Thermodynamic Fun Quick Review System vs. Surroundings In thermochemistry, the universe is divided into two parts: The tem: The physical process or chemical reaction in which we are interested. We can define the tem anyway we like. The oundings: Everything else in the universe. Surroundings System Surroundings Universe System + Surroundings 1
Spontaneous vs. Nonspontaneous processes In thermochemistry a spontaneous process is one that occurs under certain set of conditions. A process that is nonspontaneous is one that does not occur under a certain set of conditions. spontaneous nonspontaneous Processes that are exothermic are often spontaneous. For example, the following reactions are spontaneous: combustion: CH 4 (g) + O 2 (g) H 2 O(g) + CO 2 (g) ΔH = -890.4 kj/mol acid/base: H + (aq) + OH - (aq) + H 2 O(l) ΔH = -56.2 kj/mol While a negative ΔH does favor spontaneity, it is not the only factor that determines it. spontaneous above 0 C nonspontaneous above 0 C H 2 O (s) H 2 O(l) H 2 O (l) H 2 O(s) ΔH = 6.01 kj/mol ΔH = -6.01 kj/mol Entropy S Entropy is the measure of how disperse the energy of a tem is. Entropy can also be defined as the amount of randomness in a tem. Barrier in place Less entropy Barrier Removed More entropy An increase in temperature increases entropy. A decrease in temperature decreases entropy. This is true even when comparing a substance in the same state. Increase in entropy Increase in entropy decrease in entropy decrease in entropy 2
Entropy and Molecules Translational motion Rotational motion bond rotation Vibrational motion stretching bending Qualitatively Determining Entropy How can we qualitatively determine which tem has more entropy? Temperature: C(s) @ 25 C C(s) @ 5 C Volume (gas): 1 mol of O 2 (g) in 1L 1 mol of O 2 (g) in 2L Phase: Molar Mass: Molecular Complexity: O 2 (g) He(g) C 2 H 6 (g) O 2 (l) Ne(g) CH 4 (g)
Dissolution and Entropy So, we know that when we change the state of a substance, it s entropy changes. But what about dissolving a substance into a liquid to create a solution? CH OH CH OH Cl 2 Cl 2 CH OH Cl 2 CH OH Cl 2 Increase in entropy Decrease Dissolution and Entropy So, we know that when we change the state of a substance, it s entropy changes. But what about dissolving a substance into a liquid to create a solution? NaCl FeCl Na + Fe + Cl - Cl - Cl - Cl - NaCl(s) NaCl(aq) Increase in entropy FeCl (s) FeCl (aq) Decrease in entropy There is an overall decrease in entropy because the water molecules become more ordered to form the complex ion. 4
Thermodynamic Standard State H S The (naught) denotes enthalpy under standard states A gas at 1 atm An aqueous solution (1 M conc.) at a pressure of 1 atm Pure liquids and solids Standard States The most stable form of elements at 1 atm and 25 o C (298 K) Change in entropy (DS) calculations Remember from CHM151: Similar calculation for the change in entropy of a tem: DH o rxn = SnDH o f (products) SnDH o f (reactants) DS o rxn = SnS o (products) SnS o (reactants) C 2 H 5 OH (l) + O 2 (g) 2CO 2 (g) + H 2 O(g) [(2mol)(21.6 J/K mol) + (mol)(188.7 J/K mol)] [(1 mol)(161 J/K mol) + ( mol)(205.0 J/K mol)] 217 J/K Ds rxn >0 -- entropy of the tem increasing S (J/K mol) C 2 H 5 OH(l): 161 O 2 (g): 205.0 CO 2 (g): 21.6 H 2 O(g): 188.7 H 2 O(l): 69.9 H 2 (g) 11.0 O 2 (g) 205.0 2H 2 (g) + O 2 (g) 2H 2 O(l) (2mol)(69.9 J/K mol)-[(2 mol)(11.0 J/K mol)+(1 mol)(205.0 J/K mol)] -27.2 J/K Ds rxn <0 -- entropy of the tem decreasing 5
