1. Heat and Enthalpy changes.

2. Heat absorbed by the reactants in endothermic reactions becomes the added energy of the products
The heat released in an exothermic reaction is the extra energy that reactants have compared to the energy of the products
The heat absorbed or released in a reaction depends on the difference in a quantity called enthalpy
Enthalpy is represented by H

3. The energy change in a reaction is very close to the enthalpy change but we use this new term because it takes into account more than just the change in heat. There are also kinetic changes due to pressure as well if the pressure remains constant then these 2 would be the same – but this is not what happens in everyday experiments so scientists use enthalpy

4. Enthalpy change , ∆ H
Enthalpy change for a chemical reaction is equal to the heat absorbed or gained during the reaction
The enthalpy change for a reaction is written as the enthalpy of the products minus the enthalpy of the reactants.
Enthalpy changes are represented by the symbol ∆H read delta H ∆ is used to mean a change or a difference
∆ H = H products - H reactants
Look at diagrams on pg 385

5. The heat absorbed in the products compared with the enthalpy of the reactants
H products is greater than H reactants so ∆H for an endothermic reaction is always positive
In an exothermic reaction the enthalpy is negative
Figure 12-5 summarizes the relationship between the sign of ∆H and the direction of heat flow in the chemical process

6. The amount of heat that a reaction absorbs or releases depends on the conditions in which it is carried out
Conditions like temp, pressure and physical states of the reactants and products.
To make comparing enthalpy easier chemists chose 1 atmosphere as standard pressure and conventionally use 25o C as the temp for reporting enthalpy changes. In addition and important condition is that the reactants and products are in their standard states ..

7. The standard state is the most stable form of the element under standard conditions C = graphite ( more stable than diamond )
The enthalpy change that is measured when reactants in their standard states change to products in their standard states is called - standard enthalpy change = ∆Ho

8. So how can a reaction like the combustion of propane be carried out at 25oC as soon as the reaction starts it will raise the temp - The answer is that if the reaction occurs at a temp other than 25oC then the standard enthalpy change takes into account the heat involved in restoring the products to standard conditions after the reaction is complete

9. C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4H2O (g) + 2043 kJ
Could be written instead As
C3H8 (g) O2 (g)  3 CO2 (g) H2O (g)
ΔHo = kJ

10. C (s) + H2O (g) + 113 kJ  CO (g) + H2 (g)
Could be written as
C (s) H2O (g)  CO (g) + H2 (g)
ΔHo = kJ

11. Using Enthalpy changes
Chemistry problems using enthalpy are similar to Stoichiometry problems from ch 11
The amount of heat absorbed or released in a reaction depends on the number of moles of reactants involved
For example according to the equation above 113 kJ is absorbed when 1 mole of C reacts with 1 mole of water

12. How much heat will be absorbed if 2 moles of C combine with 2 moles of water ( 226kJ)
C (s) H2O (g)  CO (g) + H2 (g)
ΔHo = kJ

13. The enthalpy change for a reaction is proportionately smaller or larger depending on the quantities of the reactants and products involved.

14. How much heat will be released if 1
How much heat will be released if 1.0 g of Hydrogen peroxide decomposes in a bombardier beetle to produce a steam spray?
2 H2O­2 (l)  2 H2O (l) + O2 (g) ΔHo = kJ

15. First you must convert the grams you have been given into moles
1.0 g H2O2 ] 1 mol H2O = mol H2O2
] g H2O2
Then use the conversion factor you are given in the problem
2 moles of H2O2 = kJ
1.0 g H2O2 ] 1 mol H2O2 ] kJ = kJ
] g H2O2 ] 2 mol H2O2
The minus sign tells you that heat is released in the reaction