Thermodynamic Fun

Thermodynamic Fun Quick Review – System vs. Surroundings In thermochemistry, the universe is divided into two parts: The system: The physical process or chemical reaction in which we are interested. We can define the system anyway we like. The surroundings: 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: CH4 (g) + O2 (g) → H2 O(g) + CO2(g) ΔH° = -890.4 kJ/mol acid/base: H+ (aq) + OH-(aq) + H2 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 H2O (s) → H2O(l) ΔH° = 6.01 kJ/mol nonspontaneous above 0° C H2O (l) → H2O(s) ΔH° = -6.01 kJ/mol Entropy – S Entropy is the measure of how disperse the energy of a system is. Entropy can also be defined as the amount of randomness in a system . Barrier in place Barrier Removed Less entropy 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 system has more entropy? Temperature: C(s) @ 25°C C(s) @ 35°C Volume (gas): 1 mol of O2(g) in 1L 1 mol of O2(g) in 2L Phase: O2(g) O2(l) Molar Mass: He(g) Ne(g) C2H6(g) CH4(g) Molecular Complexity: 3 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? Cl2 CH3OH CH3OH Cl2 CH3OH Cl2 CH3OH Cl2 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 FeCl3 Fe3+ Na+ Cl- NaCl(s)→ NaCl(aq) Increase in entropy Cl- ClCl- FeCl3 (s)→ FeCl3 (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 Standard States – A gas at 1 atm – An aqueous solution (1 M conc.) at a pressure of 1 atm – Pure liquids and solids – The most stable form of elements at 1 atm and 25oC (298 K) Change in entropy (DS) calculations Remember from CHM151: Similar calculation for the change in entropy of a system: DHorxn = SnDHof (products) – SnDHof (reactants) DSorxn = SnSo (products) – SnSo (reactants) C2H5OH (l) + 3O2(g) → 2CO2(g) + 3H2O(g) [(2mol)(213.6 J/K·mol) + (3mol)(188.7 J/K·mol)] – [(1 mol)(161 J/K·mol) + (3 mol)(205.0 J/K·mol)] 217 J/K Dsrxn >0 -- entropy of the system increasing S° (J/K·mol) C2H5OH(l): 161 O2(g): 205.0 CO2(g): 213.6 H2O(g): 188.7 H2O(l): 69.9 H2(g) 131.0 O2(g) 205.0 2H2(g) + O2(g) → 2H2O(l) (2mol)(69.9 J/K·mol)-[(2 mol)(131.0 J/K·mol)+(1 mol)(205.0 J/K·mol)] -327.2 J/K Dsrxn <0 -- entropy of the system decreasing 5 Thermodynamic Laws 1st -- Energy is always conserved. 2nd -- For a reaction to be spontaneous: Δssys + Δssurr > 0 In other words: ΔSuni > 0 Entropy and Spontaneity A reaction is spontaneous if the ΔS of the surroundings + system is > 0. How does a reaction change the entropy of its surroundings? system Heat is released to the surroundings, increasing its entropy Δ Ssurr  + ΔH system Entropy S - ΔH surroundings Entropy S surroundings Heat is absorbed from the surroundings, decreasing its entropy  Δ Hsys T Reactions that release heat to the surroundings increase the entropy of the surroundings. This is why exothermic reactions are often spontaneous. However, if the total for ΔS of the surroundings + system is < 0, even exothermic reactions can be nonspontaneous 6 Determining the Spontaneity of a reaction A reaction is spontaneous if the ∆Suniverse is > 0. Remember: ∆Suniverse = ∆Ssys + ∆Ssurr C2H5OH (l) + 3O2(g) → 2CO2(g) + 