THERMOCHEMISTRY

8.4 Born-Haber CycleTHERMOCHEMISTRY 9.3 : Hesss Law

Hesss Law

Hesss Law

Hesss Law ApplicationsAlgebraic MethodEnergy Cycle Method

Hesss Law

Hesss Law states that when reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in the series of steps.

Hesss Law

Germain Henri Hess

x Y H1

A H2 H3

B H4

C H5

H6

H1 = H2 + H3= H4 + H5 + H6

Algebraic method

Energy cycle method

Methods to calculate H:

Normally we use algebraic method,energy cycle method is suitable for simple reaction only!

Given the following information, calculate the unknown H ? CO(g) + O2(g) CO2(g) H = 283.0 kJ N2(g) + O2(g) 2NO(g) H = 180.6 kJ

Two gaseous pollutants that form in auto exhaust are CO and NO. An environmental chemist is studying ways to convert them to less harmful gases through the following equation:

CO(g) + NO(g) CO2(g) + N2(g) H = ?

EXAMPLE 6 :

CO(g) + NO(g) CO2(g) + N2(g) H = ?Target equation:

Rearrange equations given:

H = 283.0 kJ + ( 90.3 kJ) = 373.3 kJ

CO(g) + O2(g) CO2(g) H = 283.0 kJNO(g) N2(g) + O2(g) H = 180.6 kJ 2

CO(g) + NO(g) CO2(g) + N2(g)

By using Algebraic Method :

From the following data, C(graphite) + O2(g) CO2(g) Hrxn = 393.5 kJ H2(g) + O2(g) H2O(l) Hrxn = 285.8 kJ 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(l) Hrxn = 3119.6 kJ

Calculate the standard enthalpy of formation of ethane C2H6(g) by using energy cycle method. The equation reaction given:

2C(graphite) + 3H2(g) C2H6(g)

Example 7

By using Algebraic Method :

Target equation:2C(graphite) + 3H2(g) C2H6(g) Hf = ?

Rearrange equations :

Equation 1 :

2 C(graphite) + 2O2(g) 2CO2(g) H1 = 787 kJ

Equation 2 :

3H2(g) + O2(g) 3H2O(l) H2 = 857.4 kJMultiply by 3

23

Multiply by 2

Equation 3 : Reverse & divide by 2

H3 = 1559.8 kJ2CO2(g) + 3H2O(l) C2H6(g) + O2(g)27

2 C(graphite) + 2O2(g) 2CO2(g) H1 = 787 kJ3H2(g) + O2(g) 3H2O(l) H2 = 857.4 kJ

23

2CO2(g) + 3H2O(l) C2H6(g) + O2(g)27 H3 = 1559.8 kJ

2C(graphite) + 3H2(g) C2H6(g)

Hf = -787 kJ + (-857.4 kJ) + 1559.8 kJ= -84.6 kJ mol -1

Draw the energy cycle and apply Hesss Law to calculate the unknown value.

Step 1

H3 = 1559.8 kJH1 = -787 kJ H2 = -857.4 kJ

= ?

By using Energy Cycle Method :

Step 2 Apply Hesss Law to calculate the unknown value.

= - 84.6 kJmol-1= - 786 kJ + (-857.4 kJ) + 1559.8 kJ

The compound is more stable than its components elements

Most Hf < 0o

EXAMPLE:

C(graphite) 0C(diamond) 1.9CO(g) 110.5CO2(g) 393.5CS2(l) 87.9

Hf OF COMPOUNDSo

aA + bB cC + dD

DETERMINING Hrxn FROM Hf o oHorxn = ?

Horxn = sum of Hf ofall of the products sum of Hof ofall of the reactants

Nitric acid (HNO3), whose worldwide annualproduction is about 8 billion kg, is used to make many products including fertilizers, dyes, and explosive. The first step in industrial production process is the oxidation of ammonia:4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) Calculate Ho Ho of NH3(g) = 45.9 kJ/molHo of NO(g) = 90.3 kJ/molHo of H2O(g) = 241.8 kJ/mol

fff

rxn

EXAMPLE 8 CHAPTER 9.3

END OF CHAPTER 9.3END OF CHAPTER 9.3