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Chapter 15 Thermodynamics

Chapter 15 Thermodynamics. Thermodynamics is the study of “heat” “change”, specifically that accompany physical and chemical processes. Often referred to as thermochemistry . .

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Chapter 15 Thermodynamics

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  1. Chapter 15 Thermodynamics

  2. Thermodynamics is the study of “heat” “change”, specifically that accompany physical and chemical processes. Often referred to as thermochemistry.

  3. We will look at the energy that accompanies chemical changes, not just the heat, so there might be kinetic and potential energy invovled. Typically these are classified as exothermic and endothermic reactions.

  4. An exothermic reaction releases energy (usually as heat). This means the products have less energy than the reactants, according to the law of conservation… Σ(Ereactants) > Σ(Eproducts) but Σ(Ereactants) = Σ(Eproducts) + Ereleased

  5. An endothermic reaction absorbs energy (usually as heat). This means the products have more energy than the reactants, according to the law of conservation… Σ(Ereactants) < Σ(Eproducts) but Σ(Ereactants) + Eabsorbed= Σ(Eproducts)

  6. Often energy diagrams are used to picture what is happening during these reactions. Activation energy is the energy required to start a chemical reaction.

  7. It is important to understand how energy is conserved during chemical reactions, even when it doesn’t seem like it is. The First Law of Thermodynamics states that the total energy of the universe is constant, so energy cannot be created or destroyed, just changed from one form to another or moved from one place to another.

  8. Enthalpy

  9. To account for the energy of a chemical reactions, chemists look at the enthalpy of the reaction. Enthalpy (H) is considered the energy content of the reaction and includes (but not limited to) heat released or absorbed during chemical reactions. (It is often assumed the pressure stays constant – atmospheric.)

  10. Enthalpy is a state function, so only the starting and ending conditions matter when calculating the enthalpy change (ΔH). (This is good news, it makes it easier.) ΔH = Hproducts – Hreactants If ΔH = -, reaction is exothermic If ΔH = +, reaction is endothermic Why?

  11. A balanced chemical equation along with the ΔH is a thermochemical equation, for example: C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)ΔH = 1367 kJ/mole rxn Or 2CO2(g) + 3H2O(l) → C2H5OH(l) + 3O2(g)ΔH = -1367 kJ/mole rxn Why?

  12. Notice: it is important to specify the state because state changes also include energy changes. For example H2O(s) → H2O(l) would have a +ΔH. Notice: enthalpy of reaction is per mole of reaction

  13. When 2.61 grams of CH3OCH3 is burned (at constant pressure), 82.5 KJ of heat is given off. What is the ΔH of the reaction?

  14. How much heat is released by the complete oxidation of 24.2 grams of aluminum? ΔH= -3352 kJ/mol rxn

  15. Standard Enthalpy

  16. Standard Enthalpies (H°) “°” is a zero, said naught, and refers to zero starting position (think number line) Standard enthalpies are the natural conditions of the substance at 1 bar (≈ 1 atm) and 25°C Standard states can be any phase of matter (see periodic table) and include diatomic elements. For solutions it is 1 M. To determine the standard enthalpy change (ΔH°) requires all reactants and products are in their standard states!

  17. Standard Molar Enthalpy of Formation (ΔHf°), sometimes called the “heat of formation”, is the enthalpy for making 1 mole of a substance in it’s standard state from it’s constituent elements in their standard state The ΔHf° for any element already in it’s standard state (also diatomic) is declared zero For example: Br2(l) is ΔHf° = 0, but Br2(g) is ΔHf° = 30.91 kJ/mol

  18. What about this? H2(g) + Br2(g) → 2HBr(g)ΔHf° = -72.8 kJ/ mol rxn Is this the heat of formation for HBr? Why?

  19. What about this? H2(g) + Br2(g) → 2HBr(g)ΔHf° = -72.8 kJ/ mol rxn Is this the heat of formation for HBr? Why? Remember, heats of formation are for one mole of the substance formed!

  20. Hess’s Law

  21. Hess’s Law states that the enthalpy for a reaction is the same whether it occurs by one step or by many. For example, what is the ΔHf° of CH4 if C(graphite)+ H2(g) → CH4(g) The problem is that this reaction would never occur, so we cannot measure the ΔHf° to put it in a chart, so how can we find ΔHf°if all we could measure is ΔH°?

  22. We can use Hess’s Law and reactions that would occur and be measureable to calculate the ΔHf° for CH4. C(graphite) + O2(g) → CO2(g) ΔH° = -393.5 kJ/mol H2(g) + ½O2(g) → H2O(l)ΔH° = -295.8 kJ/mol CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)ΔH° = -890.3 kJ/mol We need to make the above add up to C(graphite) + H2(g) → CH4(g) How do we manipulate the equations? Start by getting the reactants and products on the correct side of the arrow. Then multiply so unwanted parts cancel. Don’t forget to do to ΔH° what you do to the equation. Add up the equations and ΔH° values.

  23. C(graphite) + O2(g) → CO2(g) ΔH° = -393.5 kJ/mol H2(g) + ½O2(g) → H2O(l)ΔH° = -295.8 kJ/mol CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)ΔH° = -890.3 kJ/mol

  24. What is the heat of reaction for C2H4(g) + H2O(l) → C2H5OH(l) if C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) ΔH° = -1367 kJ/mol C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l) ΔHf° = -1411 kJ/mol Is this a ΔHf° or just a ΔH° ?

  25. Another use of Hess’s law allows us to calculate the heat of a reaction if all the heats of formation are known (or can be looked up). ΔH°reaction = Σn ΔHf°products - Σn ΔHf°reactants Multiply by the number of moles before summing, and don’t forget the ΔHf° of an element in it’s standard state is zero.

