Changes in Energy
for chemical and physical changes
Heat vs Thermal Energy
Thermal Energy and Temperature
First Law of Thermodynamics: ∆E = q + w
You will see this symbol (∆) all over this chapter. This symbol means "change". The equation above means that the change in energy is always a combination of work produced and heat (q). Obviously, heat, work and energy are all in the same units: the joule is the unit we will use the most (calories used to be used for heat and still are and horsepower was and is used for work). This topic in physics and chemistry has not only numbers but sign as well. When energy increases, the energy is positive and when it is negative, energy is decreasing. Think graphically (see figures below).
This topic is very "universal" meaning that you think not only about the "system" you are studying but also its "surroundings". If we looking at changes in energy for a metal chair on a sunny deck, that situation does not involve only the chair (if it did, the chair would not get hot) but also its surroundings. Energy is transferred in this example from the surroundings to the system (the chair). This same analogy would apply to a chemical reaction that required an input of energy as shown in the figure. The surroundings are losing energy (negative ∆E; work and heat are negative too) and the system is gaining energy (positive ∆E; work and heat are positive too). Having said all of that, we will focus on the system.
1. Calculate ∆E and determine whether the process is endothermic or exothermic for the following cases:
(a) a balloon is cooled by removing 0.655 kJ of heat. It shrinks on cooling, and the atmosphere does 38 J of work on the balloon.
(b) A 100.0 g bar of gold is heated from 25° C to 50 °C during which it absorbs 322 J of heat. Assume the volume of the gold bar remains constant.
(c) the surroundings do 1.44 kJ of work compressing a gas in a perfectly insulated environment.
2. A gas is confined to a cylinder under constant atmospheric pressure, as illustrated in figure 5.4. When 0.49 kJ of heat is added to the gas, it expands and does 214 J of work on the surroundings. What are the values of ∆H and ∆E for this process?
Thermochemistry-the balance between KE and PE
There are two basic forms of energy: kinetic energy (energy of motion) and potential energy (energy of position). Whenever energy is transformed (e.g., electrical from chemical (fuel)); you are turning potential into kinetic or vice versa. Potential energy is locked up in chemical bonds and can be released when those bonds are broken. When a molecule forms, potential energy drops (think ball on an incline). This is really the only reason molecules do form-the system rests at a lower energy. When chemical reactions occur, energy is put into breaking the bonds (ball has to be pushed back up the hill). Once the bonds break, the ball rolls down the hill again. There is a balance between this constant switching from potential to kinetic and that is what thermochemistry is all about.
Heats of reactions or enthalpies can also be calculated from other kinds of data (besides calorimetry). When reactants react to form products, bonds are broken and then reformed into new arrangements. It takes energy to break bonds (endothermic) and energy is released when bonds form (exothermic). This endothermic and exothermic business occurs in every chemical reaction. The balance between the two dictates whether a chemical reaction is overall exothermic (bond forming step is bigger) or endothermic (bond breaking step is bigger). If a reaction is endothermic, then enthalpy can be thought of as entering the reaction on the reactant side and if a reaction is exothermic, heat can be thought of as exiting the reaction on the product side.
C (s) + O2 (g) → CO2 (g) + 393.5 kJ
H2O (l) + 285.8 kJ → H2 (g) + 1/2 O2 (g)
When one needs to calculate the energy change from a thermochemical equation such as those above, a direct proportion is used. One side of the proportion comes from the thermochemical equation and the other side from any new amount introduced in the problem. For example, the top equation above liberates (exothermic) 393.5 kJ for each substance written in the equation. This means for one mole of carbon solid, one mole of oxygen gas and one mole of carbon dioxide gas. If we wanted to know the energy change for 0.5 mol of carbon dioxide, we would set it up as follows:
-393.5 kJ/1 mol CO2 = x kJ/0.5 mol CO2
Solving for x (cross multiply and divide) = -196.75 kJ
As long as both sides agree in the proportion, it does not matter which value is on top and bottom.
Which is endothermic above? Exothermic?
Take the enthalpy out of the equations above and it could be written separately from the reaction. If it is, the sign of the enthalpy value tells you whether the reaction is endothermic or exothermic. If ∆Hrxn is negative, the reaction is exothermic. If ∆Hrxn is positive, the reaction is endothermic.
