1. 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.

2. Slide 2 of 57 Bomb Calorimetry Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7 FIGURE 7-5 A bomb calorimeter assembly q rxn = -q cal q cal = q bomb + q water + q wires +… Define the heat capacity of the calorimeter: q cal =  m i c i  T = C cal  T all i heat

3. 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 measurement.

4. Bomb Calorimeter Example

5. 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?

6. Exothermic Reaction!! Houston. We have a chemical reaction!

7. 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 – 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.

8. Formation of FeCl 3 (s) 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 Note that ∆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.

9. Example above: If ΔH o 2 and ΔH o 3 are known by experiment we could combine the corresponding thermochemical equations to calculate ΔH 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!

10. Chemical Routes to Iron (III) Chloride FeCl 2 (s) + 1/2 Cl 2 (g) Fe(s) +3/2 Cl 2 (g) FeCl 3 (s)

11. 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.

12. 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.

13. Heats of Formation Standard heat of formation, ΔH 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 states (P = 1.000 bar). 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.

14. 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.

15. Phases for Elements at 298K Liquids:Hg(l) and Br 2 (l) Gases: 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).

16. 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 C 6 H 5 COOH.