View
24
Download
0
Category
Preview:
Citation preview
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
Recommended