Transcript 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.
q rxn = -q cal q cal = q bomb + q water + q wires +… heat
Define the heat capacity of the calorimeter
: q cal = m i c i T = C cal T all i FIGURE 7-5 •
A bomb calorimeter assembly
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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 O 2 (g) → MgO(s) C 9 H 20 (l) + 14 O 2 (g) → 9 CO 2 (g) + 10 H 2 O(l) CH 3 CH 2 COOH(l) + 5/2 O 2 (g) →3 CO 2 (g) + 3 H 2 O(l) Bomb calorimeters are constant volume devices and initially give us a
∆U or internal energy change
• • •
Bomb Calorimeter Example
The combustion of a rocket fuel, methylhydrazine (CH 6 N 2 (l)) is described by
CH 6 N 2 (l) + 5/2 O 2 (g) →N 2 (g) + 3 CO 2 (g) + 3 H 2 O(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 o C to 31.124 o C. Determine a value for ∆ Combustion of methyl hydrazine. (Elton John wants to know?)
• • • •
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 o C what was the pressure of oxygen gas if V bomb = 476 mL?
Houston. We have a chemical reaction!
Enthalpy and State Functions
• • Isothermal energy changes can occur either at
ΔH or enthalpy change
) or at
ΔU or internal energy change
). Often ΔH ~ ΔU for processes which do not liberate or consume gases.
Standard enthalpy change
– ΔH o T 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 FeCl
FeCl 3 (s) can be formed by the reaction of Cl 2 (g) with either Fe(s) or FeCl 2 (s). At 298K:
Fe(s) + 3/2 Cl 2 (g) → FeCl 3 (s) ∆H o 1 = - 399.5 kJ Fe(s) + Cl 2 (g) → FeCl 2 (s) ∆H o 2 = - 341.8 kJ FeCl 2 (s) + 1/2 Cl 2 (g) → FeCl 3 (s) ∆H o 3 = -57.7 kJ
∆H o 1 = ∆H o 2 + ∆H o 3
“Aside”: The 1st and 2 nd thermochemical equations specify the
heats of formation
of FeCl 2 (s) and FeCl 3 (s) respectively.
• • • • • • Example above: If ΔH
o 2 and ΔH o 3
are known by calculate
ΔH we could combine the corresponding thermochemical equations to o 3 . This obviates the need for a time consuming experiment.
Calculation: ΔH o 1 = ΔH o 2 + ΔH o 3 = (-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 Cl 2 (g) FeCl 3 (s) FeCl 2 (s) + 1/2 Cl 2 (g)
• • • We say that H is a
- having a unique value for a particular P and T. We specify state functions using capital letters.
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
measurements in the lab.
. Often we can combine two or more thermochemical equations to obtain a ΔH value for another chemical (or
without doing any
Path Dependant Functions
• • Path dependant functions – especially
(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
(not a state function). For all routes the
Displacement (a function of initial and final coordinates) is a is the same.
Heats of Formation
• • Standard heat of formation, ΔH states (P = 1.000 bar). o f : 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 ΔH o f,298K (elements) = 0 by definition ! We can use tabulated ΔH o f data for compounds to calculate enthalpy changes (constant P heats of reaction) for “any” chemical reaction.
Writing Important Thermochemical Equations
• Given heats of formation,
ΔH o f ’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 appropriate.
Phases for Elements at 298K
• • • Liquids:
Hg(l) and Br 2 (l)
He(g), Ne(g), Ar(g), Kr(g), Xe(g), Rn(g), N 2 (g), O 2 (g), F 2 (g), Cl 2 (g)
Solids: “everything else”. For some nonmetals molecular formulas are important. Thus, for elemental sulfur, one normally writes
S 8 (s)
and for phosphorus
P 4 (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 constituent of raspberries, is C 6 H 5 -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 COOH.