C-H = +413kJ mol-1

– O=O = +498kJ mol-1

– C=O = +805kJ mol-1

– H-O = +464kJ mol-1

• ΔH2 = (4 x +413) + (2 x +498)

• ΔH3 = -((2 x +805) + (4 x +464))

• Therefore, overall energy change in burning methane is ΔH2 + ΔH3 or -818kJ mol-1

+2,648kJ mol-1

-3,4466J mol-1

Important

• The reason we get a different answer than before:

– ΔHc (CH4) = -75kJ mol-1 (using reactants and products)

– ΔHc (CH4) = -818kJ mol-1 (using bond enthalpies)

• is because in bond enthalpies everything has to be in a gaseous state and is therefore no to the standard state used in reactants and products