Chemical Thermodynamics PDF

Chemical Thermodynamics 01006727 General Chemistry 01006708 Chemistry Brown Chemistry:The Central Science 13th Edition Chapter 19 (c) 2014  First Law of Thermodynamics Energy cannot be created or destroyed. Therefore, the total energy of the universe is a con...

Chemical Thermodynamics 01006727 General Chemistry 01006708 Chemistry Brown Chemistry:The Central Science 13th Edition Chapter 19 (c) 2014  First Law of Thermodynamics Energy cannot be created or destroyed. Therefore, the total energy of the universe is a constant. Energy can, however, be converted from one form to another or transferred from a system to the surroundings or vice versa. 2  Enthalpy/Entropy Enthalpy is the heat absorbed/released by a system during a constant-pressure process. Entropy is a measure of the randomness in a system. Both play a role in determining whether a process is spontaneous. 3  Spontaneous Processes Spontaneous processes proceed without any outside assistance. The gas in vessel A will spontaneously effuse into vessel B, but it will not spontaneously return to vessel A. Processes that are spontaneous in one direction are nonspontaneous in the reverse direction (i.e. un-breaking an egg) 4  Experimental Factors Affect Spontaneous Processes Temperature and pressure can affect spontaneity. An example of how temperature affects spontaneity is ice melting or freezing. 5 Reversible and Irreversible Processes Reversible process: The system changes so that the system and surroundings can be returned to the original state by exactly reversing the process. This maximizes work done by a system on the surroundings. Irreversible processes cannot be undone by exactly reversing the change to the system or cannot have the process exactly followed in reverse. Also, any spontaneous process is irreversible! 6  2nd Law of Thermodynamics Entropy can be thought of as a measure of the randomness of a system. It is a state function like ∆H: – It can be found by heat transfer from surroundings at a given temperature (i.e. add heat to melt ice – T constant at 273 K): The entropy of the universe increases in any spontaneous processes. – This results in the following relationships: 7  Third Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero is 0. Consider all atoms or molecules in the perfect lattice at 0 K; there will only be one microstate. 8  Entropy Predict whether ∆S is positive or negative Predict whether ∆S is positive or negative for for each process, assuming each occurs at each process, assuming each occurs at constant temperature: constant temperature: (a) H2O(l)→H2O(g) +ve (b) Ag+(aq) + Cl-(aq) → AgCl(s) -ve (c) 4 Fe(s) + 3 O2(g) → 2 Fe2O3(s) -ve (d) N2(g) + O2(g) → 2 NO(g) ~0 In each pair, choose the system that has greater entropy and explain your choice: (a) 1 mol of NaCl(s) or 1 mol of HCl(g) at 25 °C 1 mol of HCl(g) (b) 2 mol of HCl(g) or 1 mol of HCl(g) at 25 °C 2 mol HCl(g) (c) 1 mol of HCl(g) or 1 mol of Ar(g) at 298 K. 1 mol of HCl(g) 9  Entropy Elemental mercury is a silver liquid at room temperature. Its normal freezing point is -38.9 °C, and its molar enthalpy of fusion ∆Hfusion= 2.29 kJ/mol. What is the entropy change of the system when 50.0 g of Hg(l) freezes at the normal freezing point? 𝑄𝑄𝑟𝑟𝑟𝑟𝑟𝑟 ∆Hfusion ∆Sfusion= = 𝑇𝑇 𝑇𝑇 50.0 g q= *(-2.29 kJ/mol) = -0.570 kJ 200.59 g/mol T= (-38.9 + 273)K = 234.4 K −0.570 kJ ∆Sfusion= 234.4 K = -0.00244 kJ/K = -2.44 J/K 10  Statistical Thermodynamics Thermodynamics looks at bulk properties of substances (the big picture). – We have seen what happens on the molecular scale. So how do they relate? We use statistics (probability) to relate them. The field is called statistical thermodynamics. More “microstates” at higher – Microstate: A single possible temperature – more disorder arrangement of position and kinetic energy of molecules 11  Molecular Motions Molecules exhibit several types of motion. Translational: Movement of the entire molecule from one place to another Vibrational: Periodic motion of atoms within a molecule Rotational: Rotation of the molecule about an axis Note: More atoms means more possible molecular motions – and more microstates. 12  Entropy on the Molecular Scale The number of microstates and, therefore, the entropy tend to increase with increases in  temperature  volume  the number of independently moving molecules. 