Lecture 8 (Slides) September 17

Bomb Calorimeters – Big Budget!
• Bomb calorimeters are big and expensive
devices. They have large heat capacities and
often contain water - which serves to keep the
temperature rise resulting from highly
exothermic reactions at a moderate level.
Bomb calorimeters are suited for studying
reactions where the starting materials might
be volatile and, without a sealed calorimeter,
both starting materials and heat might escape
from the calorimeter.
Bomb Calorimetry
qrxn = -qcal
qcal = q bomb + q water + q wires +…
Define the heat capacity of the
qcal = allm
iciT = CcalT
•A bomb calorimeter assembly
Copyright © 2011 Pearson
Canada Inc.
General Chemistry: Chapter 7
Slide 2 of 57
Typical Bomb Calorimeter Reactions
• Bomb calorimeters are especially suited to the
study of combustion reactions. In such studies an
excess of oxygen is normally employed. Why?
Typical reactions:
• Mg(s) + ½ O2(g) → MgO(s)
• C9H20(l) + 14 O2(g) → 9 CO2(g) + 10 H2O(l)
• CH3CH2COOH(l) + 5/2 O2(g) →3 CO2(g) + 3 H2O(l)
• Bomb calorimeters are constant volume devices
and initially give us a ∆U or internal energy
change measurement.
Bomb Calorimeter Example
• The combustion of a rocket fuel,
methylhydrazine (CH6N2(l)) is described by
• CH6N2(l) + 5/2 O2(g) →N2(g) + 3 CO2(g) + 3 H2O(l)
• In a calorimeter having a heat capacity of
8.004 kJ.K-1 the combustion of 2.018 g of
methyl hydrazine caused the temperature of
the calorimeter to rise from 24.001 oC to
31.124 oC. Determine a value for ∆U Combustion
of methyl hydrazine. (Elton John wants to
Combustion of Methyl Hydrazine
• Asides for the previous problem:
• 1. Does the gas pressure in the bomb increase
or decrease as the methyl hydrazine burns?
• 2. How many moles of oxygen are needed for
all of the methyl hydrazine to burn?
• 3. If a fivefold excess of oxygen gas was placed
in the bomb initially at a temperature of
24.001 oC what was the pressure of oxygen
gas if Vbomb = 476 mL?
Exothermic Reaction!!
Houston. We have a chemical reaction!
Enthalpy and State Functions
• Isothermal energy changes can occur either at
constant P ( ΔH or enthalpy change ) or at
constant V ( ΔU or internal energy change ).
Often ΔH ~ ΔU for processes which do not
liberate or consume gases.
• Standard enthalpy change – ΔHoT defines the
enthalpy change when all reactants and products
are in their standard states (most stable states at
1.000 bar (P) and a particular T.
Formation of FeCl3(s)
• FeCl3(s) can be formed by the reaction of
Cl2(g) with either Fe(s) or FeCl2(s). At 298K:
• Fe(s) + 3/2 Cl2(g) → FeCl3(s) ∆Ho1 = - 399.5 kJ
• Fe(s) + Cl2(g) → FeCl2(s)
∆Ho2 = - 341.8 kJ
• FeCl2(s) + 1/2 Cl2(g) → FeCl3(s) ∆Ho3 = -57.7 kJ
• Note that ∆Ho1 = ∆Ho2 + ∆Ho3
• “Aside”: The 1st and 2nd thermochemical
equations specify the heats of formation of
FeCl2(s) and FeCl3(s) respectively.
• Example above: If ΔHo2 and ΔHo3 are known by
experiment we could combine the
corresponding thermochemical equations to
calculate ΔHo3. This obviates the need for a
time consuming experiment.
• Calculation:
• ΔHo1 = ΔHo2 + ΔHo3
= (-341.8 + (-57.7)) kJ∙mol-1
= -399.5 kJ∙mol-1
• This type of thermodynamic calculation can
be undertaken for state functions only!
Chemical Routes to Iron (III) Chloride
Fe(s) +3/2 Cl2(g)
FeCl2(s) + 1/2 Cl2(g)
• We say that H is a state function - having a
unique value for a particular P and T. We specify
state functions using capital letters.
• In contrast, path dependent functions such as
work (w) and heat (q) are specified using lower
case letters and do not have a unique value for a
particular T and P.
• The equations given on the previous slide are
called thermochemical equations. Often we can
combine two or more thermochemical equations
to obtain a ΔH value for another chemical (or
physical) change without doing any
measurements in the lab.
Path Dependant Functions
• Path dependant functions – especially q (heat)
and w (work) – have values which depend on
the “route chosen” for a specific change.
• Example: trips from C3033 to Cabot Tower.
Many routes possible. Distance travelled is
path dependant (not a state function). For all
routes the displacement is the same.
Displacement (a function of initial and final
coordinates) is a state function.
Heats of Formation
• Standard heat of formation, ΔHof: the enthalpy
change at 298K (usually!) when one mole of a
pure substance in its standard state is formed
from its constituent elements in their standard
states (P = 1.000 bar).
• ΔHof,298K(elements) = 0 by definition! We can
use tabulated ΔHof data for compounds to
calculate enthalpy changes (constant P heats
of reaction) for “any” chemical reaction.
Writing Important Thermochemical
• Given heats of formation, ΔHof’s, one must be
able to write the corresponding balanced
thermochemical equations. To do this one
needs to know the phase (solid, liquid or gas)
of all elements at 298K as well as the
molecular formulas of elements where
Phases for Elements at 298K
• Liquids: Hg(l) and Br2(l)
• Gases: He(g), Ne(g), Ar(g), Kr(g), Xe(g), Rn(g),
N2(g), O2(g), F2(g), Cl2(g)
• Solids: “everything else”. For some nonmetals
molecular formulas are important. Thus, for
elemental sulfur, one normally writes S8(s) and
for phosphorus P4(s).
Heats of Formation - Examples
• In class we will write a number of balanced
thermochemical equations given heats of
formation. Some of the data will include heats of
formation for magnesium hydroxide (-924.5
kJ∙mol-1), copper (II) sulfate pentahydrate ( -771
kJ∙mol-1) and benzoic acid (-385 kJ∙mol-1). You are
expected to be able to write correct chemical
formulas for common inorganic compounds and
acids. The molecular formula for benzoic acid, a
constituent of raspberries, is C6H5COOH.

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