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Free Energy and Temperature

Free Energy and Temperature. Free energy decreases (becomes more negative) as temperature At low T, G m for solid phase is lower than that of liquid or vapour, so the solid phase is prevalent

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Free Energy and Temperature

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  1. Free Energy and Temperature • Free energy decreases (becomes more negative) as temperature • At low T, Gm for solid phase is lower than that of liquid or vapour, so the solid phase is prevalent • As we increase T to Tfus and higher, the liquid state has a lower Gm, so it is the phase that prevails • As we increase T further to Tb, the gas phase has the lowest value of Gm

  2. 7.13: Gibbs Free Energy of Reaction • To determine the spontaneity of a reaction, we use the change in the Gibbs Free Energy, G, or the Gibbs Free Energy of Reaction We’ve seen something like this before somewhere…

  3. Standard Gibbs Free Energy of Formation, Gf° • Gf° = The standard Gibbs Free Energy of reaction per mole for the formation of a compound from its elements in their most stable form. • Most stable form? • Hydrogen = ? • Oxygen = ? • Iodine = ? • Sodium = ? • Gf°is for the formation of 1 mole of product • Different amounts of reactants may be used…Be vigilant! 

  4. Gf°: What Does it Mean? • Compounds with Gf° > 0 are Thermodynamically Unstable • Compounds with Gf° < 0 are Thermodynamically Stable

  5. Gibbs Free Energy and Nonexpansion Work • we = ‘Extra work’ • Nonexpansion work is any kind of work other than that done against an opposing pressure • Stretching a spring, moving a rope, importing a sugar molecule into a cell are all examples of nonexpansion work • All cellular processes are examples of nonexpansion work • How are the Gibbs Free Energy and we related?

  6. Gibbs Free Energy and we • G = we • If we know the change in free energy, we know how much nonexpansion work can be done • What does this mean? • Let’s look at the combustion of glucose.

  7. G° and the Combustion of Glucose C6H12O6(s) + 6 O2(g) --> 6 CO2 (g) + 6 H2O (l) • The G° of the reaction is -2879 kJ For 1 mole of glucose, we get 2879 kJ of energy Or For 180 g of glucose, we get 2879 kJ of energy • To make one mole of peptide bonds, 17 kJ of work must be done. • If we get 2879 kJ of energy from one mole of glucose, we should be able to make 170 moles of peptide bonds One molecule of glucose will provide enough energy to add 170 amino acids to a growing protein (in actuality, you can only add 10 amino acids)

  8. The Effect of Temperature on G° • Remember that H° or S° is the sum of the individual enthalpies or entropies of the products minus those of the reactants • If we change the temperature, both are affected to the same extent, so the H° and S° values don’t significantly change • This is not the case with G°. Why? G° = H° - TS°

  9. The Effect of Temperature on G°

  10. The Effect of Temperature on G°

  11. The Effect of Temperature on G°

  12. The Effect of Temperature on G°

  13. Thermodynamics Review Let’s Look at some of the most important equations we’ve covered over the past 2 chapters…

  14. The First Law of Thermodynamics • Up until now, we have only considered the changes in the internal energy of a system as functions of a single change: either work or heat • However, these changes rarely occur singly, so we can describe the change in internal energy as: U = q + w (The 1st Law) • The change in internal energy is dependent upon the work done by the system and the heat gained or lost by the system

  15. Heat Capacity q = CT = mCsT nCmT

  16. Coffee Cup Calorimeter System: Solution and chemicals that react Surroundings: Cup and the world around it! Assumptions: We use 2 cups to prevent energy transfer to the surroundings (we assume that it works as designed) Expected Changes: • As the chemical reaction occurs, the potential energy in the reactants will be released as heat or the solution can supply heat to allow formation of a product with a higher potential energy • The solution will absorb or release energy during the reaction. We will see this as a temperature change qr + qsolution = 0

  17. Constant Pressure Calorimetry: An Example We place 0.05g of Mg chips in a coffee cup calorimeter and add 100 mL of 1.0M HCl, and observe the temperature increase from 22.21°C to 24.46°C. What is the ΔH for the reaction? Mg(s) + 2HCl (aq) --> H2(g) + MgCl2(aq) Assume: Cp of the solution = 4.20 J/gK Density of HCl is 1.00 g/mL

