…HEAT

!

A simpler problem:

How much heat is given off when 1.6 g of CH4 are burned in an excess of oxygen if Hcomb = -802 kJ/mol?

Step 1: Write the reaction equation.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Step 2: Calculate molar amount involved

combusted 4CH moles 0.100g/mol 16.042

4CH g 1.6

Step 3: Calculate amount of heat given off

DHrxn = (-802 kJ/mol)(0.100 mol CH4) = -80.2 kJ

Q: Is this an exothermic or endothermic reaction?

2 AlBr3 + 3 Cl2 2 AlCl3 + 3 Br2

Energy

2 AlBr3 + 3 Cl2

2 AlCl3 + 3 Br2

DHrxn = Heat content of products – heat content reactants

DHrxn < 0 Reaction is exothermic

But how do we determine the heat content in the first

place?

Heat of formation, DHf

• The DHf of all elements in their standard state equals zero.

• The DHf of all compounds is the molar heat of reaction for synthesis of the compound from its elements

DHf (AlBr3):2 Al + 3 Br2 2 AlBr3

DHrxn = 2DHf(AlBr3)

DHrx

n2DHf(AlBr3) =

• Since the DHrxn can be used to find DHf, this means that DHf can be used to find DHrxn WITHOUT having to do all of the calorimetric measurements ourselves!!

The Law of Conservation of Energy strikes again!!

Hess’s Law: DHrxn = S DHf(products) – S DHf(reactants)

6 CO2 (g) + 6 H2O (l) C6H12O6 (s) + 6 O2 (g)

DHrxn = [DHf(C6H12O6) + 6 DHf(O2)] – [6 DHf(CO2) + 6 DHf(H2O)]

From DHf tables: DHf(C6H12O6) = -1250 kJ/mol

DHf(CO2) = -393.5 kJ/mol DHf(H2O) = -285.8 kJ/mol

DHrxn = [-1250 kJ/mol] – [6(-393.5 kJ/mol) + 6(-285.8 kJ/mol)]

DHrxn = +2825.8 kJ/molDHrxn = +2825.8 kJ/mol

Using Hess’ Law with DHrxn

What is the DHcomb for ethane?

C2H4 (g) + 3 O2 (g) → 2 CO2 (g) + 2 H2O (g) -1323 kJ

C2H4 (g) + H2 (g) → C2H6 (g) -137 kJ

2 C2H6 (g) + 7 O2 (g) → 4 CO2 (g) + 6 H2O (g)Target rxn:

Given:

Rxn 1:

Rxn 2:

1) Rxn 1 doesn’t have enough oxygens, so multiply by 2

2(C2H4 (g) + 3 O2 (g) → 2 CO2 (g) + 2 H2O (g))

DHrxn

2(-1323 kJ)

2) Rxn 2 is going the wrong direction and doesn’t have enough C6H6. Reverse reaction and multiply by 2.

2(C2H6 (g) → C2H4 (g) + H2 (g)) 2(+137 kJ)

Note: when you reverse a reaction, change the sign on the H

Rxn 3:2 H2 (g) + O2 (g) → 2 H2O (g) -242 kJ

3) Combine the two reaction equations:

2 C2H4 + 6 O2 → 4 CO2 + 4 H2O 2(-1323 kJ)

+ 2 C2H6 → 2 C2H4 + 2 H2 + 2(+137 kJ)

2 C2H4 + 2 C2H6 + 6 O2 → 4 CO2 + 4 H2O + 2 C2H4 + 2 H2

2 C6H6 (g) + 7 O2 (g) → 4 CO2 (g) + 6 H2O (g)Target rxn:

4) Still don’t have enough O2 and need to get rid of H2. Add in Rxn 3.

+ 2 H2 + O2 → 2 H2O + (-242 kJ)

2 C2H6 + 2 H2 + O2 + 6 O2 → 4 CO2 + 2 H2O + 4 H2O + + 2 H2

Final rxn: 2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

Math: 2(-1323 kJ) + 2(+137) + (-242) =-2614 kJ = Hcomb