Determining Heats of Reaction Experimentally (Calorimetry)
- how are the heats of reaction determined?
- a method to measure the amount of energy consumed or released when a reaction occurs must be
devised
Calorimetry
the measurement of heat changes in reactions
uses a device called a calorimeter – a device that will absorb or release heat from/to the reaction
types of calorimeters:
1. Simple (solution) calorimeter
- made from a styrofoam cup (or two stacked) filled with H 2O(l)
- placing a lid on the top will improve the “isolatedness” of the system
- must assume any heat from the reaction is absorbed by the water – not by the cup
- as long as there is not too much T, the results are fairly accurate
- can also be constructed from a tin (or aluminum) can –since metals readily absorb and
transfer heat energy, you can’t assume no heat is absorbed by the calorimeter, and its
specific heat capacity and energy absorbed/released MUST be included in the calculation
- used for heats of solution, fusion, neutralization, reaction (double replacement, single replacement),
specific heat capacity determination
2. Flame calorimeter
- used to measure heast of combustion (the amount of energy released when a mole of a
substance is burned)
- also used for “heats of combustion” for foods – “calorie content”
- the fuel is burned under a can, and heats the water in the can (** the can cannot be
excluded from the calorimeter – the amount of heat absorbed by the calorimeter = the
amount of heat absorbed by the water + the amount of heat absorbed by the can)
3. Bomb calorimeter
- provides as close to an isolated system as possible, containing all reactants and products in a sealed
container that is well insulated from the outside environment
- usually used to measure heats of combustion, but could also be used for other heats of reaction
4. Human calorimeter
- used to determine the amount of energy expended in physical exercise (used to predict calorie
consumption so it can be included with exercise equipment, etc.)
- in a closed, insulated room, have a person perform activities, measure the change in temperature
Calorimetry Problems
All calorimetry problems apply the Law of Conservation of Energy
Heat Lost = - Heat Gained
*** notice the negative sign (since heat lost + heat gained = 0, one is equal to the negative of the other)
to do calormetry problems, always divide them into the two parts
reaction side
H – combustion, reaction, solution, fusion…
mc T – for specific heat capacity or mixture problems
calorimeter side
ALWAYS mc T (or c T if the capacity of the
calorimeter doesn’t include the mass)
** may have two steps if the calorimeter is made of
metal
Chemistry 30 Notes – Unit 1 Part 1
page 1
-
the only connection between the two sides of the calorimeter is the amount of heat transferred.
i.e. – if the reaction side lost 1.9 kJ (Q= -1.9 kJ), the calorimeter side must have gained 1.9 kJ (Q=+1.9
kJ)
*** PLEASE DON’T FORGET THAT IF THE REACTION IS ENDOTHERMIC (gains energy), IT HAD TO GET
THAT ENERGY FROM THE CALORIMETER (so the calorimeter loses energy – is EXOTHERMIC)
ex: (calculating the heat of fusion) A 70.0 g sample of cesium is sealed in a glass vial and is lowered into 250
mL of water at 90.00C. When the cesium had melted, the temperature of water had fallen to 88.98C.
Determine the molar heat of fusion (Hfus) for cesium.
-
separate the parts of the calorimeter and reaction
find Q for whichever side of the question you have enough information for
determine what is asked for in the question
reaction side (cesium)
m = 70.0 g
Hfus=?
molar mass = 132.91 g/mol
calorimeter side (water only)
m=250 mL (250 g)
c = 4.19 J/gC
T = -1.02C
You have enough info
to solve the calorimeter
side first
Q  mc T
Q=+1.07kJ (gained)
4.19 J
)( 1.09 C)
gC
Q  1.07kJ(lost )
Q  (250 g)(
Hfus  n  Q
Q
n
132 .91g
Hfus  ( 1.07kJ)(
)( 70 .0g)
mol
Hfus  2.03kJ / mol
Hfus 
Once you have Q, change the sign
(heat lost by the calorimeter is
gained by the reaction) and move it
to the reaction side.
ex: What is the heat of combustion of an unknown fuel (in J/g) if burning 0.75 g of fuel increased the
temperature of 175 mL of water in a 12.0 g metal calorimeter (c=0.700 J/gC) from 10C to 45C?
reaction side (combustion)
m = 0.75 g
Hcomb=?
Q=-26kJ (lost)
If the water gained energy
(endothermic), then the reaction
must be exothermic
Hcomb  m  Q
Q
m
Hcomb  ( 26kJ)(
Hcomb 
1
)
0.75 g
Hcomb  35kJ / g
Chemistry 30 Notes – Unit 1 Part 1
Heat of combustion is
usually measured in kJ/mol,
but if the fuel is unknown
(or for complex foods),
comb is usually in kJ/g
or kJ/serving
calorimeter side (water & metal)
m=175 mL (175 g)
c = 4.19 J/gC
T = 35C
m=12.0 g
c = 0.700 J/gC
T = 35C
Q  mc T  mc T
4.19 J
0.700 J
Q  (175 g)(
)( 35 C)  (12 .0g)(
)( 35 C)
gC
gC
Q  26kJ(gained )
You have enough info to solve the
calorimeter side first – the
calorimeter consists of the water &
the metal can
page 2
ex: (Finding the final temperature of a mixture) If 75 mL of water with an initial temperature of 100C is mixed
with 135 mL of water with an initial temperature of 30C, find the final temperature
reaction side (water - mcT)
m = 75 mL (75 g)
c = 4.19 J/gC
Ti= 100C
You don’t have enough information
to solve for Q on either the
calorimeter or the reaction side, so
you have to use the original equation
Q lost = - Q gained
If the two parts of the mixture
are the same substance, the
specific heat capacities(c)
cancel making the math much
simpler!
calorimeter side (water - mcT)
m=135 mL (135 g)
c = 4.19 J/gC
Ti = 30C
Qlost  Qgained
mc (Tf  Ti )  mc (Tf  Ti )
4.19 J
4.19 J
(75 g)(
)( Tf  100 C)  (135 g)(
)( Tf  30 C)
gC
gC
75(Tf  100 C)  135 (Tf  30 C)
75 Tf  7500 C  135 Tf  4050 C
210 Tf  11550 C
Tf  55 C
Be careful with your signs –
if you get a final
temperature that doesn’t
make sense, check that you
didn’t miss the negative
sign!
ex: When solutions of an acid and a base are mixed, heat is released. Using a coffee cup calorimeter, 100 mL
of 1.00 mol/L hydrochloric acid are mixed with 100 mL of 1.00 mol/L sodium hydroxide solution. The initial
temperature of the solutions was 22.6C, and the temperature after mixing was 29.5C. Calculate the molar
enthalpy of solution for the reaction:
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
reaction side (neutralization)
Hneut=?
Q=-5.78kJ (lost)
If the water gained energy
(endothermic), then the reaction
must be exothermic
calorimeter side (water)
m=200mL (200g)
c = 4.19 J/gC
T = 6.9C
Hneut  n  Q
Q  mc T
Q
Hneut 
n
Q  (200 g)(
1L
)( 0.100L )
1.00mol
Hneut  57 .8kJ / mol
Hneut  ( 5.78kJ)(
The mass of the water is the mass of the
solutions – a solution is made of two
parts – the solvent (the calorimeter) and
the solutes – the reactants
4.19 J
)( 6.9C)
gC
Q  5.78kJ(gained )
Use the concentration and
volume of the solution to
find moles – the molar
enthalpy for this reaction
will be per mol of NaOH (or
equally, per mol of HCl)
Chemistry 30 Notes – Unit 1 Part 1
page 3