Enthalpy and Calorimetry
Chapter 5 part 2
Enthalpy
• H is heat under
constant pressure
or
• H=qP
• H=E+PV
• And therefore
• ΔH= ΔE+P ΔV
• ΔH=Hfinal-Hinitial
Example:
• When 1 mole of methane is
burned at a constant pressure.
890kJ of energy is released as
heat. Calculate the ΔH for a
process in which a 5.8g sample of
methane is burned at constant
pressure.
Answer
• ΔH= -890kJ/mol
• CH4 is methane, MM = 16 g/mol
• 5.8g/16g mol-1 = 0.36 mol
• 0.36 mol CH4 x -890kJ/mol CH4 = -320kJ
Calorimetry
• The science of
measuring heat.
• The device used to
measure heat
changes associated
with chemical
reaction is a
calorimeter
Heat Capacity
• Specific heat capacity
• Heat capacity of an
is the heat capacity
object is the heat
per gram of a
absorbed by the change substance or
in temperature or
• Csp=C/gram
• C=(heat absorbed)/(ΔT)
• Molar heat capacity is
the heat capacity per
mole of a substance
or
• Cmol=C/mole
Calorimeters
• Constant pressure
calorimeter
• Calculations (per
gram)
• Csp* mass*ΔT= ΔH
Remember:
• the specific heat and
the mass are of the
same object.
• The measurement is
of the surrounding
water, not the actual
system so the sign is
reversed.
Example:
• When 1.00 L of 1.00M Ba(NO3)2 solution at
25.0 °C is mixed with 1.00 L of 1.00 M
Na2SO4 solution at 25 °C in a calorimeter,
the white solid BaSO4 forms and the
temperature of the mixture increases to
28.1 °C . Assuming that the calorimeter
absorbs only a negligible amount of heat
and the specific heat of the solution is
4.18J/°Cg and the density of the solution is
1.00g/mL, calculate the enthalpy change
per mole of BaSO4 formed.
• Ba2+ (aq) + SO4 2- (aq)→BaSO4 (s)
• Mass of solution =2 liters = 2000
grams
• Csp= 4.18J/°Cg
• ΔT=28.1-25.0 =3.1 °C
• qsurroundings = 2.6 x104J
• qsystem= -2.6 x104J
Bomb Calorimeter
• This is also known as a
constant volume
calorimeter.
• Since it is under
constant volume, but
there is a change in
temperature, the
pressure is not
constant.
Bomb Calorimeter
• Note: Since there is
no change in volume,
then no work is done.
• The constant volume
calorimeter measures
the system directly.
• The specific heat is of
the calorimeter itself.
Example
• In comparing potential fuels a bomb
calorimeter with a specific heat of 11.3kJ/
°C was employed. When a 1.50 g sample
of methane was burned with an excess of
oxygen in the calorimeter, the temperature
increased by 7.3 °C . When a 1.15 g
sample of hydrogen gas was burned with
an excess of oxygen, the temperature
increase was 14.3 °C. Calculate the
energy of combustion per gram for these
two fuels.
• Methane:
released energy
from 1.5 gram.
• =(11.3 kJ/g
°C)(7.3 °C)
• =83kJ
• Per gram
• 83kj/1.5g=55kJ/g
• Hydrogen: released
energy from 1.15 g.
• =(11.3 kJ/g
°C)(14.3 °C)
• =162 kJ
• Per gram
• 162kJ/1.15 g= 141
kJ/g
Hess’s Law
• Since enthalpy is a
state function, then
• In going from a
particular set of
reactants to a
particular set of
products, the change
in enthalpy is the
same whether the
reaction takes place in
one step or a series of
steps.
Hess’s Law