Transcript Hess’s Law

Hess’s Law (4-4)
• Hess suggested that the sum
of the enthalpies (ΔH) of the
steps of a reaction will equal
the enthalpy of the overall
reaction.
1
Why Does It Work?
• If you turn an equation around, you change
the sign:
• If H2(g) + 1/2 O2(g) H2O(g) DH=-241.8 kJ
• then,
H2O(g) H2(g) + 1/2 O2(g) DH =+241.8 kJ
• also,
• If you multiply the equation by a number,
you multiply the heat by that number:
• 2 H2O(g) H2(g) + O2(g) DH =+483.6 kJ
2
• Find the heat of formation (∆H) of:
2 Al (s) + 3 CuO (s) → 3 Cu (s) + 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 (s) ∆H = -157.3kJ
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=-1676kJ
3CuO (s)→3Cu(s) + 3/2 O2 (g) ∆H=(3)157.3kJ
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2 Al (s) + 3 CuO (s) → 3 Cu (s) + Al2O3 (s)
∆H = - 1204.1 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
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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 #1
b) Multiplying equation #2 by 4
c) 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