enthalpy diagrams

Exothermic Reactions

Method 1

A+B C+D+100KJMethod 2

A+B C+D ∆H=-100KJMethod 3

C+D

∆H=-100KJ

A+B

Reaction ProgressEnthalpy H

Endothermic Reactions

A+B+100KJ C+DMethod 1

Method 2

A+B C+D ∆H=+100KJMethod 3

A+B

Reaction ProgressEnthalpy H

∆H=+100KJ

C+D

Molar EnthalpyEnthalpy per 1 mole of a substance

2A+BC+3D+400KJ

Molar enthalpy of A=-400/2=-200KJ/mol

Molar enthalpy of D=-400/3=-133KJ/mol

Q1. Consider the following reaction.N2(g) + O2(g) → 2NO(g) ΔH°rxn = +180.6 kJ(a) Rewrite the thermochemical equation, including the standardenthalpy 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 ofnitrogen monoxide?

Answer:(a)N2(g) + O2(g) +180.6 kJ → 2NO(g)

Reaction Progress

Enthalpy H

∆H=+180.6KJ

2NO

N2+O2

(b) (c) 2 mol of NO= +180.6 KJ1 mol of NO=?

+180.6KJ________2

=+90.3 KJ

2. The reaction of iron with oxygen is very familiar. You can see theresulting rust on buildings, vehicles, and bridges. You may be surprised,however, at the large amount of heat that is produced by thisreaction.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

Enthalpy H

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 enthalpychange 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

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