Unit 14 – Thermodynamics The study of energy relationships

1. Unit 14 – ThermodynamicsThe study of energy relationships Chapter 19

2. Warm-Ups • Complete ONE Page of the Chemical Equations Review per day • ONE minute per question!!!

3. Spontaneous Reactions • Naturally occurring reactions (happens on its own) that favor the formation of products • Do not favor product formation nor provide substantial product yield. *Does NOT refer to Rate! Reactions can have parts of both. *Temperature and Pressure have great impacts on spontaneous reactions Nonspontaneous Reactions

4. Free Energy, ∆G • Energy available to do work • No reaction is 100% efficient • Energy can only be obtained if the reaction occurs (spontaneous) CO2 (g)  C(s) + O2 (g) • Measure of the disorder of a system • Law of Disorder: Processes will spontaneously move in the direction of maximum disorder or randomness Entropy, ∆S : Chemical Chaos

5. Entropy and Phase of Matter • Entropy of gases is greater than the entropy of a liquid or solid • Liquid entropy is greater than a solid’s entropy • Entropy for a reaction will increase when a solid changes into a liquid or a gas. • Entropy also increases when a substance is divided into parts. • Entropy increases when the total # of product molecules is larger than that of the # of reactant molecules • Temp Increase = Entropy Increase

8. The Meaning of the Signs of Thermodynamic Properties

10. Example H2O (s)  H2O (l) • When is this process spontaneous? • At temperatures greater than 0oC • When is this process non-spontaneous? • At temperatures less than 0oC • It all depends on the TEMPERATURE!

11. Calculating Entropy, ∆S • Quantitative measurement of Disorder of a system • Symbol = ∆S • Units = J/K or J/(K*mol) • Standard Entropy = ∆So at 25 oC and 101.3kPa • A perfect crystal at 0 K would have ∆S=0 • ∆So = ΣSoproducts- ΣSoreactants

12. Example (use Appendix p. 558)H2 (g) + Cl2 (g)  2HCl (g) ∆So = (2x186.7) – (130.6+223.0) ∆So = 373.4 – 353.6 = 19.8 J/(K*mol) FORWARD: ∆So = (2*69.94) - (2*130.6 + 205.0) ∆So = 139.88 - 466.2 = -326.32 J/(K*mol) REVERSE: ∆So = (2*130.6 + 205.0) – (2*69.94) ∆So = 466.2 – 139.88 = 326.32 J/(K*mol) Calculate the ∆S for both the forward and reverse reactions for the synthesis of water.

13. Free Energy Calculations • Gibbs Free Energy Change, ∆G • Maximum amount of energy that can be coupled to another process to do useful work • Relates to Entropy and Enthalpy changes ∆G = ∆H - T∆S enthalpy temp. Entropy in K change • If ∆G is negative, the process is Spontaneous • If ∆G is positive, the process is Nonspontaneous • If ∆G is 0, the process is at equilibrium

14. Example: Calculate the Free Energy for the following reaction to determine if it is spontaneous at 25oC C(s) + O2(g)  CO2(g) Watch units!!! Convert J to kJ Then Calculate ∆ H and ∆ S

15. ∆So =∆Soproducts - ∆Soreactants = 0.214 – (0.0057 + 0.205) = 0.003 kJ/(K*mol) ∆Ho = ∆Hoproducts - ∆Horeactants = -393.5 kJ/mol – (0.00 + 0.00)kJ/mol = -393.5 kJ/mol

16. ∆Go = ∆Ho – T∆So T = 25oC = 298.15 K ∆Go = ∆Ho – T∆So = -393.5 kJ/mol – (298.15 K * 0.003kJ/(K*mol) = -394.4 kJ/mol Reaction is Spontaneous because ∆Go is negative! Predict whether the equilibrium lies to the left or to the right

17. Free Energy and Equilibrium ∆Go = ∆Ho – T∆ So can be used algebraically to solve for any unknown quantity ∆Ho, T, or ∆ So ∆G = ∆Go + RT lnQ Free energy at Reaction Quotient Non-standard (compare to Keq ) Conditions At standard conditions, Q = 1 ∆Go = - RT lnKeq ∆Go = negative, Keq > 1 ∆Go = 0, Keq = 1 ∆Go = positive, Keq < 1

18. Calculating Enthalpy, ∆H • Calorimetry: q = mC∆T • Hess’s Law : Add reactions • Hess’s Law : Σproducts – Σ reactants • Bond Energies : Bonds Broken – Bonds Formed • ∆Go = ∆Ho – T∆ So • ∆Go = -nF€o • ∆Go = - RT lnKeq Calculating Free Energy, ∆G

19. Example: Calculate ∆H, ∆S, and ∆Go for the following reaction at 298K: Use the appendix in your book to find the individual values. 2SO2(g) + O2(g)  2SO3(g)

20. Example: Calculate ∆H, ∆S, and ∆Go for the following reaction at 298K: Use the appendix in your book to find the individual values. 2SO2(g) + O2(g)  2SO3(g) ∆ H = 2(-396) – 2(297) = -198 kJ/mol ∆ S = 2(257) – [2(248) + 205] = -187 j/(K*mol) = -0.187 kJ/(K*mol) ∆ G = ∆ H – T∆S = -198 –(298)(-0.187) =-142 kJ/mol

21. Example: Consider the ammonia synthesis reaction: N2(g) + 3H2(g)  2NH3(g) • Where ∆ G = -33.3 kJ/mol of N2 consumed at 25oC • Calculate the value for the equilibrium constant. ∆Go = - RT lnKeq

22. Example: Consider the ammonia synthesis reaction: N2(g) + 3H2(g)  2NH3(g) • Where ∆ G = -33.3 kJ/mol of N2 consumed at 25oC • Calculate the value for the equilibrium constant. ∆Go = - RT lnKeq ∆Go = -33.3 kJ/mol = -33,300 J/mol R = 8.3145 J/(K*mol) -33,300 = -(8.3145)(298K) (lnKeq) 13.4 = ln K e13.4 = elnK K = 686780 (no units for Keq) Does this favor products or reactants?

23. Example: Calculate the value of ∆Go for the following reaction at 389K where the [NH3] = 2.0 M, [H2] = 1.25 M, and [N2] = 3.01 M • N2(g) + 3H2(g)  2NH3(g)

24. Example: Calculate the value of ∆Go for the following reaction at 389K where the [NH3] = 2.0 M, [H2] = 1.25 M, and [N2] = 3.01 M N2(g) + 3H2(g)  2NH3(g) 3.01 1.25 2.0 Keq = (2.0)2 . (3.01)(1.25)3