Science SS1 Thermochemistry 4 th Class Entropy and Free Energy

Copyright © Cengage Learning. All rights reserved 2 Thermodynamic s vs. Kinetics Domain of Kinetics – Rate of a reaction depends on the pathway from reactants to products. Thermodynamics tells us whether a reaction is spontaneous based only on the properties of reactants and products.

Copyright © Cengage Learning. All rights reserved 3 Spontaneous Processes and Entropy Thermodynamics lets us predict the direction in which a process will occur but gives no information about the speed of the process. A spontaneous process is one that occurs without outside intervention.

Copyright © Cengage Learning. All rights reserved 4 Entropy, ΔS The driving force for a spontaneous process is an increase in the entropy of the universe. A measure of molecular randomness or disorder.

Copyright © Cengage Learning. All rights reserved 5 The Expansion of An Ideal Gas Into an Evacuated Bulb

Positional Entropy A gas expands into a vacuum to give a uniform distribution because the expanded state has the highest positional probability of states available to the system. Therefore: S solid < S liquid << S gas

Entropy ΔS°reaction = ΣnpS°products – ΣnrS°reactants ΔG=ΔH-tΔS

Entropy ΔS J/t × mole of Reaction (Watch Units) Negative ----Going to order (Products more ordered than reactants) Positive -----Going to disorder (Product more disorder than reactants)

ΔS surr ΔS surr = +; entropy of the universe increases ΔS surr = -; process is spontaneous in opposite direction ΔS surr = 0; process has no tendency to occur Copyright © Cengage Learning. All rights reserved 9

10 Predict the sign of ΔS for each of the following, and explain: a)The evaporation of alcohol b)The freezing of water c)Compressing an ideal gas at constant temperature d)Heating an ideal gas at constant pressure e)Dissolving NaCl in water + – – + + CONCEPT CHECK!

ΔS surr The sign of ΔS surr depends on the direction of the heat flow. The magnitude of ΔS surr depends on the temperature. Copyright © Cengage Learning. All rights reserved 11

ΔS surr Heat flow (constant P) = change in enthalpy = ΔH Copyright © Cengage Learning. All rights reserved 12

Free Energy ΔG°reaction = ΣnpG°products – ΣnrG°reactants ΔG° = ΔH° – TΔS° Equil ΔG° = -RTlnk Electro ΔG° = -nfe

Free Energy ΔG KJ/mole of Reaction Negative = spontaneous reaction Positive = non-spontaneous reaction (will not occur at current conditions) Zero = System is in equilibrium

Describe the following as spontaneous/non-spontaneous/cannot tell, and explain. A reaction that is: a)Exothermic and becomes more positionally random Spontaneous b)Exothermic and becomes less positionally random Cannot tell a)Endothermic and becomes more positionally random Cannot tell a)Endothermic and becomes less positionally random Not spontaneous Explain how temperature affects your answers. CONCEPT CHECK!

Free Energy (G) A process (at constant T and P) is spontaneous in the direction in which the free energy decreases. – Negative ΔG means positive ΔS univ. Copyright © Cengage Learning. All rights reserved 16

Relation between ΔG, ΔH, and ΔS ∆S ∆H Spontaneous High temp Always spontaneous Never spontaneous Spontaneous Low temp -∆H -∆S Based on the formula ΔG=ΔH-TΔS Δ GΔ HΔSΔS Spontaneous at high temp positvepositive Always spontaneous negativepositive Spontaneous at low temp negative Never spontaneous Positivenegative

A liquid is vaporized at its boiling point. Predict the signs of: w q ΔH ΔS ΔS surr ΔG Explain your answers. Copyright © Cengage Learning. All rights reserved 18 – – 0 CONCEPT CHECK!

Third Law of Thermodynamics The entropy of a perfect crystal at 0 K is zero. The entropy of a substance increases with temperature. Copyright © Cengage Learning. All rights reserved 19

A stable diatomic molecule spontaneously forms from its atoms. Predict the signs of: ΔH° ΔS°ΔG° Explain. Copyright © Cengage Learning. All rights reserved 20 – – – CONCEPT CHECK!

The Meaning of ΔG for a Chemical Reaction A system can achieve the lowest possible free energy by going to equilibrium, not by going to completion. Copyright © Cengage Learning. All rights reserved 21

The equilibrium point occurs at the lowest value of free energy available to the reaction system. ΔG = 0 = ΔG° + RT ln(K) ΔG° = –RT ln(K) Copyright © Cengage Learning. All rights reserved 22

Change in Free Energy to Reach Equilibrium Copyright © Cengage Learning. All rights reserved 23

Copyright © Cengage Learning. All rights reserved 24

Question 1 Which of the following is a graph that describes the pathway of reaction that is exothermic and has high activation energy? A. B. C. D.

Question 1 Breakdown Correct answer is C What can you say about A? What can you say about B? What can you say about D? Label Activation energy on each graph Label ∆H for each graph

Question 2 When solid NH 4 SCN is mixed with solid Ba(OH) 2 in a closed container, the temperature drops and a gas is produced. Which of the following indicates the correct signs for ΔG, ΔH, and ΔS for the process? ΔG ΔH ΔS A) – – – B) – + – C) – + + D) + – +

Question 2 break down Correct answer is C G is negative because it happens spontaneously S is positive since going from solid to gas H is positive because it feels cold (endothermic)

Question 3 N 2 (g) + 3 H 2 (g) → 2 NH 3 (g) The reaction indicated above is thermodynamically spontaneous at 298 K, but becomes nonspontaneous at higher temperatures. Which of the following is true at 298 K? A) ΔG, ΔH, and ΔS are all positive. B) ΔG, ΔH, and ΔS are all negative. C) ΔG and ΔH are negative, but ΔS is positive. D) ΔG and ΔS are negative, but ΔH is positive. E) ΔG and ΔH are positive, but ΔS is negative.

