Chapter 12: Heat in Chemical
Reactions
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
• Thermochemistry – study of the changes
in heat in a chemical reaction.
• Exothermic reactions – reactions that
RELEASE heat
• Endothermic reactions – reactions that
ABSORB heat
Exothermic reactions:
• Burning of a camp stove
• C3H8 + 5O2  3CO2 + 4H2O + 2043 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.
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.
Calculations
• Enthalpy change ∆H
• ∆H = Hproducts – Hreactants
• Standard Temp and pressure
• 1 atmosphere is p 25C is T
• Sign ∆H
• +
• -
Process
Endo
Exo
Heat
Absorbed
Released
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.
• 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 = 0.029 mol H2O2
•
34 g
• 0.029 mol H2O2 x -190kJ = -2.8KJ
2 mol H2O2
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.
•
•
•
•
N2 + 2O2  2NO2
N2 + O2  2NO ∆H = +181KJ
2NO + O2  2NO2 = ∆H2 = -113KJ
N2 + 2O2 + 2NO  2NO + 2NO2
• ∆Hnet = ∆H1 + ∆H2
• ∆H = 181 + -113.
• ∆H = 68KJ
• 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.
• Calculate ∆H for the reaction that
produces SO2
• S + O2  SO2
• 2SO2 + O2  2SO3 ∆H = -196KJ
• 2S + 3O2  2SO3 ∆H = -790 KJ
•
•
•
•
•
•
•
•
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
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
• Specific heat – the heat capacity of 1 gram
of a substance
•
Water = 4.184 J/g C
• To raise the temp of 1 gram of water 1
degree C you need 4.184J
• Determine ∆H for the rxn.
• NaOH  Na+ + OH• The calorimeter is filled with 75.0g water.
The initial temp is 19.8C. A 0.050-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
•
•
•
•
•
•
•
•
•
•
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
• 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-
•
•
•
•
•
1st calculate qsur
Qsur = m x C x ∆T
Qsur = 60.0g x 4.184J/gC x (16.9-21.0)
Qsur = -1.0 x 103 J
Qrxn = -qsur = + 1.0 x 103 J/1000 = 1 KJ
• 4.25g NH4OH x 1 mole NH4OH
•
• = + 8.5KJ
1.0 KJ
35 g NH4OH 0.121mNH4OH