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T h e r m o c h e m i s t r y

T h e r m o c h e m i s t r y. The study of energy transfers and chemical reactions Section 3 (lm). Thermochemical Equations. Balanced equations that include physical states of reactants & products and the energy change, expressed as Δ H. Ex . Cellular respiration (Exothermic)

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T h e r m o c h e m i s t r y

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  1. Thermochemistry The study of energy transfers and chemical reactions Section 3 (lm)

  2. Thermochemical Equations • Balanced equations that include physical states of reactants & products and the energy change, expressed as ΔH. • Ex. Cellular respiration (Exothermic) C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l) ΔH=-2808 kJ

  3. Enthalpy (heat) of Combustion (ΔHcomb)The enthalpy change for the complete burning of one mole of a substance. Standard conditions (NOT STP): 1 atm & 298K / 25oC

  4. Changes of State (review) • Molar Enthalpy (heat) of vaporization (ΔHvap): the heat required to vaporize one mole of a liquid. • Molar enthalpy (heat) of fusion (ΔHfus): the heat required to melt one mole of a solid substance.

  5. Hess’s Law States: If you can add two or more thermochemical equations to produce a final equation for a reaction, then the sum of the enthalpy changes for the individual reactions is the enthalpy change for the final reaction.

  6. Translation: • The amount of energy transferred is the same regardless of the chemical pathway • Recall that a reaction reversed will have opposite enthalpy signs. • If A + B → C ; Δ H = + X J/mol, then • C → A + B; Δ H = - X J/mol

  7. Sample Problem

  8. That’s Pretty Abstract:Let’s Make it REAL What’s the enthalpy change for burning 2 mols of carbon in oxygen to produce 2 mols of carbon monoxide? 2C(s) + O2(g) → 2CO(g)

  9. 2C(s) + O2(g) → 2CO(g); ΔH = ? But we only know that 2C(s) + 2O2(g) → 2CO2(g) ΔH = -787 kJ And 2CO(g) + O2(g) → 2CO2(g) ΔH = -566 kJ We examine both reactions and arrange them in such a way that summing them produces our given reaction above, so

  10. Rearranging 2C(s) + 2O2(g) → 2CO2(g) ΔH = -787 kJ 2CO2(g) → 2CO(g) + O2(g) ΔH = +566 kJ Before we sum these reactions, we cancel like terms on both sides of the arrow, as if the arrow were an equals sign in an algebra problem. Notice the changed sign when we reversed the second equation.

  11. Cancelling “like terms” 2C(s) + 2O2(g) → 2CO2(g)ΔH = -787 kJ 2CO2(g) → 2CO(g) + O2(g)ΔH = +566 kJ Summing, we produce 2C(s) + O2(g) → 2CO(g) Hess’s law states that ΔH = +566 kJ – 787kJ = -221 kJ Because enthalpy is NEGATIVE the reaction is exothermic!

  12. Standard Enthalpy (heat) of Formation (ΔHfo) • The change in enthalpy that accompanies the formation of one mole of the compound in its standard state from its elements in their standard states.

  13. Standard Enthalpy (heat) of Formation (ΔHfo) • Is always zero for elements in their “standard states”. • Hess’s law can be used with text tables 15.5 (page 538) and R-11 (p.975) • Or we can use the “summation equation” to solve them (next up)

  14. Sample Problem # 11 p 987 Text • Use standard enthalpies of formation to calculate ΔHrxnfor 2HF (g) → H2(g) + F2(g) • Table R-11 says that ΔHfo for HF is -273.3kJ/mol. Since we have 2 mols, we will double that amount. • That means that H2(g) + F2(g) → 2HF (g) ΔHrxn= -557 kJ

  15. AND, importantly • We must also consider that this is the “reverse” reaction so we will change the sign, so, if • H2(g) + F2(g) → 2HF (g) ΔHrxn= -557 kJ, then • 2HF (g) → H2(g) + F2(g) ΔHrxn= + 557kJ

  16. The Summation Equation ΔHrxno = ∑ΔHfo(products) – ∑ΔHfo(reactants) In English, we say the Enthalpy change for any reaction, is equal to the sum of the enthalpy changes or the products minus the sum of the enthalpy changes of the reactants. ∑ means “sum of”

  17. An Example of the Summation Equation in Use ΔHrxno = ∑ΔHfo(products) – ∑ΔHfo(reactants) Use this equation to determine the enthalpy change for the following reaction:

  18. Again • Use the tables in your Text • 15.5 (page 538) and R-11 (p.975)

  19. Most commonly made errors by beginning HS students: • Instead of products – reactants (CORRECT), students write and compute reactants – products (WRONG!), because that’s how we read…left to right. • Forgetting to use the coefficients as multipliers in the problem. • So…DON’T make these mistakes!

  20. Entropy • A measure of the disorder of a system: • More disorder = higher entropy • Less disorder = lower entropy • Entropy has the symbol S • A messy room has more disorder than a neat room

  21. More formal idea of entropy • A measure of the number of possible ways that the energy of a system can be distributed. • Related to the freedom of a system’s particles to move • And the number of ways they can be arranged.

  22. Reaction Spontaneity • We need to know if reactions will occur under the conditions that are written. • That’s what spontaneity means: • The reaction will occur as written • How do we know?

  23. Reaction Spontaneity, more formally… • Any physical or chemical change that once begun, occurs with no outside intervention is a spontaneous process. • Changes in enthalpy and entropy determine whether a process is spontaneous.

  24. Natural Tendencies: • Reactions tend to occur where the enthalpy is decreased. • Reactions tend to release energy as they occur-going from higher to lower energy. Enthalpy Decreases

  25. Natural Tendencies: • Reactions tend to occur where the entropy increases. • Reactions tend to go from more organized to less organized. Entropy Increases

  26. Enthalpy is not the only factor in determining spontaneity! Entropy, (S) • Freedom of the system’s particles to move (disorder or randomness). • Entropy increases when volume increases, when energy increases, when # particles increases, or when the particle’s freedom of movement increases.

  27. Entropy TIPS • G > L > S • Entropy can be predicted by examining the products and reactants with that in mind. • Example: • Liquid water vaporizing, the entropy is increasing; Evaporation tends to be spontaneous

  28. More phase entropy • Solutions: • Gas dissolving in a liquid decreases entropy; • But what about dissolving a solid in a liquid? • Hint: • Does it happen by itself? • Do you stir your sugar in your coffee? • Do you need to? (only if you’re in a hurry!)

  29. Chemical Reactions • That produce more gaseous products than reactants tend to increase entropy.

  30. Reaction or NOT? • Some general tendencies, if you consider the two previous statements

  31. If those are true, then

  32. That’s where the Gibbs works!

  33. Rust As shown below, enthalpy is negative, so tends to be spontaneous. The reverse reaction, positive enthalpy is NOT spontaneous.

  34. Gibbs Free Energy Equation TIPS • Allows us to compute whether reaction is spontaneous or not. • Symbol G • Uses S and H and T • Guess what? T is in K, just like the gas laws.

  35. Gibb’s Free Energy (Gsystem)The energy available to do work. Variables: ΔG: free energy Δ H: enthalpy T: in K Δ S: change in entropy

  36. More Tips • The S and H terms use dis-similar units for energy, so we have to convert one or the other • There are more challenging problems and simpler problems to tackle. • You’ll be given Gibbs equation and all enthalpies and entropies you need for your test.

  37. Let’s go do some problems! • The end…or the beginning?

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