• Chemical reactions involves the breaking and the making of bonds.

• Energy is needed to break down a bond.• Energy is released when a bond is formed.• If more energy is released than is absorbed, the

reaction will be Exothermic.• If more energy is needed to break the bonds than

is given when new bonds are formed, the reaction will be endothermic.

Energy level

Reaction progress

Activation energy

Energy given out

by reactionUsing a catalyst might lower the

activation energy

Enthalpy

• The energy contained in a chemical bond that can be converted into heat is known as enthalpy.

• Enthalpy is given the symbol H.• Enthalpy can not be measured

directly, but we can measure the enthalpy change in a reaction, ΔH

• A process is endothermic when H is positive.

• A process is endothermic when H is positive.

• A process is exothermic when H is negative.

• EXOTHERMIC – more energy is given out than is taken in (burning, respiration)

• ENDOTHERMIC – energy is taken in but not necessarily given out. (photosynthesis)

• H changes sign when a process is reversed. • Therefore, a cyclic process has the value H = 0.

Chapter 6: Thermochemistry 11

Same magnitude; different signs.

Using H

12

Values are measured experimentallyNegative values indicate exothermic reactionsPositive values indicate endothermic reactionsChanges sign when a process is reversed.

Therefore, a cyclic process has the value H = 0For problem-solving, one can view heat being absorbed in an endothermic reaction as being like a reactant and heat being evolved in an exothermic reaction as being like a product

• We measure heat flow using calorimetry.• A calorimeter is a device used to make

this measurement.• A “coffee cup” calorimeter may be used

for measuring heat involving solutions.

13

A “bomb” calorimeter is used to find heat of combustion; the “bomb” contains oxygen and a sample of the material to be burned.

14

Specific heat of a substance is the amount of heat required to raise the temperature of one gram by 1 o C or by 1 Kelvin.

Specific heat = C = q/mT units of C: J g–1 oC–1 or J g–1 K–

1

Chapter 6: Thermochemistry 15EOS

16

The standard enthalpy of reaction (Ho) is the enthalpy change for a reaction in which the reactants in their standard states yield products in their standard states

The standard enthalpy of formation (Ho

f) of a substance is the enthalpy change that occurs in the formation of 1 mol of the substance from its elements when both products and reactants are in their standard states

•Is the enthalpy change that occurs during the complete combustion of one mole of a substance.•The enthalpy of combustion is defined in terms one mole of reactants, where the enthalpy of formation is defined in terms of one mole of products.•the symbol Ho

C is used to represent standard enthalpy change os combustion.

Chapter 6: Thermochemistry 20EOS

The heat of a reaction is constant, regardless of the number of steps in the process

Hoverall = H’s of individual reactions

When it is necessary to reverse a chemical equation, change the sign of H for that reaction

When multiplying equation coefficients, multiply values of H for that reaction

H is well known for many reactions, and it is inconvenient to measure H for every reaction in which we are interested.

• However, we can estimate H using H values that are published and the properties of enthalpy.

The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants:

H = Hproducts − Hreactants

Hess’s law states that “If a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.”

• Imagine this as occurringin 3 steps:

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

C3H8 (g) 3 C(graphite) + 4 H2 (g)

3 C(graphite) + 3 O2 (g) 3 CO2 (g)

4 H2 (g) + 2 O2 (g) 4 H2O (l)

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

C3H8 (g) 3 C(graphite) + 4 H2 (g)

3 C(graphite) + 3 O2 (g) 3 CO2 (g)

4 H2 (g) + 2 O2 (g) 4 H2O (l)

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

• The sum of these equations is:

We can use Hess’s law in this way:

H = nHf(products) - mHf(reactants)

where n and m are the stoichiometric coefficients.

C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)H= [3(-393.5 kJ) + 4(-285.8 kJ)] - [1(-103.85 kJ) + 5(0 kJ)]

= [(-1180.5 kJ) + (-1143.2 kJ)] - [(-103.85 kJ) + (0 kJ)]= (-2323.7 kJ) - (-103.85 kJ)= -2219.9 kJ