  18. Constant Pressure Calorimetry: An Example To solve this: ΔT = (24.46°C – 22.21°C) = (297.61K – 295.36K)=2.25K Mass of solution = Now, let’s calculate qsolution: qsolution = mCmΔT = (100.05g)(4.20 J/gK)(2.25K) = 945.5 J Now, let’s calculate qr: qr = -qsolution = -945.5 J

  19. Enthalpy • In a constant volume system in which no work is done (neither expansion nor non-expansion), we can rearrange the first law to: U = q + w (but w=0) U = q • Most systems are constant pressure systems which can expand and contract • When a chemical reaction takes place in such a system, if gas is evolved, it has to push against the atmosphere in order to leave the liquid or solid phase • Just because there’s no piston, it doesn’t mean that no work is done!

  20. Enthalpy • Let’s look at an example: • If we supply 100J of heat to a system at constant pressure and it does 20J of work during expansion, the U of the system is +80J (w=-20J) • We can’t lose energy like this • Enthalpy, H, is a state function that we use to track energy changes at constant pressure H=U + PV • The change in enthalpy of a system (H) is equal to the heat released or absorbed at constant pressure

  21. Enthalpy • Another way to define enthalpy is at constant pressure: H = q Enthalpy is a tricky thing to grasp, but we can look at it this way: • Enthalpy is the macroscopic energy change (in the form of heat) that accompanies changes at the atomic level (bond formation or breaking) • Enthalpy has the same sign convention as work, q and U • If energy is released as heat during a chemical reaction the enthalpy has a ‘-’ sign • If energy is absorbed as heat from the surrounding during a reaction, the enthalpy has a ‘+’ sign

  22. The 2nd Law of Thermodynamics The entropy of an isolated system increases in the course of any spontaneous change • We can summarize this law mathematically as:

  23. Entropy Change as a Function of Temperature at Constant Volume • If T2 > T1, then the logarithm is ‘+’ and entropy increases • Makes sense since we are raising the temperature and thermal motion will increase • The greater (higher) the heat capacity, the higher the entropy change

  24. Entropy Change as a Function of Changing Volume • We can use a similar logic to derive the change in entropy when the volume changes: • When V2 > V1, the entropy increases • Note: Units are still J/K 

  25. Entropy Change as a Function of Pressure • Remember Boyle’s Law? • We can substitute this relationship into the equation for entropy change as a function of volume to get: Entropy decreases for a samples that has been compressed isothermally (P1>P2) 

  26. Boiling Water and Entropy Let’s get 3 facts straight: At a transition temperature (Tf or Tb), the temperature remains constant until the phase change is complete At the transition temperature, the transfer of heat is reversible Because we are at constant pressure, the heat supplied is equal to the enthalpy

  27. Water Boiling and Entropy (at the boiling temperature) • We use the ‘ º ’ superscript to denote the standard entropy or the entropy at 1 bar of pressure

  28. Ice Melting and Entropy • We use the same logic to determine the entropy of fusion, Sfus

  29. S = k lnW The Boltzmann Formula Where: k = Boltzmann’s constant = 1.381 x10-23 J/K W=# of ways atoms or molecules in the system can be arranged and still give the same total energy • W is a reflection of the ensemble, the collection of molecules in the system • This entropy value is called the statistical entropy

  30. Calculating the Entropy of a Reaction • Sometimes we can’t always use our judgement and we need to calculate the entropy • In order to do this, we need the standard molar entropies of the products and the reactants as well as the number of moles of each 

  31. The Surroundings • IfSTot is positive, the reaction is spontaneous • If the Ssystem is negative, the reaction will still be spontaneous if SSurr is that much more positive STot = SSystem + SSurr

  32. Gibbs Free Energy G = H - TS (at constant T) G = -TS (at constant T and P) • A negative value of G indicates that a reaction will spontaneously occur • Large negative H values (like we’d have in a combustion reaction) would probably give you a large negative G • If TS is large and positive, the value of G may be large and negative

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