  26. What is the heat of reaction for • C2H4(g) + H2O(l) → C2H5OH(l) • If • ΔHf° C2H4(g) = 52.3 kJ/mol • ΔHf°H2O(l) = -285.8 kJ/mol • ΔHf°C2H5OH(l) = -277.7 kJ/mol

  27. What is the ΔHf° for PbO(s, yellow)? • PbO(s, yellow) + CO(g) → Pb(s) + CO2(g) ΔH° = -65.69 kJ • ΔHf° for CO2(g) = -393.5 kJ/mol ΔHf° for CO(g) = -110.5 kJ/mol

  28. Bond Energy

  29. Bond Energy is the amount of energy needed to break one mole of bonds of a covalent substance in gaseous state. Bond energies increase as the strength of the bonds increase, so it can indicate the amount of bonds (why?) or the stability of the bond (what’s that mean?).

  30. For example: • H2(g) → 2H(g) • To break this H-H bond takes 436 kJ/mol • (see page 610 for a lot of bond energies) • Why is it positive? • Why is it per mole? • What is the ΔH° of this reaction? • Is this ΔHf°?

  31. So bond energies could tell us the ΔHf° in some particular instances. • Another example: • H2(g) + Br2(g) → 2HBr(g) • What has to occur for this to happen? • Which steps should have positive bond energies and which should have negative? • Would the B. E. be related to ΔH°? To ΔHf°?

  32. H2(g) + Br2(g) → 2HBr(g) • H2(g) → 2H(g) B.E. = 436 kJ/mol • Br2(g) → 2Br(g) B.E. = 193 kJ/mol • H(g) + Br(g) → HBr(g) B.E. = -366 kJ/mol • What is the ΔH°rxn of HBr? • What would be the ΔHf°?

  33. Internal Energy

  34. Internal Energy (E) represents all the energy contained within a substance, and is a state function. • The internal energy of a system includes both heat (q) and the mechanical work of the system (w), thus • ΔE = q + w • Why is this another representation for the first law of thermodynamics?

  35. ΔE = q + w • q is + if heat is absorbed into the system • q is - if heat is given off by the system • w is + if work is done to the system • w is - if work is done by the system • thus • E is + (absorbs energy) if q and w are both + • E is - (releases energy) if q and w are both - • E is + or - if q and w are different in signs, depending on which is larger

  36. ΔE = q + w • q can be: • q = m c ΔT • q = m Hf • q = m Hv • w can be: • w = f Δd (like a piston moving) • w = -P ΔV (like a balloon inflating) • because P is always +, w depends on • the sign of ΔV

  37. More about work: • If the volume remains constant (sealed vessel) then -P ΔV = 0, so w = 0, so ΔE = qv only • Also w = -P ΔV = -Δn R T, where Δn is the change in moles of the gases (R and T are always positive, so w depends on Δn)

  38. Predict the sign of w in each of these examples, and then state what that means: • 2NH4NO3(S) → 2N2(g) + 4H2O(g) + O2(g) • H2(g) + Cl2(g) → 2HCl(g) • 2SO2(g) + O2(g) → 2SO3(g)

  39. Enthaply and Internal Energy

  40. To correctly determine the enthalpy of a reaction, we need the entire energy of the system (E) and we need to account for any work it did (PV). • For example, the enthalpy for the combustion of one mole of ethanol at constant pressure is -1367 kJ. The change in internal energy of ethanol when combusted is only -1365 kJ. What happens to the other 2 kJ of energy?

  41. Thus, ΔH = ΔE + P ΔV • and ΔH = q + w + P ΔV • and ΔH = q + -P ΔV + P ΔV at constant T and P • so ΔH = qp(at constant pressure) • also, ΔH = ΔE + Δn R T (constant T and P)

  42. A system receives 93 J of electrical work, performs 227 J of pressure-volume work, and releases 155 J of heat. What is the change to the internal energy of the system?

  43. What is the amount of work done when one mole of brown nitrogen dioxide is converted into colorless dinitrogentetroxide at 10.0° C?

  44. Spontaneity

  45. One of the biggest concerns of thermodynamics is predicting whether a particular process would occur under a certain set of conditions. In other words, would the reactants spontaneously form products, or just stay as reactants? • If the reaction will proceed, then it is spontaneous, or the products are thermodynamically favored. • Otherwise it is nonspontaneous, or the products are not thermodynamically favored.

  46. Note: the speed of a reaction has nothing to do with whether or not it is thermodynamically favorable • For example, C(diamond) → C(graphite), is thermodynamically favorable, but takes so long you would have to wait millions of years for the diamond in your ring to turn to graphite. • The rate at which a thermodynamically favored reaction proceeds will be a topic for later (chapter 16).

  47. Typically two factors will make a reaction thermodynamically favorable (spontaneous): • a decrease in the energy of the products • Is that endothermic or exothermic? • an increase in the disorder, or spread-out-ed-ness • The second factor can be so strong a driving force that is it our Second Law of Thermodynamics - the disorder of the universe is always increasing (although within a system is can decrease).

  48. Entropy

  49. Entropy (S) is a measure of the disorder of a system, and is a state function. The greater the disorder, the greater the entropy. • If the entropy of a system increases, the reaction is likely to be spontaneous, but it does not have to be spontaneous to get an increase in entropy. • If the entropy of a system decreases, the entropy of the surroundings must increase by a greater amount so the entropy of the universe still increases.

  50. When entropy increases, ΔS = + • When entropy decreases, ΔS = - • ΔSuniverse must always be + • Can you cool your house by leaving the refrigerator door open? • To clean up your room is to make it more orderly, how does the ΔSuniverse still increase? • Unfortunately we cannot measure the ΔSuniverse, so it is understood all talk of ΔS is ΔSsystem.

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