TCq calrxn

• Reaction carried out under constant volume.

• Use a bomb calorimeter.• Usually study combustion.

Bomb Calorimetry (Constant Volume Calorimetry)

Example

A 0.5269 g sample of octane is placed in a bomb calorimeter with a heat capacity of 11.3 kJ °C-1. The original temperature of the water in the calorimeter is 21.93°C. After combustion, the temperature increases to 24.21°C. Determine the enthalpy of combustion per mole and per gram of octane.

Example

A 0.5269 g sample of octane is placed in a bomb calorimeter with a heat capacity of 11.3 kJ °C-1. The original temperature of the water in the calorimeter is 21.93°C. After combustion, the temperature increases to 24.21°C. Determine the enthalpy of combustion per mole and per gram of octane.

-48.9 kJ g-1

-5590 kJ mol-1

• Hess’s law: if a reaction is carried out in a number of steps, H for the overall reaction is the sum of H for each individual step.

• For example:

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = -802 kJ

2H2O(g) 2H2O(l) H = -88 kJ

CH4(g) + 2O2(g) CO2(g) + 2H2O(l) H = -890 kJ

Note that: H1 = H2 + H3

Example

Given:

Fe2O3 (s) + 3 CO (g) 2 Fe (s) + 3 CO2 (g) ΔH = -23 kJ

3 Fe2O3 (s) + CO (g) 2 Fe3O4 (s) + CO2 (g) ΔH = -39 kJ

Fe3O4 (s) + CO (g) 3 FeO (s) + CO2 (g) ΔH =+18 kJ

What is ΔH for the following?

FeO (s) + CO (g) Fe (s) + CO2 (g)

Example

Given:

Fe2O3 (s) + 3 CO (g) 2 Fe (s) + 3 CO2 (g) ΔH = -23 kJ

3 Fe2O3 (s) + CO (g) 2 Fe3O4 (s) + CO2 (g) ΔH = -39 kJ

Fe3O4 (s) + CO (g) 3 FeO (s) + CO2 (g) ΔH =+18 kJ

What is ΔH for the following?

FeO (s) + CO (g) Fe (s) + CO2 (g)

-11 kJ

Example

The bombardier beetle uses an explosive discharge as a defensive measure. The chemical reaction involved is the oxidation of hydroquinone by hydrogen peroxide to produce quinone and water:

C6H4(OH)2 (aq) + H2O2 (aq) C6H4O2 (aq) + 2 H2O (l)

C6H4(OH)2 (aq) C6H4O2 (aq) + H2 (g) ΔH = +177.4 kJ

H2 (g) + O2 (g) H2O2 (aq) ΔH = -191.2 kJ

H2 (g) + ½ O2 (g) H2O (g) ΔH = -241.8 kJ

H2O (g) H2O (l) ΔH = -43.8 kJ

C6H4(OH)2 (aq) + H2O2 (aq) C6H4O2 (aq) + 2 H2O (l)

ΔH = ???

C6H4(OH)2 (aq) C6H4O2 (aq) + H2 (g) ΔH = +177.4 kJ

H2 (g) + O2 (g) H2O2 (aq) ΔH = -191.2 kJ

H2 (g) + ½ O2 (g) H2O (g) ΔH = -241.8 kJ

H2O (g) H2O (l) ΔH = -43.8 kJ

C6H4(OH)2 (aq) + H2O2 (aq) C6H4O2 (aq) + 2 H2O (l)

ΔH = ???

-202.6 kJ

• If 1 mol of compound is formed from its constituent elements, then the enthalpy change for the reaction is called the enthalpy of formation, Ho

f .

• Standard conditions (standard state): 1 atm and 25 oC (298 K).

• Standard enthalpy, Ho, is the enthalpy measured when everything is in its standard state.

• Standard enthalpy of formation: 1 mol of compound is formed from substances in their standard states.

• If there is more than one state for a substance under standard conditions, the more stable one is used.

• Standard enthalpy of formation of the most stable form of an element is zero.

Examples – write formation reactions for each:

1. H2SO4

2. C2H5OH

3. NH4NO3

Examples – write formation reactions for each:

1. H2SO4 H2 (g) + S (s) + 2O2 (g) → H2SO4

2. C2H5OH 2C (s) + 3H2 (g) + 1/2O2 (g) → C2H5OH

3. NH4NO3 N2 (g) + 2H2 (g) + 3/2O2 (g) → NH4NO3

Enthalpies of FormationEnthalpies of Formation

Using Enthalpies of Formation of Calculate Enthalpies of Reaction

• For a reaction

reactantsproductsrxn ff HmHnH

Examples

Use the tables in the back of your book to calculate ΔH for the following reactions:

1. 4 CuO (s) 2 Cu2O (s) + O2 (g)

2. C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)

3. NH3 (g) + HCl (g) NH4Cl (s)

Examples

Use the tables in the back of your book to calculate ΔH for the following reactions:

1. 4 CuO (s) 2 Cu2O (s) + O2 (g) 283 kJ

2. C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l) -2220 kJ

1. NH3 (g) + HCl (g) NH4Cl (s) -175.9 kJ

1. Vaporize the metal (enthalpy of vaporization)• Na (s) Na (g)

2. Break diatomic nonmetal molecules (if applicable) (bond enthalpy)

• ½ Cl2 (g) Cl-

3. Remove electron(s) from metal (ionization energy)• Na (g) Na+ (g) + e-

4. Add electron(s) to nonmetal (electron affinity)• Cl (g) + e- Cl- (g)

5. Put ions together to form compound (lattice energy)• Na+ + Cl- NaCl (s)

Born-Haber CycleBorn-Haber CycleApplication of Hess’s LawApplication of Hess’s Law

• Overall Reaction:

– Na (s) + ½ Cl2 (g) NaCl (s)

• This is useful because all quantities are directly measurable except lattice energy. The Born-Haber cycle can be used to calculate lattice energy from the other values.

Step Energetics

1

2

3

4

5

Step Energetics

1 endothermic

2 endothermic

3 endothermic

4 Exothermic

5 EXOTHERMIC (highly)