Heat in
Chemical
Reactions
10.1 Chemical Reactions
that Involve Heat
1. Heat: Energy (symbol - q) that is transferred
from one object to another due to a difference in
temperature. Measured in Joules (symbol - J)
2. Thermochemistry: The study of heat changes in
a chemical reaction.
3. Types of Chemical Reactions
a. Exothermic Reactions: release heat into their
surroundings. Heat is a product of the reaction and
temperature increases. This occurs during bond
formation.
surroundings
Exothermic
Reaction
ΔH =
-
ΔT = +
Combustion reactions are exothermic: burning
propane
C3H8 + 5O2 → 3CO2 + 4H2O + 2043 kJ
Exothermic Reaction
b. Endothermic Reactions: Heat is absorbed by the
reactants and stored in the chemical bonds of the
products. Heat acts as a reactant and temperature
decreases. This occurs during bond breaking.
Electrolysis of water requires electrical energy.
2H2O + 572kJ → 2H2 + O2
surroundings
Endothermic
Reaction
ΔH=+
ΔT= -
Exothermic/Endothermic
Reactions
10.2 Heat and Enthalpy Changes
1. Enthalpy: The heat content of a system at
constant pressure (symbol is H ).
2. Enthalpy Change: The heat absorbed or released
during a reaction (symbol is ΔH ).
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4. When reactions take place at standard temperature and
pressure, q = H.
5. Standard Enthalpy Change (H): Enthalpy change that
occurs when reactants in their standard states (most stable
form) change to products in their standard states. STP
Standard Temperature and Pressure are: 0 C and 1 atm.
The H is listed after the equation. If the H is positive
the reaction is endothermic and heat was absorbed. If the
H is negative, the reaction is exothermic and heat was
released.
6. The amount of heat absorbed or released in a reaction
depends upon the number of moles of reactants.
7. Enthalpy Changes in Stoichiometry Problems:
Ex) How much heat will be released if 5.0 g of H2O2
decomposes?
2H2O2 → 2H2O + O2
ΔH = -190 kJ
 1 mol H 2O2  -190 kJ 
5.0 g of H 2O2 

 = -14 kJ or 14 kJ released
 34 g of H 2O2  2 mol H 2O2 
Ex) How much heat is transferred when you eat a
10. g Jolly rancher which is made of glucose
(C6H12O6)? It reacts in your body with oxygen
according to the following equation. If 4.184kJ = 1
Cal, how many Calories are in the Jolly Rancher?
C6H12O6 + 6O2 → 6CO2 + 6H2O ΔH = -2803 kJ
 1 mol C6 H12O6  -2803 kJ 
 1 Cal 
10. g of C6 H12O6 
=
-160
kJ