Question 3 breakdown G = H - T S Spontaneous means negative G. S is negative since you are going from 4 gases to 2 gases (disorder is leaving) H must be negative if G is negative. Correct answer is B.

Alternate process Question 3 G is negative since it is spontaneous G is becoming more positive as Temp increases. Only S is affected by the temp. S must be negative since TS is positive. Since TS is positive H must be negative if G is negative.

Question 4 Of the following reactions, which involves the largest decrease in entropy? A) 2 CO(g) + O 2 (g) → 2 CO 2 (g) B) Pb(NO 3 ) 3 (s) + 2 KI(s) → PbI 2 (s) + 2 KNO 3 (s) C) C 3 H 8 (g) + O 2 (g) → 3 CO 2 (g) + 4 H 2 O(g) D) 4 La(s) + 3 O 2 (g) → 2 La 2 O 3 (s)

Question 4 breakdown Correct answer D. Going from gas to solid Which choice is largest increase? Why?

Question 5 Assume the data graphed was collected at a constant pressure of 0.97 atm and represents four different temperature samples of pure neon gas. Which of the following temperatures most likely corresponds to the data graphed for sample “D”? A) 273 K B) 298 K C) 305 K D) 338 K

Question 5 breakdown KE= ½ mv 2 Small molecules move faster than larger molecules. The high point of each curve hits the x axis in order from A to D. A is the slowest and biggest. D is the smallest and fastest. Temperature means molecules are moving faster therefore D is the highest temperature

Question 6 At 298 K, as the salt MX dissolves spontaneously to form an aqueous solution, ∆S and ∆H are positive. Which describes the value of ∆G and the absolute values of its components, T∆S and ∆H? A) ∆G |∆H| B)∆G < 0; |T∆S| < |∆H| C) ∆G > 0; |T∆S| > |∆H| D) ∆G > 0; |T∆S| < |∆H|

Breakdown 6 Spontaneous means G is negative or less than zero If both H and S are positive then S is the driving force of the spontaneity therefore if the reaction is spontaneous it must mean that TS is greater than H. Correct answer is A.

Free response 1 C 6 H 5 OH(s) + 7 O 2 (g) → 6 CO 2 (g) + 3 H 2 O(l) When a gram sample of pure phenol, C6H5OH(s), is completely burned according to the equation above, kilojoules of heat is released. Use the information in the table below to answer the questions that follow. substanceDelta H Delta S Joules/mol CO 2 − H2H H2OH2O − O2O C 6 H 5 OH ?144.0

(a) Calculate the molar heat of combustion of phenol in kilojoules per mole at 25°C. 2.00/94.12=.021 moles of phenol.021moles/64.98 KJ= 1mole/X X= KJ/mole = delta H (b) Calculate the standard heat of formation, ΔH°f, of phenol in kilojoules per mole at 25°C. H= prod- react = -3058=(6(−393.5)+3(−285.85))-X X=-161 J/K*mol

(c) Calculate the value of the standard free-energy change, ΔG° for the combustion of phenol at 25°C. ∆ S= prod-react=(6(213.6)+3(69.91)) – (7(205.0) ) S= J/mol G= H-TS = – 298( ) G = KJ/mol

Free Response 2 O 3 (g) + NO(g) → O 2 (g) + NO 2 (g) Consider the reaction represented above. (a) Referring to the data in the table below, calculate the standard enthalpy change, ΔH, for the reaction atv25°C. Be sure to show your work. KJ/mol O3O3 NONO 2 Std. enthalpy of formation ∆Hf at 25◦C

A. H= 33- (90+143)= -200 KJ/mol (b) Make a qualitative prediction about the magnitude of the standard entropy change, ΔS°, for the reaction at25°C. Justify your answer. 2 moles of gas to 2 moles of gas very small magnitude (very close to zero) (c) On the basis of your answers to parts (a) and (b), predict the sign of the standard free-energy change, ΔG°, for the reaction at 25°C. Explain your reasoning. H is negative and TS is very small after converting to KJ therefore G is negative.

Free Response 3 C 2 H 2 (g) + 2 H 2 (g) ---> C 2 H 6 (g) a) If the value of the standard entropy change, ΔS°, for the reaction is joules per mole Kelvin, calculate the standard molar entropy, S°, of C2H6 gas. (b) Calculate the value of the standard free-energy change, ΔG°, for the reaction. What does the sign of ΔG° indicate about the reaction above? (c) Calculate the value of the equilibrium constant, K, for the reaction at 298 K. SubstanceS° (J/mol K)ΔH°f (kJ/mol) C 2 H 2 (g) H 2 (g) C 2 H 6 (g)

S = prod –react = = X – ( (130.7)) X = J/mol B. H = prod –react= -84.7–226.7= KJ/mol G = H – TS = – 298(-.2372) G = KJ/mol C. G = -RTlnK = -(8.31)(298)lnK ln K = 97.7 K = 3x 10 42

Formulas tell you what to do. ∆E = q (heat) + w(work) q=mC∆T (water’s specific heat is J/g °C) ∆H, S, G = (sum of products) – (sum of reactants) ∆H bond energy = (sum of reactants) – (sum of products) ∆G = H-TS (watch for units on S) ∆G=-RTlnK (watch for units on G) ∆G=-nFE Hess’s law match the equation and add up ∆H rxn (changes affect equation and ∆H)