Enthalpy diagrams

1. Exothermic Reactions
Method 1
A+B C+D+100KJ
Method 2
A+B C+D ∆H=-100KJ
Method 3
C+D
∆H=-100KJ
A+B
Reaction Progress
EnthalpyH

2. Endothermic Reactions
A+B+100KJ C+D
Method 1
Method 2
A+B C+D ∆H=+100KJ
Method 3
A+B
Reaction Progress
EnthalpyH
∆H=+100KJ
C+D

3. Molar Enthalpy
Enthalpy per 1 mole of a substance
2A+BC+3D+400KJ
Molar enthalpy of A=-400/2=-200KJ/mol
Molar enthalpy of D=-400/3=-133KJ/mol

4. Q1. Consider the following reaction.
N2(g) + O2(g) → 2NO(g) ΔH°rxn = +180.6 kJ
(a) Rewrite the thermochemical equation, including the standard
enthalpy of reaction as either a reactant or a product.
(b) Draw an enthalpy diagram for the reaction.
(c) What is the enthalpy change for the formation of one mole of
nitrogen monoxide?
Answer:
(a)N2(g) + O2(g) +180.6 kJ → 2NO(g)
Reaction Progress
EnthalpyH
∆H=+180.6KJ
2NO
N2+O2
(b)
(c) 2 mol of NO= +180.6 KJ
1 mol of NO=?
+180.6KJ________
2
=+90.3 KJ

5. 2. The reaction of iron with oxygen is very familiar. You can see the
resulting rust on buildings, vehicles, and bridges. You may be surprised,
however, at the large amount of heat that is produced by this
reaction.
4Fe(s) + 3O2(g) → 2Fe2O3(s) + 1.65 × 103
kJ
(a) What is the enthalpy change for this reaction?
(b) Draw an enthalpy diagram that corresponds to the thermochemical
equation.
(a) ∆H=-1.65 × 103
kJ
2Fe2O3(s)
∆H=-1.65 × 103
kJ
4Fe(s) + 3O2(g)
Reaction Progress
EnthalpyH

6. 3. Tetraphosphorus decoxide, P4O10, is an acidic oxide. It reacts with
water to produce phosphoric acid, H3PO4, in an exothermic reaction.
P4O10(s) + 6H2O() → 4H3PO4(aq) ΔH°rxn = −257.2 kJ
(a) Rewrite the thermochemical equation, including the enthalpy
change as a heat term in the equation.
(b) How much energy is released when 5.00 mol of P4O10 reacts with
excess water?
(a)P4O10(s) + 6H2O() → 4H3PO4(aq)+257.2 kJ
(b)1 mol of P4O10= -257.2 kJ
5 mol of P4O10= ?
5x -257.2 kJ=-1286 KJ

7. 4.
(a) -483.6/2= -241.8KJ/mol
(b) -1134.4/4= -283.6KJ/mol
(c) +163.2/2= +81.6KJ/mol
(d) -1118.4/3= -372.8KJ/mol