3H2O(g) Δ Ssurr   Δ Hsys T ∆S °sys= 217 J/K [(2mol)(-393.5 kJ/mol) + (3mol)(-241.8 kJ/mol)] – [(1mol)(-277.7 kJ/mol) + (3 mol)(0 kJ/mol)] ∆Hsys = -1234.7 kJ ∆H° (kJ/mol) C2H5OH(l): -277.7 CO2(g): -393.5 H2O(g): -241.8 H2O(l): -285.8 What about the ∆S of the surroundings? Δ Ssurr   Δ Hsys T 3 Δ Ssurr  1234.7x10 J  4141.2 J/K 298.15K Since both the system and surroundings increase in entropy, this reaction will be spontaneous at all temperatures ∆S System: ∆S Surroundings: ∆S Universe: 217 J/K 4141.2 J/K 4358 J/K SPONTANEOUS Determining the Spontaneity of a reaction Can a reaction that decreases the entropy of the system be spontaneous? 2H2(g) + O2(g) → 2H2O(l) ∆S °sys= -327.2 J/K ∆H° (kJ/mol) H2O(g): -241.8 H2O(l): -285.8 (2mol)(-285.8 kJ/mol) – [(2 mol)(0kJ/mol) + (1 mol)(0 kJ/mol)] ∆Hsys = -571.6.kJ Δ Ssurr   Δ Hsys T Δ Ssurr  571.6x10 3 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? Same value opposite sign of ∆S System 571.6x10 3 J Δ Ssurr   327.2 J/K x ∆S System: ∆S Surroundings: ∆S Universe: -327.2 J/K 1917 J/K 1590. J/K SPONTANEOUS x = 1747 K At 1747 K (and higher) the reaction will be nonspontaneous Although entropy of the system is decreased, this is offset by the change in entropy of the surroundings 7 Determining the Spontaneity of a phase change H2O(s) → H2O (l) ΔS°sys = (1 mol)(69.9J/K·mol) – (1 mol)(47.9 J/K·mol) = 22.0 J/K ΔH°sys = (1 mol)(-285.81 kJ/mol) – (1 mol)(-291.8 kJ/mol) = 6.01 kJ Δ Ssurr  6010 J  20.2 J/K 298.15K Is the process of ice melting always spontaneous? Based on the value of ∆Ssys what value of ∆Ssurr would we need to give us negative values for ∆Suni? Δ Ssurr  S° (J/K·mol) H2O (l): 69.9 H2O (s): 47.9 ΔH° (kJ/mol) H2O (l): -285.8 H2O (s): -291.81 ∆S System: ∆S Surroundings: 22.0 J/K -20.2 J/K ∆S Universe: 1.8 J/K SPONTANEOUS - 6010 J  22.0 J/K x x = 273K…normal melting point of water is 273.15K. Anything lower than 273K this process is non-spontaneous, as expected At 0°C, ΔSuniv= 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°sys = (1 mol)(57.56 J/K·mol) – (1 mol)(51.05 J/K·mol) = 6.51 J/K ΔH°sys = (1 mol)(2.41 kJ/mol) – (1 mol)(0 kJ/mol) = 2.41 kJ Δ Ssurr  - 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 Ssurr  - 2410 J  6.51 J/K x x = 370. K or 97.0 °C – close to the normal melting point of Na, 98 °C S° (J/K·mol) Na(s): 51.05 Na(l): 57.56 ΔH° (kJ/mol) Na(l) 2.41 Na(s) ? ∆S System: ∆S Surroundings: 6.51 J/K -8.08 J/K ∆S Universe: -1.57 J/K NON-SPONTANEOUS 8 Determining the +ΔS Spontaneity of a reaction using Gibbs Free Energy +ΔS -ΔH +ΔH endothermic (+∆Hsys ) exothermic (-∆Hsys ) increase in entropy (+∆Ssys) increase in entropy (+∆Ssys) Spontaneous at higher temperatures Always spontaneous -ΔS -ΔS +ΔH -ΔH exothermic (-∆Hsys ) endothermic (+∆Hsys ) decrease in entropy (-∆Ssys) decrease in entropy (-∆Ssys) Spontaneous at lower temperatures Never spontaneous Change in Free Energy – ∆G 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 system? D Suniverse  D Ssys  D Ssurr D Suniverse  D Ssys  D Hsys T  T D Suniverse  T D Ssys  D Hsys The term -T∆Suniverse becomes ∆G, Gibb’s Free Energy ∆G = ∆H - T∆S In this equation, everything is defined in terms of the system C2H5OH (l) + 3O2(g) → 2CO2(g) + 3H2O(g) 2H2(g) + O2(g) → 2H2O(g) ∆H°sys = -1234.7 kJ ∆Hsys = -571.6 kJ ∆S °sys= 217 J/K ∆S °sys= -327.2 J/K UNITS! UNITS! ∆G° = -1234700 J - 298.15K(217 J/K) = -1299.4 kJ ∆Suniverse 4258 J/K ∆G° = -571600 J - 298.15K(-327.2 J/K) = -474.0 kJ Both methods agree that rxns are spontaneous ∆Suniverse 1590. J/K 9 Gibb’s Free Energy of formation (∆Gf) While we can use ∆H and ∆S to determine ∆G, we can also take another short cut… Remember from CHM151: DHorxn = SnDHof (products) – SnDHof (reactants) DGorxn = SnDGof (products) – SnDGof (reactants) Gf° (kJ/mol) C2H5OH(l): -174.9 CO2(g): -394.0 H2O(g): -228.6 H2O(l): -237.2 C2H5OH (l) + 3O2(g) → 2CO2(g) + 3H2O(g) DGo rxn = [(2mol)(-394.0 kJ/mol)+(3mol)(-228.6 kJ/mol)]-[(1mol)(-174.9 kJ/mol)+ (3 mol)(0 kJ/mol)] = -1298.9 kJ Compare to answer using other equation ∆G° = ΣnΔGof (products) – Σn Δ Gof (reactants) = -1298.9 kJ ∆G° = ΔH – T∆S = -1299 kJ 2H2(g) + O2(g) → 2H2O(l) DGorxn = (2mol)(-237.2 kJ/mol)-[(2mol)(0kJ/mol)+ (1 mol)(0 kJ/mol)] = -474.4 kJ ∆G° = ΣnΔGof (products) – Σn Δ Gof (reactants) = -474.4 kJ Compare to answer using other equation ∆G° = ΔH – T∆S = -474.0 kJ Determining the Spontaneity of a reaction using Gibbs Free Energy ∆G = ∆H - T∆S +ΔS -∆G -ΔH +ΔS +ΔH +∆G -∆G Lower temp exothermic (-∆Hsys ) increase in entropy (+∆Ssys) Always spontaneous -ΔS +ΔH increase in entropy (+∆Ssys) Spontaneous at higher temperatures -ΔS -ΔH +∆G endothermic (+∆Hsys ) decrease in entropy (-∆Ssys) Never spontaneous Higher temp endothermic (+∆Hsys ) +∆G -∆G Higher temp Lower temp exothermic (-∆Hsys ) decrease in entropy (-∆Ssys) Spontaneous at lower temperatures 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? H2O(s) → H2O (l) ΔS°sys = 22.0 J/K ΔG° = 6.01x103 J – (298.15K)(22.0J) = -549 J ΔH°sys = 6.01 kJ Spontaneous at room temp When does a spontaneous reaction become non-spontaneous? Earlier we mentioned that a reaction becomes non spontaneous if ∆Suniv = 0. What about in terms of ΔG? +∆G ΔG = 6.01x103 J – (T)(22.0J) = 0 T = 273.18K non-spontaneous When ΔG crosses 0 -∆G Non-spontaneous at 0°C and lower temperatures 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 N2(g) + 3H2(g) → 2NH3(l) Q ∆G° = -33.0x103J PN2= PH2 = 10.0atm @ standard conditions New conditions 1 1   1.00x10 4 (PN 2 ) (PH 2 )3 (10.0atm) (10.0atm)3 ∆G = -33.0x103J + (8.314 J/K mol)(298.15K)(-9.210) = -55.8x103 J PN2= PH2 = 0.001atm New conditions Q 1 1   1.00x10 12 (PN 2 )(PH 2 )3 (0.00100atm)(0.00100atm)3 ∆G = -33.0x103J + (8.314 J/K mol)(298.15K)(27.631) = 3.5x104 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 ∆G°= - RT lnK 0 = ∆G° + RT lnK Whether a reaction ultimately reactant or product favored is based on ΔG°. ΔG° = -549 J H2O(s) → H2O (l) -549 J = - (8.314 J/K∙mol)(298.15K)(lnK) 0.2215 = lnK K= 1.25 C2H5OH (l) + 3O2(g) → 2CO2(g) + 3H2O(g) ∆G° = -1298.9 kJ -1298.9x103 J = - (8.314 J/K∙mol)(298.15K)(lnK) 524 = lnK K= 3.72x10227 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