1. Thermochemistry

2. Thermochemistry Endothermic Takes in heat
has positive change in energy
+∆H
Exothermic
Heat is released
Has negative change in energy
-∆H

3. Heat of reaction
Heat change when the number of moles of reactants indicated in the balanced equation for the reaction react completely
Must always indicate the phsyical state of reactant and products

4. Heat of Combustion
Heat change when one mole of the substance is completely burned in excess oxygen
C(s) + 1/2O2(g) = CO(g) ∆H=-111kj/mol
This is not heat of combustion because not burned in excess oxygen
C(s) + O2(g) = CO2(g) ∆H=greater

5. Bomb Calorimeter

6. Hydrocarbons and Heat Most hydrocarbons are used as fuels.
Knowing how much energy a fuel provides, can tell us if it is useful for a certain application.
For example, the amount of energy a food releases when burned, can tell us about it’s caloric content (fats release lots of energy).
Heat energy released during combustion can be measured with a calorimeter.
A “bomb calorimeter” is shown. It includes water in a heavily insulated container, a stirrer, valve, bomb chamber, ignition wires, & a thermometer (pg327).

7. Why do we always mention the state of each compounds in the equations
C(s) + 1/2O2(g) = CO2(g)
Because changing the state of either reactants/ products can alter the heat change (ΔH)

8. Kilogram Calorific value
Of a fuel is the heat energy produced when 1kg of the fuel is completely burned in excess oxygen

9. Bond Energy CH4(s) +2O2(g) = CO2(g) + 2H2O(g) ∆H=-890kj/mol
Energy is required to break a bond and energy is released to form a bond
BOND ENERGY is the energy required to break one mole of covalent bonds and to separate the neutral atoms completely from each other
Calculated by studying spectra and heats of reaction values

10. Bond energy values are approximate
The energy to break the first C-H bond varies depending on what molecule it is in
CH4 ∆H=+425kj/mol
CH3 ∆H=+470kj/mol
CH2 ∆H=+416kj/mol
CH ∆H=+335kj/mol
Therefore average bond energy
( )/4=411.5

11. Heat of Neutralisation
Is the heat change when one mole of H+ ions from an acid and one mole of OH- ions from a base react completely
All strong acids and bases have similar heat of neutralisation ∆H=-57.1
Weak acids and bases are less because they don’t dissociate fully
Less than ∆H= -57

12. Heat liberated = m×c×(t2-t1)
Polystyrene to reduce heat loss & lid
Ensure same temperature before mixing
Use thermometer 0.1 °C
Stir at all times

13. Heat of formation
Of a compound is the heat change that takes place when one mole of a compound in its standard state is formed from its elements in their standard states

14. Hess’s law
States that if a chemical reaction takes place in a number of stages, the sum of the heat changes in the separate stages is equal to the heat change if the reaction is carried out in one stage

15. Law of conservation of energy
States that energy cannot be created or destroyed but can only be converted from one form of energy to another

16. Exothermic and Endothermic changes
In either case, the change in heat of the water tells us about the reaction of the chemicals.
An increase in water temperature indicates that the chemicals released energy when they reacted. This is called an “exothermic” reaction.
In an “endothermic” reaction, water temperature decreases as the chemicals absorb energy.
We will see that heat is measured in Joules (J) or kiloJoules (kJ). Before we do any heat calculations, you should know several terms …

17. Specific heat capacity
The heat needed to  the temperature of 1kg of a substance by 1 C. Symbol: c, units: J/(gC).
Heat capacity
The heat needed to  the temperature of an object by 1 C. Symbol: C (=c x m), units: J/C
Heat of reaction
The heat released during a chemical reaction. Symbol: none, units: J.
Specific heat (of reaction)
The heat released during a chemical reaction per gram of reactant. Symbol: h, units: J/g.
Molar heat of reaction
The heat released during a chemical reaction per mole of reactant. Symbol: H, units: J/mol.

18. Heat Calculations Sample problem: water = 4180 J/Kg/K
To determine the amount of heat a substance produces or absorbs we often use ∆H = c X m X T(t2-t1)
∆H: heat in J,
c: specific heat capacity in J/Kg/K, must change to joule
m: mass in kg, must change to kg
T: temperature change in K,
Sample problem: water = 4180 J/Kg/K
When 12 g of a food was burned in a calorimeter, the 100 cm3 of water in the calorimeter changed from 20C to 33C. Calculate the heat released.
∆H =c.m.T
= 4.18 J/(gC)
x 0.1Kg
x 13K
= 5.4 kJ