Introduction to enthalpy and calorimetry

1. Lesson
Exothermic and
Endothermic Reactions
and Calorimetry
IB Chemistry Power Points
Topic 05
Energetics
www.pedagogics.ca

2. Great thanks to
JONATHAN HOPTON & KNOCKHARDY PUBLISHING
www.knockhardy.org.uk/sci.htm
Much taken from
ENTHALPY
CHANGES

3. Background and Review
First Law of Thermodynamics (Law of Energy Conservation)
Energy can be neither created nor destroyed but it can be converted from one
form to another
Energy Changes in Chemical Reactions
All chemical reactions are accompanied by some form of energy change
Exothermic Energy is given out
Endothermic Energy is absorbed
Examples Exothermic combustion reactions
neutralization (acid + base)
Endothermic photosynthesis
thermal decomposition of calcium carbonate

4. The heat content of a chemical system is called
enthalpy (represented by H).
Key Concept
We cannot measure enthalpy directly, only
the change in enthalpy ∆H i.e. the amount of
heat released or absorbed when a chemical
reaction occurs at constant pressure.
∆H = H(products) – H(reactants)
∆Ho (the STANDARD enthalpy of reaction) is the
value measured when temperature is 298 K and
pressure is 100.0 kPa.

5. If ∆H is negative, H(products) < H(reactants)
There is an enthalpy decrease and heat is
released to the surroundings.
Enthalpy Diagram -Exothermic Change
enthalpy

6. If ∆H is positive, H(products) > H(reactants)
There is an enthalpy increase and heat is
absorbed from the surroundings.
Enthalpy Diagram - Endothermic Change
enthalpy

7. Exothermic reactions release heat
Example: Enthalpy change in a chemical reaction
N2(g) + 3H2(g)  2 NH3(g)
∆H = -92.4 kJ/mol
The coefficients in the
balanced equation
represent the number of
moles of reactants and
products.

8. N2(g) + 3H2(g)  2 NH3(g)
∆H = -92.4 kJ/mol
State symbols are ESSENTIAL as changes of state
involve changes in thermal energy.
The enthalpy change is directly proportional to the
number of moles of substance involved in the reaction.
For the above equation, 92.4 kJ is released
- for each mole of N2 reacted
- for every 3 moles of H2 reacted
- for every 2 moles of NH3 produced.

9. The reverse reaction
2 NH3(g)  N2(g) + 3H2(g)
∆H = +92.4 kJ/mol
Note: the enthalpy
change can be read
directly from the
enthalpy profile diagram.

10. Thermochemical Standard Conditions
The ∆H value for a given reaction will depend
on reaction conditions.
Values for enthalpy changes are standardized :
for the standard enthalpy ∆Ho
-Temperature is 298 K
- Pressure is 100 kPa
- All solutions involved are 1 M concentration

11. Calorimetry – Part 1
Specific Heat Capacity
The specific heat capacity of a substance is a physical
property. It is defined as the amount of heat (Joules)
required to change the temperature (oC or K) of a unit
mass (g or kg) of substance by ONE degree.
Specific heat capacity (SHC) is measured in
J g-1 K-1 or kJ kg-1 K-1 (chemistry)
J kg-1 K-1 (physics)

12. Calorimetry – continued
Heat and temperature change
Knowing the SHC is useful in thermal chemistry. Heat
added or lost can be determined by measuring
temperature change of a known substance (water).
Q = mc∆T
heat = mass x SHC x ∆Temp

13. Calorimetry – Part 2
Applications
A calorimeter is used to
measure the heat
absorbed or released in
a chemical (or other)
process by measuring
the temperature change
of an insulated mass of
water.

14. Sample Problem 1
When 3 g of sodium carbonate are added to 50 cm3 of
1.0 M HCl, the temperature rises from 22.0 °C to 28.5 °C.
Calculate the heat required for this temperature change.

15. Sample problem 2: 50.0 cm
3
of a 1.00 mol dm
-3
HCl solution is mixed with 25.0
cm
3
of 2.00 mol dm
-3
NaOH. A neutralization reaction occurs. The initial
temperature of both solutions was 26.7
o
C. After stirring and accounting for
heat loss, the highest temperature reached was 33.5
o
C. Calculate the enthalpy
change for this reaction.
NaOH HCl
both 26.7o
.
26.7o  33.5o

16. After writing a balanced equation, the molar quantities and
limiting reactant needs to be determined.
Note that in this example there is exactly the right amount of
each reactant. If one reactant is present in excess, the heat
evolved will associated with the mole amount of limiting reactant.
Sample problem 2: 50.0 cm
3
of a 1.00 mol dm
-3
HCl solution is mixed with 25.0
cm
3
of 2.00 mol dm
-3
NaOH. A neutralization reaction occurs. The initial
temperature of both solutions was 26.7
o
C. After stirring and accounting for
heat loss, the highest temperature reached was 33.5
o
C. Calculate the enthalpy
change for this reaction.

17. Next step – determine how much heat was released.
There are some assumptions in this calculation
- Density of reaction mixture (to determine mass)
- SHC of reaction mixture (to calculate Q)
Sample problem 2: 50.0 cm
3
of a 1.00 mol dm
-3
HCl solution is mixed with 25.0
cm
3
of 2.00 mol dm
-3
NaOH. A neutralization reaction occurs. The initial
temperature of both solutions was 26.7
o
C. After stirring and accounting for
heat loss, the highest temperature reached was 33.5
o
C. Calculate the enthalpy
change for this reaction.

18. Final step – calculate ΔH for the reaction
Sample problem 2: 50.0 cm
3
of a 1.00 mol dm
-3
HCl solution is mixed with 25.0
cm
3
of 2.00 mol dm
-3
NaOH. A neutralization reaction occurs. The initial
temperature of both solutions was 26.7
o
C. After stirring and accounting for
heat loss, the highest temperature reached was 33.5
o
C. Calculate the enthalpy
change for this reaction.

19. Sample problem 3: to determine the enthalpy of combustion for ethanol (see
reaction), a calorimeter setup (below) was used. The burner was lit and allowed
to heat the water for 60 s. The change in mass of the burner was 0.518 g and
the temperature increase was measured to be 9.90
o
C.
What is the big
assumption made
with this type of
experiment?

20. First step – calculate heat evolved using calorimetry
Last step – determine ΔH for the reaction
Sample problem 3: to determine the enthalpy of combustion for ethanol (see
reaction), a calorimeter setup (below) was used. The burner was lit and allowed
to heat the water for 60 s. The change in mass of the burner was 0.518 g and
the temperature increase was measured to be 9.90
o
C.

21. Sample problem 4: 100.0 cm
3
of 0.100 mol dm
-3
copper II sulphate solution is
placed in a styrofoam cup. 1.30 g of powdered zinc is added and a single
replacement reaction occurs. The temperature of the solution over time is
shown in the graph below. Determine the enthalpy value for this reaction.
First step
Make sure you understand
the graph.
Extrapolate to determine
the change in
temperature.
The extrapolation is necessary to compensate for heat loss while the reaction
is occurring. Why would powdered zinc be used?

22. Determine the limiting reactant
Calculate Q
Calculate the enthalpy for the reaction.
Review Exercise 2
Sample problem 4: 100.0 cm
3
of 0.100 mol dm
-3
copper II sulphate solution is
placed in a styrofoam cup. 1.30 g of powdered zinc is added and a single
replacement reaction occurs. The temperature of the solution over time is
shown in the graph below. Determine the enthalpy value for this reaction.