TOPIC 5ENERGETICS/THERMOCHEMISTRY

5.2HESS’S LAW

ESSENTIAL IDEA

In chemical transformations energy can neither be created nor destroyed

(the first law of thermodynamics).NATURE OF SCIENCE (2.4)

Hypotheses – based on the conservation of energy and atomic theory, scientists can

test the hypothesis that if the same products are formed from the same initial

reactants then the energy change should be the same regardless of the number of steps.

UNDERSTANDING/KEY IDEA 5.2.A

The enthalpy change for a reaction that is carried out in a series of steps is equal to the sum of the enthalpy changes for the individual steps.

Hess’s law states that the enthalpy change for any chemical reaction is independent of the route, provided the starting conditions and final conditions, and reactants and products, are the same.

• Hess’s Law is designed so that you will manipulate a series of reactions (called step equations) to reach the required reaction which is most always given.

1. If you reverse a reaction, reverse the sign of ΔH.

2. If you multiply or divide a step equation by an integer, you must also multiply or divide your ΔH by the same factor.

• Hess’s law is a statement of the Law of Conservation of Energy.

• It is used to measure enthalpy changes which cannot be measured directly in the laboratory.

Example Problem• Consider the standard enthalpy of

formation of methane. C(s) + 2H2(g) CH4(g)

• Notice that one mole of methane was formed from its elements in their standard states.

• This looks like a simple enough equation where we could react carbon with hydrogen to form methane and determine the heat change.

• However, it is difficult to react carbon with hydrogen and practically impossible to do so experimentally.

• We can easily burn carbon, hydrogen and methane to determine the standard enthalpy of combustion ΔHcº directly for each of the three substances.

• This would give us our step equations with enthalpy values that we could manipulate to determine the ΔH for the ΔHfº for methane.

• C(s) + O2(g) CO2(g) ΔHcº = -393 kJ mol-1

• H2(g) + 1/2 O2(g) H2O(l) ΔHcº = -286 kJ mol-1

• CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔHcº = -890 kJ mol-1

• The main equation again is C(s) + 2H2(g) CH4(g)

• I need CH4 on the right and 2 – H2’s on the left.• I will have to reverse the 3rd equation so I change

the sign of ΔH.• I will have to double the 2nd equation because I

need 2H2’s in the final equation so I will multiply the ΔH by 2.

• C(s) + O2(g) CO2(g) ΔHcº = -393 kJ mol-1

• 2H2(g) + 2/2 O2(g) 2H2O(l) ΔHcº = 2(-286) kJ mol-1

• CO2(g) + 2H2O(l) CH4(g) + 2O2(g) ΔHcº = +890 kJ mol-1

• Now cross out anything that is the same on both sides.

C(s) + O2(g) CO2(g) ΔHcº = -393 kJ mol-1

• 2H2(g) + 2/2 O2(g) 2H2O(l) ΔHcº = 2(-286) kJ mol-1

• CO2(g) + 2H2O(l) CH4(g) + 2O2(g) ΔHcº = +890 kJ mol-1

• Adding the equations will now give us the main equation.

• C(s) + 2H2(g) CH4(g)

• ΔHfº = +890 kJ mol-1 + 2(-286) kJ mol-1 + -393 kJ mol-1

= -75 kJ mol-1

APPLICATION/SKILLS

Be able to calculate the ΔH reactions using ΔHf

◦ data.

• The standard enthalpy of formation (ΔHf

º ) is defined as the change in

enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states.

• The degree symbol means that the process was carried out under standard conditions.

GUIDANCE

Enthalpy of formation data can be found in the data booklet in section 12.

GUIDANCE

Be able to use the following equation:ΔH reaction = Σ(ΔHf

◦products) - Σ(ΔHf

◦reactants)

Calculating ΔHf⁰

ΔHf⁰ ∑ ΔHf

⁰(products) ∑ ΔHf

⁰(reactants)

You can calculate the standard heat of reaction (ΔH⁰) for any reaction using the above equation.

• You will be given values in a table, plug them into the above equation and multiply these values by the number of moles in the equation. Then solve.

• The standard enthalpy of formation (ΔHf

⁰) for an element is zero.

• There is no chemical change and so no enthalpy change when an element is formed from itself.

Sample Problem

• Calculate the enthalpy change for the following reaction given the standard enthalpy changes of formation.

• C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)

• ΔHf⁰ C3H8(g) = -104 kJ mol-1

• ΔHf⁰ CO2(g) = -394 kJ mol-1

• ΔHf⁰ H2O(l) = -286 kJ mol-1

• C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)• -104 0 3(-394) 4(-286)• • ΔHf

⁰ ∑ ΔHf⁰(products) ∑ ΔHf

⁰(reactants)

• = [3(-394) + 4(-286)] – [(-104) + (0)]• = -2220 kJ mol-1

How many ways to calculate ΔH so far

• 1. Thermochemical equations• 2. Calorimetry experiments (5)• 3. Hess’s law• 4. ΔHf

⁰ ∑ ΔHf⁰(products) ∑ ΔHf

⁰(reactants)