Entropy S 6/17/2014 Thermodynamic Laws 1 st -- Energy is always conserved. 2 nd -- For a reaction to be spontaneous: Δs + Δs > 0 In other words: ΔS uni > 0 Entropy and Spontaneity A reaction is spontaneous if the ΔS of the oundings + tem is > 0. How does a reaction change the entropy of its oundings? oundings oundings - ΔH tem + ΔH tem Entropy S Heat is released to the oundings, increasing its entropy Δ S ΔH T Heat is absorbed from the oundings, decreasing its entropy Reactions that release heat to the oundings increase the entropy of the oundings. This is why exothermic reactions are often spontaneous. However, if the total for ΔS of the oundings + tem is < 0, even exothermic reactions can be nonspontaneous 6
Determining the Spontaneity of a reaction A reaction is spontaneous if the S universe is > 0. Remember: S universe = S + S Δ S ΔH T C 2 H 5 OH (l) + O 2 (g) 2CO 2 (g) + H 2 O(g) S = 217 J/K [(2mol)(-9.5 kj/mol) + (mol)(-241.8 kj/mol)] [(1mol)(-277.7 kj/mol) + ( mol)(0 kj/mol)] H = -124.7 kj What about the S of the oundings? H (kj/mol) C 2 H 5 OH(l): -277.7 CO 2 (g): -9.5 H 2 O(g): -241.8 H 2 O(l): -285.8 Δ S ΔH 124.7x10 J Δ S 4141.2 J/K T 298.15K Since both the tem and oundings increase in entropy, this reaction will be spontaneous at all temperatures S System: 217 J/K S Surroundings: 4141.2 J/K S Universe: 458 J/K SPONTANEOUS Determining the Spontaneity of a reaction Can a reaction that decreases the entropy of the tem be spontaneous? 2H 2 (g) + O 2 (g) 2H 2 O(l) S = -27.2 J/K (2mol)(-285.8 kj/mol) [(2 mol)(0kj/mol) + (1 mol)(0 kj/mol)] H (kj/mol) H 2 O(g): -241.8 H 2 O(l): -285.8 H = -571.6.kJ Δ S ΔH T Δ S 571.6x10 J 1917 J/K 298.15K Are there any conditions where the reaction can be non-spontaneous? Under what conditions would change in entropy in the universe be zero? Δ S Same value opposite sign of S System 571.6x10 J 27.2 J/K x x = 1747 K At 1747 K (and higher) the reaction will be nonspontaneous S System: S Surroundings: S Universe: SPONTANEOUS -27.2 J/K 1917 J/K 1590. J/K Although entropy of the tem is decreased, this is offset by the change in entropy of the oundings 7
Determining the Spontaneity of a phase change Δ S 6010 J 20.2 J/K 298.15K H 2 O(s) H 2 O (l) ΔS = (1 mol)(69.9j/k mol) (1 mol)(47.9 J/K mol) = 22.0 J/K ΔH = (1 mol)(-285.81 kj/mol) (1 mol)(-291.8 kj/mol) = 6.01 kj Is the process of ice melting always spontaneous? S System: S Surroundings: S (J/K mol) H 2 O (l): 69.9 H 2 O (s): 47.9 ΔH (kj/mol) H 2 O (l): -285.8 H 2 O (s): -291.81 22.0 J/K -20.2 J/K Based on the value of S what value of S would we need to give us negative values for S uni? S Universe: 1.8 J/K Δ S - 6010 J 22.0 J/K x SPONTANEOUS x = 27K normal melting point of water is 27.15K. Anything lower than 27K this process is non-spontaneous, as expected At 0 C, ΔS univ = 0 because the process is in equilibrium Determining the Spontaneity of a phase change Let s assume we have solid sodium and we want to melt it. Phase changes, just like chemical reactions have an equation Na(s) Na(l) ΔS = (1 mol)(57.56 J/K mol) (1 mol)(51.05 J/K mol) = 6.51 J/K ΔH = (1 mol)(2.41 kj/mol) (1 mol)(0 kj/mol) = 2.41 kj Δ S -2410 J 8.08 J/K 298.15K To be spontaneous, the S Universe must be positive. Think about at what point does the value become positive..as it crosses 0. Δ S -2410 J 6.51 J/K x x = 70. K or 97.0 C close to the normal melting point of Na, 98 C S System: S Surroundings: S Universe: S (J/K mol) Na(s): 51.05 Na(l): 57.56 ΔH (kj/mol) Na(l) 2.41 Na(s)? NON-SPONTANEOUS 6.51 J/K -8.08 J/K -1.57 J/K 8
Determining the Spontaneity of a reaction using Gibbs Free Energy +ΔS -ΔH +ΔS +ΔH exothermic (- H ) increase in entropy (+ S ) Always spontaneous endothermic (+ H ) increase in entropy (+ S ) Spontaneous at higher temperatures -ΔS +ΔH -ΔS -ΔH endothermic (+ H ) decrease in entropy (- S ) Never spontaneous exothermic (- H ) decrease in entropy (- S ) Spontaneous at lower temperatures Change in Free Energy G DS Although the spontaneity of a process be derived from knowing the change in entropy of the universe, do we have a way to of a process to just things we can measure in the tem? universe DS DS D S universe D S DH T T DS universe T DS The term -T S universe becomes G, Gibb s Free Energy DH G = H - T S In this equation, everything is defined in terms of the tem C 2 H 5 OH (l) + O 2 (g) 2CO 2 (g) + H 2 O(g) 2H 2 (g) + O 2 (g) 2H 2 O(g) H = -124.7 kj S = 217 J/K H = -571.6 kj S = -27.2 J/K UNITS! UNITS! G = -124700 J - 298.15K(217 J/K) = -1299.4 kj G = -571600 J - 298.15K(-27.2 J/K) = -474.0 kj S universe 4258 J/K Both methods agree that rxns are spontaneous S universe 1590. J/K 9
