Thermochem Notes Set-1 (Enthalpy

Section 2: Enthalpy and Calorimetry
• definition: H = E + PV
• since E, P and V are all state functions, then H is too
• for the following, the process is at constant P and the only
type of work allowed is PV work
∆E = qP + w = qP - P∆V 
H = E + PV 
qP = ∆ E +P∆V
∆H = ∆E+P∆V
so, qP = ∆H at constant P
• heat of reaction and change in enthalpy are
used interchangeably for a reaction at
constant P
∆H = Hproducts - Hreactants
endo: + ∆H
exo: - ∆H
• science of measuring
• calorimeter- device used to experimentally
find the heat associated with a chemical
• substances respond differently when heated
Heat Capacity
• (C) how much heat it takes to raise a
substance’s T by one °C or K
• the amount of energy depends on the
amount of substance
heat absorbed (in J)
heat capacity  C 
increase in T (in C or K)
Heat Capacity
• specific heat capacity
– (s) heat capacity per gram
– in J/C°*g or J/K*g
• molar heat capacity
– heat capacity per mole
– in J/C°*mol or J/K*mol
Constant-Pressure Calorimetry
• uses simplest calorimeter (coffee-cup
calorimeter) since it is open to the air
• used to find changes in enthalpy (heats of
reaction) for reactions occurring in a
solution since qP = ∆H
• heat of reaction is an extensive property, so
we usually write them per mole so they are
easier to use
Constant-Pressure Calorimetry
• when 2 reactants are mixed
and T increases, the
chemical reaction must be
releasing heat so it is
• the released energy from the
reaction increases the
motion of molecules, which
in turn increases the T
Constant-Pressure Calorimetry
• If we assume that the calorimeter did not
leak energy or absorb any itself (that all the
energy was used to increase the T), we can
find the energy released by the equation:
E released by rxn = E absorbed by soln
∆H = qP = sP x m x ∆T
Constant-Volume Calorimetry
• uses a bomb calorimeter
• weighed reactants are placed inside the
rigid, steel container and ignited
• water surrounds the reactant container so
the T of it and other parts are measured
before and after reaction
Constant-Volume Calorimetry
• Here, the ∆V = 0 so -P∆V = w = 0
∆E = q + w = qV for constant volume
E released by rxn = ∆T x Ccalorimeter
Example 1
• When 1 mol of CH4 is burned at constant P,
890 kJ of heat is released. Find ∆H for
burning of 5.8 g of CH4 at constant P.
• 890 kJ is released per mole of CH4
1mol eCH 4
5.8 gCH 4 
 320kJ
16.04276 gCH 4 1mol CH 4
Example 2
• When 1.00 L of 1.00 M Ba(NO3)2 solution at
25.0°C is mixed with 1.00 L of 1.00 M Na2SO4
solution at 25.0°C in a coffee-cup calorimeter,
solid BaSO4 forms and the T increases to
28.1°C. The specific heat capacity of the
solution is 4.18 J/g*°C and the density is 1.0
g/mL. Find the enthalpy change per mole of
BaSO4 formed.
Example 2
• Write the net ionic equation for the reaction:
Ba2+ (aq) + SO42- (aq)  BaSO4(s)
• Is the energy released or absorbed? What does that
mean about ∆H and q?
exothermic: -∆H and –qP
• How can we calculate ∆H or heat?
heat = q = sP x m x ∆T
• How can we find the m?
use density and volume
Example 2
• Find the mass:
D   m  D V
 (1.0 g / mL)(2.0  10 mL)  2.0  10 g
• Find the change in T:
T  28.1  25.0  3.1 C
• Calculate the heat created:
 (4.18
)(2.0  10 g )(3.1 C )  2.6 10
Example 2
• since it is a one-to-one ratio and the moles of
reactants are the same, there is no limiting
• 1.0 mol of solid BaSO4 is made so
∆H= -2.6x104 J/mol = -26 kJ/mol
Example 3
• Compare the energy released in the
combustion of H2 and CH4 carried out in a
bomb calorimeter with a heat capacity of 11.3
kJ/°C. The combustion of 1.50 g of methane
produced a T change of 7.3°C while the
combustion of 1.15 g of hydrogen produced a
T change of 14.3°C. Find the energy of
combustion per gram for each.
Example 3
• methane: CH4
H  Ccalorimeter  T
H  (11.3 )  (7.3C )  83kJ
H  83kJ / 1.5 g  55kJ / g
• hydrogen: H2
H  Ccalorimeter  T
H  (11.3 )  (14.3C )  162kJ
H  162kJ / 1.15 g  141kJ / g
• The energy released by H2 is about 2.5 times
the energy released by CH4