Hess’ LawLearning Goal: I will be able to use Hess’ Law to add enthalpy changes for the individual steps leading to a product in order to determine an overall enthalpy change for the reaction. I will also be able to use standard enthalpy (ΔH) values from tables in order to determine the standard enthalpy change for a reaction

ReviewEx 1)When white phosphorus burns in the

air, it produces phosphorus (V) oxide.

White phosphorus can be used as a weapon as it burns fiercely and can ignite cloth, fuel, ammunition and other combustibles. It is also a highly efficient smoke-producing agent, used in smoke grenades to mask movement, position or the origin of fire from the enemy.

Ex 2) Carbon disulfide burns in air, producing carbon dioxide and sulfur dioxide.

It is used as an insecticide for the fumigation of grains, nursery stock, in fresh fruit conservation and as a soil disinfectant against insects and nematodes. It is also a solvent for fats, resins, rubber, and asphalt.

•It states that for a chemical equation that can be written as the sum of two or more steps, the enthalpy change for the overall equation is the sum of the enthalpy changes for the individual steps.

Hess’s Law of Heat of Summation

A B + C H = x

B + C D H = y

A D H = ?

What will the change in altitude be if you take the path on the right? Path on the left? Is there a difference?

Enthalpy diagram illustrating Hess’s law

• For example, suppose you are given the following data:

Example # 1

kJ -297H );g(SO)g(O)s(S o22

kJ 198H );g(O)g(SO2)g(SO2 o223

• Could you use these data to obtain the enthalpy change for the following reaction?

?H );g(SO2)g(O3)s(S2 o32

•If we multiply the first equation by 2 and reverse the second equation, they will sum together to become the third.

Techniques for Manipulating Equations

1.Reverse an equation• the products become the reactants, and reactants become

the products• the sign of the ΔH value must be changed

2.Multiply each coefficient• all coefficients in an equation are multiplied by the same

integer or fraction• the value of ΔH must also be multiplied by the same

number

•What is the change in enthalpy for the reaction of 2NO2(g) N2O4(g) if:

2NO2(g)N2(g) + 2O2(g) ΔH = -84.8kJ

N2(g) + 2O2(g)N2O4(g) ΔH = 12kJ

Example 1

Standard Enthalpies of Formation

•The term standard state refers to the standard thermodynamic conditions chosen for substances when listing or comparing thermodynamic data: 1 atmosphere pressure and the specified temperature (usually 25 oC).– The enthalpy change for a reaction in

which reactants are in their standard states is denoted DHo (“delta H naught”).

• The standard enthalpy of formation of a substance, denoted DHf

o, is the enthalpy change for the formation of one mole of a substance in its standard state from its component elements in their standard state.

– Note that the standard enthalpy of formation for a pure element in its standard state is zero.

– i.e. O2(g), Na(s) these things cannot be broken down further

• The law of summation of heats of formation states that the enthalpy of a reaction is equal to the total formation energy of the products minus that of the reactants.

S is the mathematical symbol meaning “the sum of”, n is the coefficients of the substances in the chemical equation.

)reactants(Hm)products(HnH of

of

o

Example # 2• Large quantities of ammonia are used to prepare nitric

acid according to the following equation:

What is the standard enthalpy change for this reaction? (Refer to tables for data)

)g(OH6)g(NO4)g(O5)g(NH4 223

Learning CheckDetermine ∆H˚r for the following reaction using the enthalpies of formation that are provided.

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

∆H˚f of C2H5OH(l): –277.6 kJ/mol∆H˚f of CO2(g): –393.5 kJ/mol∆H˚f of H2O(l): –285.8 kJ/mol

Homework

•Pg 324 Q2, 5, 7