Gibb s Free Energy of formation ( G f ) While we can use H and S to determine G, we can also take another short cut Remember from CHM151: DH o rxn = SnDH o f (products) SnDH o f (reactants) DG o rxn = SnDG o f (products) SnDG o f (reactants) C 2 H 5 OH (l) + O 2 (g) 2CO 2 (g) + H 2 O(g) DG o rxn = [(2mol)(-94.0 kj/mol)+(mol)(-228.6 kj/mol)]-[(1mol)(-174.9 kj/mol)+ ( mol)(0 kj/mol)] = -1298.9 kj Compare to answer using other equation 2H 2 (g) + O 2 (g) 2H 2 O(l) G = ΔH T S = -1299 kj DG o rxn = (2mol)(-27.2 kj/mol)-[(2mol)(0kj/mol)+ (1 mol)(0 kj/mol)] = -474.4 kj G f (kj/mol) C 2 H 5 OH(l): -174.9 CO 2 (g): -94.0 H 2 O(g): -228.6 H 2 O(l): -27.2 G = ΣnΔG o f (products) Σn Δ G o f (reactants) = -1298.9 kj Compare to answer using other equation G = ΣnΔG o f (products) Σn Δ G o f (reactants) = -474.4 kj G = ΔH T S = -474.0 kj Determining the Spontaneity of a reaction using Gibbs Free Energy G = H - T S +ΔS -ΔH exothermic (- H ) increase in entropy (+ S ) Always spontaneous - G +ΔS +ΔH endothermic (+ H ) + G - G Lower temp Higher temp increase in entropy (+ S ) Spontaneous at higher temperatures -ΔS +ΔH endothermic (+ H ) decrease in entropy (- S ) Never spontaneous + G -ΔS -ΔH exothermic (- H ) decrease in entropy (- S ) + G - G Higher temp Spontaneous at lower temperatures Lower temp 10
Gibb s Free energy and equilibrium Earlier, we saw that when a reaction or process is in equilibrium S = 0. Well, what about G? H 2 O(s) H 2 O (l) ΔS = 22.0 J/K ΔH = 6.01 kj ΔG = 6.01x10 J (298.15K)(22.0J) = -549 J Spontaneous at room temp When does a spontaneous reaction become non-spontaneous? Earlier we mentioned that a reaction becomes non spontaneous if S univ = 0. What about in terms of ΔG? ΔG = 6.01x10 J (T)(22.0J) = 0 T = 27.18K Non-spontaneous at 0 C and lower temperatures When ΔG crosses 0 + G non-spontaneous - G spontaneous Nonstandard-State Conditions Obviously, most reactions don t happen under standard conditions. How do we determine the G for reactions under different conditions? G = G + RT lnq N 2 (g) + H 2 (g) 2NH (l) G = -.0x10 J @ standard conditions P N2 = P H2 = 10.0atm New conditions 1 1 4 Q 1.00x10 (P ) (P ) (10.0atm) (10.0atm) N2 H2 G = -.0x10 J + (8.14 J/K mol)(298.15k)(-9.210) = -55.8x10 J P N2 = P H2 = 0.001atm New conditions 1 1 12 Q 1.00x10 (P )(P ) (0.00100atm)(0.00100atm) N2 H2 G = -.0x10 J + (8.14 J/K mol)(298.15k)(27.61) =.5x10 4 J Notice under the new conditions the reaction is now non-spontaneous 11
Gibb s Free Energy and K Does Gibb s free energy tell us anything besides the spontaneity of the reaction? G = G + RT lnq 0 = G + RT lnk G = - RT lnk Whether a reaction ultimately reactant or product favored is based on ΔG. H 2 O(s) H 2 O (l) ΔG = -549 J -549 J = - (8.14 J/K mol)(298.15k)(lnk) 0.2215 = lnk K= 1.25 C 2 H 5 OH (l) + O 2 (g) 2CO 2 (g) + H 2 O(g) G = -1298.9 kj -1298.9x10 J = - (8.14 J/K mol)(298.15k)(lnk) 524 = lnk K=.72x10 227 The larger the free energy, the more product-favored a reaction Gibb s Free Energy and K As a reaction progresses, the free energy of the reaction decreases until it reaches 0 at equilibrium G reactants G products If we start a productfavored at a point where Q > K, then the reaction will NOT be spontaneous in the forward reaction. Product-favored reaction 12