Enthalpy
Enthalpy (H): heat flow for a chemical reaction.
qrxn @ constant P = DH = Hproducts – Hreactants
• exothermic: q < DH
• endothermic: q > DH
Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law
Hproducts < Hreactants
Hproducts > Hreactants
Standard Enthalpy of Formation and Reaction
Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law
Standard Enthalpy of Formation and Reaction
aA + bB → cC + dD
standard-state conditions = 25oC & 1 atm
Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law
Example 6.10 Standard Enthalpy of Reactions: Direct Method
2Al(s) + 1Fe2O3(s) → 1Al2O3(g) + 2Fe(l)
Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law
Example 6.10 (con’t)
2Al(s) + 1Fe2O3(s) → 1Al2O3(g) + 2Fe(l)
Al(s)
Fe(l)
0
kJ/mol
12.40 kJ/mol
Al2O3(s) –1669.8 kJ/mol
Fe2O3(s)
H = 0 kJ/mol for elements at 25oC
(e.g., Fe(s) = 0 kJ/mol; Fe(l)  0)
Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law
–822.2 kJ/mol
Example 6.10 (con’t)
2Al(s) + 1Fe2O3(s) → 1Al2O3(g) + 2Fe(l)
Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law
INDIRECT METHOD (Hess’s Law)
Hess’s Law is used is convenient for obtaining
values of DH for reactions that are difficult to carry
out in a calorimeter.
E.g., C(s) + ½O2(g)  CO(g)
Nearly impossible to carry out in a calorimeter
because when C is burned, CO2 is almost
always produced.
For H2 + Cl2  2HCl
DH = –185 kJ
• –185 kJ..............sign = exothermic
• coefficients represent moles
(the -185 kJ refers to ONE MOLE of reaction)
• states of matter (phases) MUST be written
• standard temp = 25oC
(not STP: 0oC)
Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law
Rules:
(For H2 + Cl2  2HCl
DH = –185 kJ
1. Magnitude of DH is directly proportional to
the amount of reactant or product.
-185 kJ
1 mol H2
-185 kJ
1 mol Cl2
-185 kJ
2 mol HCl
2. Reverse the sign for DH for the reverse
reaction.
3. DH for a reaction is the same, whether
carried out in one step or several
(State Function)
Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law
Example 6.09 Standard Enthalpy of Reactions: Hess’s Law
2C(graphite) + 1H2(g) → 1C2H2(g)
ΔH kJ/mol
(a)
C(graphite) + O2(g) →CO2(g)
–393.5
(b)
H2(g) + ½O2(g) → H2O(l)
–285.8
(c)
2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(l) –2598.8
The values for H are determined by experiment, so they need to be given to
you; either directly or in a table.
Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law
Example 6.09 Standard Enthalpy of Reactions: Hess’s Law
2C(graphite) + 1H2(g) → 1C2H2(g)
ΔH kJ/mol
(a)
C(graphite) + O2(g) → CO2(g)
–393.5
(b)
H2(g) + ½O2(g) → H2O(l)
–285.8
(c)
2C2H2(g) + 5O2(g)→ 4CO2(g) + 2H2O(l) –2598.8
Concentrate on aligning the substances in the step reactions to be on the same
sides as the original equation.
Notice: There’s 2 moles of C(graphite) in the original equation but only 1 mole
in (a)..........................
Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law
Example 6.09 Standard Enthalpy of Reactions: Hess’s Law
2C(graphite) + 1H2(g) → 1C2H2(g)
(2a)
(b)
(c)
2C(graphite) + 2O2(g) → 2CO2(g)
H2(g) + ½O2(g) → H2O(l)
ΔH kJ/mol
–787.0
–285.8
2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(l) –2598.8
Double (a) to match the original equation.
And ΔH doubles, too (–393.5 * 2 = –787.0 kJ/mol)
Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law
Example 6.09 Standard Enthalpy of Reactions: Hess’s Law
2C(graphite) + 1H2(g) → 1C2H2(g)
(2a)
(b)
(c)
2C(graphite) + 2O2(g) → 2CO2(g)
H2(g) + ½O2(g) → H2O(l)
ΔH kJ/mol
–787.0
–285.8
2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(l) –2598.8
H2 is a reactant in both the original equation and in step (b).
And, there’s 1 mole H2 in the original equation and 1 mol H2 in step (b).
Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law
Example 6.09 Standard Enthalpy of Reactions: Hess’s Law
2C(graphite) + 1H2(g) → 1C2H2(g)
(2a)
(b)
(c)
2C(graphite) + 2O2(g) → 2CO2(g)
H2(g) + ½O2(g) → H2O(l)
ΔH kJ/mol
–787.0
–285.8
2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(l) –2598.8
Align (c) C2H2 with original equation as a product by reversing step (c) ..........
Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law
Example 6.09 Standard Enthalpy of Reactions: Hess’s Law
2C(graphite) + 1H2(g) → 1C2H2(g)
(2a)
(b)
(–c)
2C(graphite) + 2O2(g) → 2CO2(g)
H2(g) + ½O2(g)→ H2O(l)
ΔH kJ/mol
–787.0
–285.8
4CO2(g) + 2H2O (l)→ 2C2H2(g) + 5O2(l) +2598.8
Reverse (c) so it on the same side as the original equation.
Include reversing the sign of ΔH, too. Shown by (–c).
However, there 2 moles of C2H2 in step (c) but only 1 mole in the original
equation. So, we need to halve step (c)..........................................................
Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law
Example 6.09 Standard Enthalpy of Reactions: Hess’s Law
2C(graphite) + 1H2(g) → 1C2H2(g)
(2a)
(b)
2C(graphite) + 2O2(g) → 2CO2(g)
H2(g) + ½O2(g) → H2O(l)
ΔH kJ/mol
–787.0
–285.8
Divide step (–c), including ΔH, by 2 to match the original equation’s 1 mole
C2H2. (shown by (–c/2)
Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law
Example 6.09 Standard Enthalpy of Reactions: Hess’s Law
(2a)
2C(graphite) + 1H2(g) → 1C2H2(g)
ΔH kJ/mol
2C(graphite) + 2O2(g) → 2CO2(g)
–787.0
(b)
1H2(g) + ½O2(g) → 1H2O(l)
O2:
2+½ →
CO2:
H2O:
–285.8
5/2
2 →
2
1 
1
H = 226.6 kJ/mol
Determine the net equation for both substances & ΔH by adding them together.
Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law
Example 6.09 Standard Enthalpy of Reactions: Hess’s Law
2C(graphite) + 1H2(g) → 1C2H2(g)
(2a)
(b)
(-c/2)
2C(graphite)
ΔH kJ/mol
–787.0
+ 1H2(g)
–285.8
→ 1C2H2(g)
2C(graphite) + 1H2(g) → 1C2H2(g)
+1299.4
+226.6 kJ/mol
Net equation of steps = original equation H = +226.6 kJ/mol
Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law
Example 6.09 Standard Enthalpy of Reactions: Hess’s Law
2C(graphite) + 1H2(g) → 1C2H2(g)
Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law
ΔH = +226.6 kJ/mol
Fin
330_06_05 Energy: Calorimetry