Exothermic or Endothermic (p 5):
3. Determine whether each process is exothermic or endothermic and indicate the sign of ∆H.
(a) dry ice subliming (solid to gas)
(b) the wax in a candle burning
(c) the wax in a candle melting as a result of the wax burning
4. The decomposition of slaked lime, Ca(OH)2 (s), into lime, CaO(s), and H2O (g) at constant pressure requires the addition of 109 kJ per mole of Ca(OH)2. (a) Write a balanced thermochemical equation for the reaction. (b) Draw an enthalpy diagram for the reaction.
5. Consider the equation for the combustion of acetone (C3H8O), the main ingredient in nail polish remover:
C3H8O (l) + O2 (g) --> CO2 (g) + H2O (g) ∆Hrxn = -1790 kJ
If a bottle of nail polish remover contains 155 g of acetone, how much heat is released by its complete combustion?
6. Without referring to tables, predict which of the following has the higher enthalpy in each case: (a) 1 mol CO2 (s) or 1 mol CO2 (g) at the same temperature, (b) 2 mol of hydrogen atoms or 1 mol of H2, (c) 1 mol H2 (g) and 0.5 mol O2(g) at 25 °C or 1 mol H2O(g) at 25 °C, (d) 1 mol N2 at 100 °C or 1 mol N2 (g) at 300 °C. (5.42)
7. At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat KClO3:
2KClO3 (s) --> 2KCl (s) + 3O2 (g) ∆H = -89.4 kJ
For this reaction, calculate ∆H for the formation of (a) 1.36 mol of O2 and (b) 10.4 g of KCl. (c) the decomposition of KClO3 proceeds spontaneously when it is heated.Do you think the reverse reaction, the formation of KClO3 from KCl and O2 is likely to be feasible under ordinary conditions? Explain. (5.46)
Examples 3, 4, 5, 8, and 9 from the practice problems. See links for other practice problem solutions.
4 ways to calculate change in heat or enthalpy
1. Calculating the amount of Heat Transfer-Calorimetry
8. Calculate the amount of heat required to heat a 3.5 kg gold bar from 21 °C to 67 °C.
9. If 57 J of heat are added to an aluminum can with a mass of 17.1 g, what is it's temperature change?
2. Hess's Law-enthalpy is a state function after all!
10. From the enthalpies of reaction
2C (s) + O2 (g) --> 2CO (g) ∆H = -221.0 kJ
2C (s) + O2 (g) + 4H2 (g) --> 2CH3OH (g) ∆H = -402.4 kJ
calculate ∆H for the reaction
CO (g) + 2H2 (g) --> CH3OH (g) (5.64)
11. Given the data
N2 (g) + O2 (g) --> 2NO (g) ∆H = +180.7 kJ
2NO (g) + O2 (g) --> 2NO2 (g) ∆H = -113.1 kJ
2N2O (g) --> 2N2 (g) + O2 (g) ∆H = -163.2 kJ
use Hess's Law to calculate ∆H for the reaction
N2O (g) + NO2 (g) --> 3NO (g) (5.66)
Another Hess' Law example:
3. Heats of formation
12. Write balanced chemical equations that describe the formation of the following compounds from elements in their standard states, and use Appendix C to obtain values for their standard enthalpies of formation, ∆Hf. (a) H2O2 (g), (b) CaCO3 (s), (c) POCl3 (l), (d) C2H5OH (l). (5.70)
13. Many portable gas heaters and grills use propane, C3H8 (g), as a fuel. Using standard enthalpies of formation, calculate the quantity of heat produced when 10.0 g of propane is completely combusted in air under standard conditions. (5.72)
14. Calcium carbide solid (CaC2) reacts with liquid water to form acetylene gas (C2H2) and solid Ca(OH)2. From the following enthalpy of reaction data and data in Appendix C, calculate ∆Hf for CaC2 (s). (5.76) The heat for this reaction is -127.2 kJ
WS1: #4: a) -2044 kJ; b) -184.6 kJ; c) -802.34 kJ; d) -502.8 kJ; e) omit #5 (answers given are wrong) a) 490.7 kJ; b) -333.4 kJ; c) -598.4 kJ; d) -1088.4 kJ; e) 753.6 kJ. WS 2: #5: -385.9 kJ; 7) b; 8) 3050 J; 9) omit; 10) c = 0.275; 11) = -166.7 kJ