1. Chapter 12: Heat in Chemical Reactions

2. 12-1 Chemical Reactions that Involve Heat
Chemical reactions involve breaking or making bonds and rearranging atoms.
Breaking bonds requires energy and making bonds releases energy.
Heat- the energy that is transferred from one object to another due to a difference in temperature

3. Thermochemistry – study of the changes in heat in a chemical reaction.
Exothermic reactions – reactions that RELEASE heat
Endothermic reactions – reactions that ABSORB heat

4. Exothermic reactions:
Burning of a camp stove
C3H8 + 5O2  3CO2 + 4H2O kJ
2043 kJ of heat are released when 1 mole of C3H8 is burned.
The energy released during forming new bonds is greater than the energy required to break the old bonds.

5. Endothermic Reactions:
A process in industry fuel called water gas, passing steam over hot coals
C + H2O + 113KJ  CO + H2
The energy released as new bonds are formed in the products is less than the energy required to break the bonds in the reactants. This energy must be provided for the reaction to take place. The energy provided is stored in the bonds of the products.

6. Calculations Enthalpy change ∆H ∆H = Hproducts – Hreactants
Standard Temp and pressure
1 atmosphere is p 25C is T
Sign ∆H Process Heat
+ Endo Absorbed
- Exo Released

7. 12-2 Heat and Enthalpy Changes
Enthalpy – the heat absorbed or gained during a chemical reaction. Accounts for the kinetic energy plus the volume and temperature of the substance.
The difference between energy and enthalpy is very small. When the pressure remains constant, the heat absorbed or released during a chemical reaction is equal to the enthalpy change for the reaction.

8. How much heat will be released if 1
How much heat will be released if 1.0g of H2O2 decomposes in a beetle to produce the steam spray
2H2O2  2H2O + O2 ∆H = -190KJ
1.0g H2O2 x 1mol H2O2 = mol H2O2
34 g
0.029 mol H2O2 x -190kJ = -2.8KJ
2 mol H2O2

9. 12-3 Hess’s Law
Allows you to find enthalpy changes of reactions that cannot be done directly
Hess’s Law- if a series of reactions are added together, the enthalpy change for the net reaction will be the sum of the enthalpy changes for the individual steps.
Consider the haze in a large city.

10. N2 + 2O2  2NO2
N2 + O2  2NO ∆H = +181KJ
2NO + O2  2NO2 = ∆H2 = -113KJ
N2 + 2O2 + 2NO  2NO + 2NO2
∆Hnet = ∆H1 + ∆H2
∆H =
∆H = 68KJ

11. Rules for applying Hess’s Law
Multiply the enthalpy change by the coefficient of the substance.
If the equation is reversed, so is the sign of ∆H.

12. Calculate ∆H for the reaction that produces SO2
S + O2  SO2
2SO2 + O2  2SO3 ∆H = -196KJ
2S + 3O2  2SO3 ∆H = -790 KJ

13. 1st switch the 1st reaction.
2SO3  2SO2 + O2 = ∆H = 196KJ
2S + 3O2  2SO3 ∆H = -790KJ
You’re left with
2S + 2O2  2SO2
∆H = (+196KJ) + (-790KJ)
∆H = -594KJ
HOWEVER, the equation is only ½ of this so you have to
∆H = ½ (-594 KJ) = -297 KJ

14. 12-4 Calorimetry
Calorimetry- the study of heat flow and heat measurement
Heat capacity – the amount of heat needed to raise the temperature of the object by 1 Celsius degree.
For example, the heat capacity of a cup of water at 18o C is the number of joules needed to make it 19.
Depends on mass and composition

15. Specific heat – the heat capacity of 1 gram of a substance
Water = J/g C
To raise the temp of 1 gram of water 1 degree C you need 4.184J

16. Determine ∆H for the rxn.
NaOH  Na+ + OH-
The calorimeter is filled with 75.0g water. The initial temp is 19.8C. A mole sample of solid NaOH is added and the temp increases to 26.7 C.
the temp increased so its exothermic.
The heat lost by the sodium is gained by the water. q is used to show heat.
so we can say qrxn = -qsur

17. qsur = m x C x (Tf – Ti)
m = mass of water
C = specific heat
temp change = T – T
qsur = (75.0g)(4.184J/gC)(26.7 – 19.8)
qsur = +2170J (J to KJ)
qrxn = -qsur = -2.17kJ
∆H = -2.17kJ
0.050 mole NaOH
-43kJ

18. When a 4. 25g sample of solid NH4OH. dissolves in 60
When a 4.25g sample of solid NH4OH dissolves in 60.0g of water, the temperature drops from 21.0C to 16.9C solve ∆H
NH4NO3  NH4+ + NO3-

19. 1st calculate qsur Qsur = m x C x ∆T
Qsur = 60.0g x 4.184J/gC x ( )
Qsur = -1.0 x 103 J
Qrxn = -qsur = x 103 J/1000 = 1 KJ
4.25g NH4OH x 1 mole NH4OH KJ
35 g NH4OH mNH4OH
= + 8.5KJ