DefinitionsStandard Enthalpy Change

the change of enthalpy that accompanies the formation of 1 mole of a substance in its standard state from its constituent elements in their standard states.

Exothermic Rxn Typical examples; combustion, neutralisation,

respiration

Endothermic Rxn Typical examples; melting, evaporating, photosynthesis.

Signs Exothermic

Endothermic

Energy Level Diagrams

Exo and Endo Energy Level Diagrams

http://www.docbrown.info/page03/3_51energy.htm#3.

EnergeticsCalculation of Enthalpy

http://www.youtube.com/watch?v=lxTmei2yrBg

ExperimentsEnthalpy of Neutralisation of HCl with NaOH.

Enthalpy of dissolution.

Enthalpy of precipitation.

DefineEnthalpy of combustion

Enthalpy of formation

Draw energy level diagram for the process of evaporation.

Hess’s LawStudents should be able to use simple enthalpy

cycles and enthalpy level diagrams and to manipulate equations. Students will not be required to state Hess’s law…..but I’ll tell you anyway

Energy can not be created or destroyed. It can only be converted from one form to another – first law of thermodynamics

Conventions -ve for exothermic

+ve for endothermic

If forward reaction is exo the reverse reaction is endo and of identical magnitude.

Hess’s Law states that the enthalpy change for a reaction is independent of the route the reaction takes.

The overall enthalpy change depends only on the initial and final stages.

Direct measurement of enthalpy is impossible.

Enthalpy typesCombustion

Energy released when one mole of a compound is burned in excess oxygen.

FormationEnergy change when one mole of a compound is

formed under standard conditions from its constituent elements.

Bond Enthalpies – we will discuss later

Hess’ Law Defined

Hess’ Law: DH for a process involving the transformation of reactants into products is not dependent on pathway. Therefore, we can pick any pathway to calculate DH for a reaction.

Shortcuts Learn Them Enthalpies of formation :

ΔH = Σproducts – Σ reactants

Enthalpies of combustion :ΔH = Σreactants – Σ products

Average Bond EnthalpiesΔ H = [bonds broken] – [bonds made]

Hess’ Law: An Example

Using Hess’ LawWhen calculating

DH for a chemical reaction as a single step, we can use combinations of reactions as “pathways” to determine DH for our “single step” reaction.

2NO2 (g)

N2 (g) + 2O2 (g)

q

2NO2 (g)N2 (g) + 2O2 (g)

Example (cont.)Our reaction of interest is:

N2(g) + 2O2(g) 2NO2(g) DH = 68 kJ

• This reaction can also be carried out in two steps:

N2 (g) + O2 (g) 2NO(g) DH = 180 kJ2NO (g) + O2 (g) 2NO2(g) DH = -112 kJ

Example (cont.)

If we take the previous two reactions and add them, we get the original reaction of interest:

N2 (g) + O2 (g) 2NO(g) DH = 180 kJ 2NO (g) + O2 (g) 2NO2(g) DH = -112 kJ

N2 (g) + 2O2 (g) 2NO2(g) DH = 68 kJ

Changes in EnthalpyConsider the following expression for a chemical process:

DH = Hproducts - Hreactants

If DH >0, then qp >0. The reaction is endothermic

If DH <0, then qp <0. The reaction is exothermic

Example (cont.)Note the important things about this

example, the sum of DH for the two reaction steps is equal to the DH for the reaction of interest.

We can combine reactions of known DH to determine the DH for the “combined” reaction.

Hess’ Law: DetailsOnce can always reverse the direction of a

reaction when making a combined reaction. When you do this, the sign of DH changes.

N2(g) + 2O2(g) 2NO2(g) DH = 68 kJ

2NO2(g) N2(g) + 2O2(g) DH = -68 kJ

Details (cont.)The magnitude of DH is directly

proportional to the quantities involved (it is an “extensive” quantity).

As such, if the coefficients of a reaction are multiplied by a constant, the value of DH is also multiplied by the same integer.

N2(g) + 2O2(g) 2NO2(g) DH = 68 kJ

2N2(g) + 4O2(g) 4NO2(g) DH = 136 kJ

Using Hess’ LawWhen trying to combine reactions to form a

reaction of interest, one usually works backwards from the reaction of interest.

Example:

What is DH for the following reaction?

3C (gr) + 4H2 (g) C3H8 (g)

Example (cont.) 3C (gr) + 4H2 (g) C3H8 (g) DH = ?

