the heat evolved or absorbed in a chemical process is the same whether the process takes

place in one or several steps.

if two or more chemical equations can be added together to produce an overall

equation, the sum of the enthalpy equals the enthalpy change of the overall equation.

This is called the Heat of Summation, ∆H

Analogy for Hess's Law

There is an old Chinese proverb which says: There are many ways to the top of a mountain, but the view from the top is always the same.

Develop an analogy for soccer and scoring a goal.

Develop an analogy for soccer and scoring a goal.

Hess’s LawRead through the whole questionPlan a Strategy Evaluate the given equations. Rearrange and manipulate the equations so

that they will produce the overall equation. Add the enthalpy terms.

H2O(g) + C(s) → CO(g) + H2(g)

Use these equations to calculate the molar enthalpy change which produces hydrogen gas.

C(s) + ½ O2(g) → CO(g) ∆H = -110.5kJH2(g) + ½ O2(g) → H2O(g) ∆H = -241.8kJ

Example 1

H2O(g) + C(s) → CO(g) + H2(g)

Use these equations to calculate the molar enthalpy change which produces hydrogen gas.

C(s) + ½ O2(g) → CO(g) ∆H = -110.5kJH2O(g) → H2(g) + ½ O2(g) ∆H = +241.8kJ

_____________________________________C(s) + H2O(g) → H2(g) + CO(g) ∆H=+131.3kJ

4C(s) + 5H2(g) → C4H10(g)Use these equations to calculate the molar enthalpy

change which produces butane gas.C4H10(g) + 6 ½ O2(g) → 4CO2(g) + 5H2O(g) ∆H= -2657.4kJ/molC(s) + O2(g) → CO2(g) ∆H= -393.5kJ/molH2(g) + ½O2(g) → H2O(g) ∆H= -241.8kJ/molRead through the whole questionPlan a Strategy Evaluate the given equations. Rearrange and manipulate the equations so that they will produce the

overall equation. Add the enthalpy terms. REWRITE THE CHANGES.

Example 2

4C(s) + 5H2(g) → C4H10(g)Use these equations to calculate the molar

enthalpy change which produces butane gas.C4H10(g) + 6 ½ O2(g) → 4CO2(g) + 5H2O(g) ∆H= -2657.4kJ/molC(s) + O2(g) → CO2(g) ∆H= -393.5kJ/molH2(g) + ½O2(g) → H2O(g) ∆H= -241.8kJ/mol

Example 2

4C(s) + 5H2(g) → C4H10(g)Use these equations to calculate the molar

enthalpy change which produces butane gas.5H2O(g) + 4CO2(g) → 6 ½ O2(g) + C4H10(g) ∆H= +2657.4kJ/molC(s) + O2(g) → CO2(g) ∆H= -393.5kJ/molH2(g) + ½O2(g) → H2O(g) ∆H= -241.8kJ/mol

Example 2

4C(s) + 5H2(g) → C4H10(g)Use these equations to calculate the molar

enthalpy change which produces butane gas.5H2O(g) + 4CO2(g) → 6 ½ O2(g) + C4H10(g) ∆H= +2657.4kJ/mol4(C(s) + O2(g) → CO2(g)) ∆H=

4(-393.5kJ/mol)H2(g) + ½O2(g) → H2O(g) ∆H= -241.8kJ/mol

Example 2

4C(s) + 5H2(g) → C4H10(g)Use these equations to calculate the molar

enthalpy change which produces butane gas.5H2O(g) + 4CO2(g) → 6 ½ O2(g) + C4H10(g) ∆H= +2657.4kJ/mol4C(s) + 4O2(g) → 4CO2(g) distribute the 4 ∆H=

4(-393.5kJ/mol)H2(g) + ½O2(g) → H2O(g) ∆H= -241.8kJ/mol

Example 2

4C(s) + 5H2(g) → C4H10(g)Use these equations to calculate the molar

enthalpy change which produces butane gas.5H2O(g) + 4CO2(g) → 6 ½ O2(g) + C4H10(g) ∆H= +2657.4kJ/mol4C(s) + 4O2(g) → 4CO2(g) distribute the 4 ∆H=

4(-393.5kJ/mol)5(H2(g) + ½O2(g) → H2O(g)) distribute the 5 ∆H=

5(-241.8kJ/mol)

Example 2

4C(s) + 5H2(g) → C4H10(g)Use these equations to calculate the molar

enthalpy change which produces butane gas.5H2O(g) + 4CO2(g) → 6 ½ O2(g) + C4H10(g) ∆H= +2657.4kJ/mol4C(s) + 4O2(g) → 4CO2(g) ∆H=

4(-393.5kJ/mol)5H2(g) + 2½O2(g) → 5H2O(g) ∆H=

5(-241.8kJ/mol)

Example 2

4C(s) + 5H2(g) → C4H10(g)Use these equations to calculate the molar enthalpy

change which produces butane gas.5H2O(g) + 4CO2(g) → 6 ½ O2(g) + C4H10(g) ∆H= +2657.4kJ/mol4C(s) + 4O2(g) → 4CO2(g) ∆H= 4(-393.5kJ/mol)5H2(g) + 2½O2(g) → 5H2O(g) ∆H= 5(-241.8kJ/mol)

_____________________________________________________

∆H = -125.6kJ/mol

Example 2