 = 38 Cal
 4.184 kJ 
 180 g of C6 H12O6  1 mol C6 H12O6 
C6H12O6 + 6O2 → 6CO2 + 6H2O + 2803 kJ
3. Enthalpy Diagrams:
#1
#2
#1
#2
a. Which has a higher enthalpy? Products or Reactants
R
P
b. Was heat absorbed or released?
R
A
Exo
Endo
d. Is ΔH for this reaction positive or negative?
-
+
e. Would the ΔH be on the left or right side of the yield sign?
R
L
Endo
Exo
c. Is this an endothermic or exothermic reaction?
f. Is the reverse reaction exothermic or endothermic?
g. Rewrite each equation with the heat term
in the reaction as a reactant or product:
#1) C3H8 + 5O2 → 3CO2 + 4H2O + 2043 kJ
#2) C + H2O + 113kJ → CO + H2
 = “change in”
H = Hproducts ─ Hreactants
H
Hproducts
Exothermic
Reaction
-
low high
Endothermic
Reaction
+
high low
Hreactants
10.3 Hess’s Law - (1802-1850)
1. The enthalpy change for a reaction is the sum of
the enthalpy changes for a series of reactions that
adds up to the overall reaction.
2. This is also called the Law of Heat of
Summation (Σ)
3. This allows you to determine the enthalpy change
for a reaction by indirect means when a direct
method cannot be done.
4. Steps for using Hess’s Law
1. Identify the compounds
2. Locate the compounds on the Heat of Reaction Table.
3. Write the reaction from the table so the compound is a
reactant or product.
4. Write appropriate ΔH for each ‘sub equation.”
• If needed, multiply equation and enthalpy change value. (coefficients)
• If you reverse the equation, change the sign of the enthalpy change.
5. Add the equations to arrive at the desired net (original)
equation.
6. Add ΔH (enthalpy changes) of each “sub equation.”
10.3 Calorimetry:
1. The Kinetic Theory states that heat results from
the motion & vibration of particles.
2. Heat: The transfer of kinetic energy from a hotter
object to a colder object. Heat is dependent on
composition and amount.
3. Temperature is a measure of how hot or cold
something is; specifically it is the measure of the
average kinetic energy (speed) of the particles in
an object. It is independent of amount.
Heat
• Temperature is not the same as heat.
• Temperature is a measure of the average kinetic
energy of the particles in an object.
• A temperature change is a result of a energy
transfer.
• Julius Sumner Miller - Physics - Heat & Temperature
• Heat vs. Temperature Animation
4. Calorimetry is the study of
heat flow and measurement.
5. Calorimetry experiments
determine the heats of
reactions (enthalpy changes)
by making accurate
measurements of temperature
changes produced in a
calorimeter.
6. A Calorimeter is an
insulated device used to
measure heat absorbed or
released in a chemical or
physical change.
7. Specific Heat (Cp): The amount of heat needed to
raise 1 g of a substance by 1C.
Formula for specific heat:
q
Cp =
mΔT
m=mass (substance)
T=change in temperature of the substance (Tf─Ti)
Specific Heat of Water = 4.184 J/g ºC
= 1 calorie or .001food Calorie
8. Measuring Specific Heat of a Metal:
Ex #1) What is the specific heat of a nickel if the
temperature of a 32.2 g sample of nickel is increased
by 3.5ºC when 50. J of heat is added.
q
+50.J


Cp =
+
mΔT  32.2g   3.5 C 
0.44 J/g C
Ex #2) How much heat is absorbed to be able to
increase the temperature of a 26.2 g sample of
aluminum (Cp = 0.897 J/gºC) from 25.3 ºC to 65.9 ºC?
q = mC pΔT
q = .897 J/g C   26.2g   40.6 C 
o
q = +954 J
+
o
9. Measuring Heat (q) of a Substance Dissolved in Water:
You can rearrange this formula to determine the heat released or
absorbed by the surroundings (solution) as the substance dissolves
based on this assumption:
q reaction = -q surroundings
ΔT = T - T
f
i
q = mCpΔT
1) Calculate q for the surroundings (solution) and determine qrxn.
2) Calculate the moles of solute dissolved in the water.
3) Calculate H = heat (q)
mol
Ex) When a 4.25 g sample of solid NH4NO3 dissolves
in 60.0 g of H2O in a calorimeter,the temperature
drops from 21.0 ºC to 16.9 ºC. Calculate H.
Rewrite the thermochemical equation with the heat
term as a reactant or product.
NH4NO3(s) + 21 kJ → NH4+(aq) + NO3-(aq) H = ?
1.
2.
qsur=mCpΔT = (64.3g)(4.184J/g°C)(-4.1°C)
qsur=-1100J
qrxn = -qsur = +1100J
 1 mol NH 4 NO3 
4.25 g NH 4 NO3 
 = 0.0531 mol
80.0
g
NH
NO
4
3

3. ΔH = q = +1100J = + 21,000 J/mol = +21 kJ/mol
mol
.0531 mol
10. Foods as Fuels:
A. Carbohydrates typically have high enthalpies;
however, the products of their combustion, CO2 and
H2O, have low enthalpies.
B. Therefore, the combustion of carbohydrates and
fats, is exothermic.
C. Sugars and Starches break down to glucose, which
reacts with O2 in a combustion reaction.
D. Nutritional information on food labels can be
gathered using a calorimeter.
Carbs = 4 Cal/g Protein = 4 Cal/g Fat = 9 Cal/g