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• Hess suggested that the sum of the enthalpies (ΔH) of the steps of a reaction will equal the enthalpy of the overall reaction.

Hess’s Law

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Why Does It Work?• If you turn an equation around, you change

the sign:• If H2(g) + 1/2 O2(g) H2O(g) H=-285.5 kJ

• then, H2O(g) H2(g) + 1/2 O2(g) H =+285.5 kJ

• also,• If you multiply the equation by a number, you

multiply the heat by that number:• 2 H2O(g) H2(g) + O2(g) H =+571.0 kJ

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• How do you get good at this?

• Find the heat of formation (∆H) of:2 Al (s) + 3 CuO (g) → 3 Cu (g) + Al2O3 (s)

Using two equations from the sheet:

2Al (s) + 3/2 O2 (g) → Al2O3 (s) ∆H=-1676kJ

Cu(s) + 1/2 O2 (g) → CuO (g) ∆H = -155kJ

Need to flip the second equation and change the sign on ∆H and multiply it by 3

2Al (s) + 3/2 O2 (g) → Al2O3 (s) ∆H=-1676kJ3CuO (g)→3Cu(s) + 3/2 O2 (g) ∆H=(3)155kJ___________________________________2 Al (s) + 3 CuO (g) → 3 Cu (s) + Al2O3 (s)

∆H = - 1211 kJ( oxygen can be cancelled because it exists on

both sides of the reactions)

• Calculating Heats of Reaction using Hess's Law

• 1) Write the overall equation for the reaction if not given.

• 2) Manipulate the given equations for the steps of the reaction so they add up to the overall equation.

• 3) Add up the equations canceling common substances in reactant and product.

• 4) Add up the heats of the steps = heat of overall reaction.

H2O(g) + C(s) → CO(g) +H2(g) ∆H=?

These are the equations chosen from the sheet

1) H2(g) + 1/2O2(g) →H2O(g) ∆H = -242.kJ

2) C(s) +1/2 O2(g) → CO(g) ∆H = -110. kJ

Note that the H2O (g) is a reactant and in step #1 it is a product. Thus step one needs to be reversed and the sign of the heat.

H2O(g) → H2(g) + ½ O2(g) ∆H = 242.kJ

C(s) + ½ O2(g) → CO(g) ∆H = -110. kJ

_________________________________H2O(g) + C(s) → CO(g) +H2(g) ∆H=132

Calculate the heat of reaction for the following equation

C3H8(g) + 5 O2 (g) ---> 3 CO2(g)+4H2O (g)

given the following steps in the reaction mechanism.

1) 3C (s)+ 4 H2 (g) -------> C3H8 (g)

2) H2 (g) + ½ O2 (g) -------> H2O (g)

3) C (s) + O2 (g) --------> CO2 (g)

• We can manipulate the equations by:

a) Reversing equation #1b) Multiplying equation #2 by 4c) Multiplying equation #3 by 3

∆H

C3H8 (g) -------> 3C (s)+ 4 H2 (g) 103.8

4 H2 (g) + 2O2 (g) -----> 4H2O (g) -967.2

3C (s) + 3O2 (g) -----> 3CO2 (g) -1180.5

C3H8(g) + 5 O2 (g) ---> 3 CO2(g)+4H2O (g)

∆H = - 2043.9 kJ

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Standard Heats of Formation• The H for a reaction that produces 1 mol of

a compound from its elements at standard conditions

• Standard conditions: 25°C and 1 atm.• Symbol is H f

0

The standard heat of formation of an element = 0This includes the diatomics

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What good are they?

• The heat of a reaction can be calculated by: – subtracting the heats of formation of the

reactants from the products

Ho = (H f0 H f

0Products) - ( Reactants)

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Examples• CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)

H f0

CH4 (g) = - 74.86 kJ/molH f

0O2(g) = 0 kJ/mol

H f0

CO2(g) = - 393.5 kJ/molH f

0H2O(g) = - 241.8 kJ/mol

H= [-393.5 + 2(-241.8)] - [-74.68 +2 (0)]H= - 802.4 kJ