• You’re given the following reactions:

C (gr) + O2 (g) CO2 (g) DH = -394 kJ

C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) DH = -2220 kJ

H2 (g) + 1/2O2 (g) H2O (l) DH = -286 kJ

Example (cont.)Step 1. Only reaction 1 has C (gr). Therefore,

we will multiply by 3 to get the correct amount of C (gr) with respect to our final equation.

C (gr) + O2 (g) CO2 (g) DH = -394 kJ

3C (gr) + 3O2 (g) 3CO2 (g) DH = -1182 kJ

Initial:

Final:

Example (cont.)Step 2. To get C3H8 on the product side of the

reaction, we need to reverse reaction 2.

3CO2 (g) + 4H2O (l) C3H8 (g) + 5O2 (g) DH = +2220 kJ

C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) DH = -2220 kJInitial:

Final:

Example (cont.)Step 3: Add two “new” reactions together to see

what is left:

3C (gr) + 3O2 (g) 3CO2 (g) DH = -1182 kJ

3CO2 (g) + 4H2O (l) C3H8 (g) + 5O2 (g) DH = +2220 kJ2

3C (gr) + 4H2O (l) C3H8 (g) + 2O2 DH = +1038 kJ

Example (cont.)

Step 4: Compare previous reaction to final reaction, and determine how to reach final reaction:

3C (gr) + 4H2O (l) C3H8 (g) + 2O2 DH = +1038 kJ

H2 (g) + 1/2O2 (g) H2O (l) DH = -286 kJ

3C (gr) + 4H2 (g) C3H8 (g)

Need to multiply second reaction by 4

Example (cont.)

Step 4: Compare previous reaction to final reaction, and determine how to reach final reaction:

3C (gr) + 4H2O (l) C3H8 (g) + 2O2 DH = +1038 kJ

4H2 (g) + 2O2 (g) 4H2O (l) DH = -1144 kJ

3C (gr) + 4H2 (g) C3H8 (g)

Example (cont.)• Step 4 (cont.): 3C (gr) + 4H2O (l) C3H8 (g) + 2O2 DH =

+1038 kJ 4H2 (g) + 2O2 (g) 4H2O (l) DH =

-1144 kJ3C (gr) + 4H2 (g) C3H8 (g) DH = -106 kJ

Changes in EnthalpyConsider the following expression for a chemical process:

DH = Hproducts - Hreactants

If DH >0, then qp >0. The reaction is endothermic

If DH <0, then qp <0. The reaction is exothermic

Another Example

Calculate DH for the following reaction:

H2(g) + Cl2(g) 2HCl(g)

Given the following:

NH3 (g) + HCl (g) NH4Cl(s) DH = -176 kJ

N2 (g) + 3H2 (g) 2NH3 (g) DH = -92 kJ

N2 (g) + 4H2 (g) + Cl2 (g) 2NH4Cl(s) DH = -629 kJ

Another Example (cont.)Step 1: Only the first reaction contains the

product of interest (HCl). Therefore, reverse the reaction and multiply by 2 to get stoichiometry correct.

NH3 (g) + HCl (g) NH4Cl(s) DH = -176 kJ

2NH4Cl(s) 2NH3 (g) + 2HCl (g) DH = 352 kJ

Another Example (cont.)Step 2. Need Cl2 as a reactant, therefore, add

reaction 3 to result from step 1 and see what is left.

2NH4Cl(s) 2NH3 (g) + 2HCl (g) DH = 352 kJ

N2 (g) + 4H2 (g) + Cl2 (g) 2NH4Cl(s) DH = -629 kJ

N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g)

DH = -277 kJ

Another Example (cont.)Step 3. Use remaining known reaction in

combination with the result from Step 2 to get final reaction.

N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g) DH = -277 kJ

( N2 (g) + 3H2(g) 2NH3(g) DH = -92 kJ)

H2(g) + Cl2(g) 2HCl(g) DH = ?

Need to take middle reaction and reverse it

Another Example (cont.)Step 3. Use remaining known reaction in

combination with the result from Step 2 to get final reaction.

N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g) DH = -277 kJ

2NH3(g) 3H2 (g) + N2 (g) DH = +92 kJ

H2(g) + Cl2(g) 2HCl(g) DH = -185 kJ

1

Changes in EnthalpyConsider the following expression for a chemical process:

DH = Hproducts - Hreactants

If DH >0, then qp >0. The reaction is endothermic

If DH <0, then qp <0. The reaction is exothermic