CALORIMETRY
calorimeter – device used to measure the amount of
heat absorbed or released during a
chemical or physical process
Law of Conservation of Energy – in any physical or chemical
process energy is neither created nor destroyed,
it is conserved
(all energy is work, heat, or stored energy)
heat lost = heat gained or
(m lost x C lost x
q (lost) = – q (gained)
T lost) = – (m gained x C gained x
Tgained)
Ex 1: If 10 g of aluminum at 109oC is added to 50 g water at 24.0oC, the
temperature of the final mix is 27.5oC. What is the specific heat of Al?
aluminum
water
m Al =10 g
TAl = Tf – Ti = – 81.5oC
CAl = ?
m H2O = 50 g
TH2O = Tf – Ti = 3.5oC
CH2O = 1cal/goC
heat lost by Al = heat gained by H2O:
(m Al x C Al x
q Al = – q H2O
T Al) = – (mH2O x C H2O x
TH2O)
(10 g) (C Al) (– 81.5 oC) = – (50 g) (1 cal/g oC) (3.5oC)
(– 815 g oC) (C Al) = – 175 cal so C Al = .215 cal/goC
Ex 2: If 250 g of silver at 50.0oC is added to 125 g of water at 23.0oC
What is the final temperature of the mixture? (C Ag = 0.0566 cal/goC)
Ag:
m Ag = 250 g
Ti Ag = 50.0oC
C Ag = 0.0566 cal/goC
qAg = - qH2O or mAg x CAg x
H2O:
m H2O =125 g
Ti H2O = 23.0oC
C H2O = 1 cal/goC
TAg = – (mH2O x CH2O x
TH2O )
250 g x 0.0566 cal/goC x (Tf – 50oC) = – 125 g x 1cal/goC x (Tf – 23oC)
(14.15 cal/oC) Tf - 707.5 cal = -125 cal/oC) Tf + 2875 cal
- 3582.5 cal = -139.15 cal x Tf
- 3582.5 cal/-139.15 cal = Tf = 25.75oC
THERMOCHEMISTRY
enthalpy – heat content of a system measured at
constant pressure
enthalpy change – quantity of heat gained or lost
during a process at constant pressure
Heat of Reaction ( H) the heat absorbed or released
during a chemical reaction
H =
Hproducts -
Hreactants
exothermic reaction – releases energy in the form of heat;
H is negative
 heat appears on the products side of the equation

H reactants >
H products
endothermic reaction – absorbs or requires energy in
the form of heat;
H is positive
 heat appears on the reactants side of the equation

H products >
H reactants
thermochemical equation - a chemical equation that shows
the amount of heat produced or absorbed by a reaction
2 H2 (g) + O2 (g)  2 H2O (g)
H = - 483.6 kJ (bonds forming = exo.)
2 H2O (g)  2 H2 (g) + O2 (g)
H = + 483.6 kJ(bonds breaking = endo.)
molar heat of formation ( Hof) quantity of heat absorbed
or released when one mole of a compound is
formed from its elements at standard state
(1 atm & 25oC)
Based on 1 mole of products:
H2 (g) + ½ O2 kJ/mol (g)  H2O (l)
Hof = - 285.8 kJ
High “-”
Hof implies a substance is more stable
than its elements
High “+”
Hof implies a substance is less stable
than its elements
heat of combustion – quantity of heat released by the
complete combustion of 1 mole of a substance
Calculating Heats of Reaction
Enthalpy changes can be calculated in 2 ways:
A.
Using standard heats of formation :
Ho =
Hof (products ) -
Hof (reactants)
Ex 1: Use the heats of formation to find the enthalpy change for the
combustion of methane gas (CH4) to form carbon dioxide gas
and liquid water.
1. Write balanced equation:
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (l)
2. Look up standard heats of formation:
CH4 (g) = - 74.81 kJ
CO2 (g) = - 393.51 kJ
O2 (g) = 0 kJ
H2O (l) = -285.83 kJ
3. Sum the
Hof of the reactants and the products taking into
account the number of moles of each:
Ho =
Hof (products ) -
Hof (reactants)
= - 393.51 kJ + (2 mol) (- 285.83 kJ) – (- 74.81 kJ)
= - 393.51 kJ – 571.66 kJ + 74.81 kJ
= - 890.36 kJ
H is negative so rxn is exoth.
Practice: Calculate the heat of reaction (enthalpy change) for the
combustion of 2 moles of sulfur dioxide gas into 2 moles of
sulfur trioxide gas.
1. Write balanced equation: 2SO2 (g) + O2 (g)  2SO3 (g)
2. Look up standard heats of formation:
SO3 (g) = - 395.72 kJ
3. Sum the
Ho =
SO2 (g) = - 296.83 kJ
Hof of the reactants and the products
Hof (products ) -
Hof (reactants)
= (2 mol) (- 395.72 kJ) – (2 mol) (- 296.83 kJ)
= - 791.44 kJ + 593.66 kJ
= -197.78 kJ
exothermic reaction
B. Using Hess’s Law of Heat Summation and a given set of equations:
Hess’s Law – the overall enthalpy change for a reaction is
equal to the sum of the enthalpychanges for
the individual steps in the process
Ex 2: What is the heat of reaction for the decomposition of solid ice to its
elements?
target equation: H2O (s)  H2 (g) + ½ O2 (g)
H = ? kJ
H2 (g) + ½ O2 (g)  H2O (l)
Hof = - 285.8 kJ
H2O (s)  H2O (l)
Hof =
Solution: H2O (s)  H2O (l)
6.0 kJ
Hof =
6.0 kJ
H2O (l)  H2 (g) + ½ O2 (g)
Hof = + 285.8 kJ
H2O (s)  H2 (g) + ½ O2 (g)
H
= + 291.8 kJ
Rules for manipulating thermochemical equations:
1. When an equation is reversed, the
Hof must also be reversed.
2. Formulas cancelled from both sides of the equation must be the
same substance in the same physical state.
3. If all the coefficients of an equation are multiplied or divided by
the same factor, Hof must also be changed by the same factor.
Practice: What is the enthalpy change for the formation silicon from
SiCl4 and Mg?
H = ? kJ
Target equation: 2Mg (s) + SiCl4 (l) Si (s) + 2 MgCl2 (s)
Si (s) + 2 Cl2 (g)  SiCl4 (l)
Hof = - 687 kJ
Mg (s) + Cl2 (g)  MgCl2 (s)
Hof = - 641 kJ
SiCl4 (l)  Si (s) + 2 Cl2 (g)
2x Mg (s) + Cl2 (g)  MgCl2 (s)
Hof = + 687 kJ
2( Hof = - 641) = - 1282 kJ
2Mg (s) + SiCl4 (l) Si (s) + 2 MgCl2 (s)
H = - 595 kJ