13  Effect of Volume and Temperature Change on the System If we increase volume, there are more positions possible for the molecules. This results in more microstates, so increased entropy. If we increase temperature, the average kinetic energy increases. This results in a greater distribution of molecular speeds. Therefore, there are more possible kinetic energy values, resulting in more microstates, increasing entropy. 14  Entropy and Physical States Entropy increases with the freedom of motion of molecules. S(g) > S(l) > S(s) Entropy of a system increases for processes where:  gases form from either solids or liquids.  liquids or solutions form from solids.  the number of gas molecules increases during a chemical reaction. 15  Standard Entropies The reference for entropy is 0 K, so the values for elements are not 0 J/mol K at 298 K. – Standard molar enthalpy for gases are generally greater than liquids and solids. (Be careful of size!) – Standard entropies increase with molar mass. – Standard entropies increase with number of atoms in a formula. 16  Entropy Changes Entropy changes for a reaction can be calculated in a manner analogous to that by which ∆H is calculated:  where n and m are the coefficients in the balanced chemical equation. ∆S° = Σn∆S°(products) – Σm∆S°(reactants) Heat that flows into or out of the system changes the entropy of the surroundings. For an isothermal process:  At constant pressure, qsys is simply ∆H° for the System. 17  Entropy & Free Energy Calculate the change in the standard entropy of the system, S°, for the synthesis of ammonia from N2(g) and H2(g) at 298 K: N2(g) + 3 H2(g) → 2 NH3(g) ∆𝑆𝑆 𝑜𝑜 = 2*S°(NH3) – [S°(N2) + 3*S°(H2)] ∆𝑆𝑆 𝑜𝑜 =2*(1192.5 J) - [1*(191.5 J/mol.K) +3*(130.6 J/mol/K)] ∆𝑆𝑆 𝑜𝑜 = -198.3 J/mol.K 18  Entropy Change in the Universe The universe is composed of the system & surroundings. Therefore, ∆Suniverse = ∆Ssystem + ∆Ssurroundings For spontaneous processes ∆Suniverse > 0 19  Total Entropy and Spontaneity ΔSuniverse = ΔSsystem + ΔSsurroundings Substitute for the entropy of the surroundings: ΔSuniverse = ΔSsystem – ΔHsystem/T Multiply by −T: −TΔSuniverse = −TΔSsystem + ΔHsystem Rearrange: −TΔSuniverse = ΔHsystem − TΔSsystem We call −TΔSuniverse the Gibbs Free Energy (ΔG): ΔG = ΔH − T ΔS 20  Gibbs Free Energy 1. If ∆G is negative, the forward reaction is spontaneous. 2. If ∆G is zero , the system is at equilibrium. 3. If ∆G is positive, the reaction is spontaneous in the reverse direction. 21  Standard Free Energy Changes Analogous to standard enthalpies of formation are standard free energies of formation, ∆Gf°: where n and m are the stoichiometric coefficients. ∆G° = Σn∆G°(products) – Σm∆G°(reactants) 22  Free Energy Changes Only reactions with ΔG < 0 are spontaneous so the sign of ΔH and ΔS are important and the magnitude of the T will affect spontaneity. ΔG = ΔH – TΔS How does ΔG change with temperature? – Increasing T will increase the spontaneity entropically favorable reactions (i.e. ΔS is +ve and -TΔS is –ve) - such as the dissolution of insoluble salts in water. – Increasing T does not increase the spontaneity of entropically unfavorable reactions 23  Free Energy Changes Calculate the standard free-energy change for the reaction below at 298 K given that H° = 180.7 kJ and S° = 24.7 J/K. Is the reaction spontaneous under these conditions? ∆G° = ∆H° - T∆S° 1 𝑘𝑘𝑘𝑘 = 180.7 kJ – (298 K)(24.7 J/K)* 1000 𝐽𝐽 N2(g) + O2(g) → 2 NO(g) = 180.7 kJ - 7.4 kJ = 173.3 kJ * The reaction is non-spontaneous at 298 K 24  ∆G° Reaction Calculate the standard free-energy change at 298K for the reaction below given the following data. What is ∆G° for the reverse of this reaction? ∆𝐺𝐺𝑓𝑓𝑜𝑜 𝑃𝑃𝑃𝑃𝑃𝑃3 = -269.6 kJ/mol ∆𝐺𝐺𝑓𝑓𝑜𝑜 𝑃𝑃4 = 24.4 kJ/mol P4(g) + 6 Cl2(g) → 4 PCl3(g) 𝑜𝑜 ∆𝐺𝐺𝑟𝑟𝑟𝑟𝑟𝑟 = 4 ∗ ∆𝐺𝐺𝑓𝑓𝑜𝑜 𝑃𝑃𝑃𝑃𝑃𝑃3 - [∆𝐺𝐺𝑓𝑓𝑜𝑜 𝑃𝑃4 + 6 ∗ ∆𝐺𝐺𝑓𝑓𝑜𝑜 𝐶𝐶𝐶𝐶2 ] = -4*269.6 kJ/mol - [4*(24.4 kJ/mol) - 6*(0 kJ/mol)] = -1102.8 kJ The reaction is spontanous in the forward direction 𝑜𝑜 The reverse reaction will have a ∆𝐺𝐺𝑟𝑟𝑟𝑟𝑟𝑟 = +1102.8 kJ 25  ∆G° Reaction Use Hess’s law to calculate ∆G° for the combustion of propane gas at 298 K: C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O(l) ∆H° = -2220 kJ ∆𝐺𝐺𝑓𝑓𝑜𝑜 𝐶𝐶𝐶𝐶2 = -394.4 kJ/mol ∆𝐺𝐺𝑓𝑓𝑜𝑜 𝐻𝐻2𝑂𝑂 =-237.13 kJ/mol ∆𝐺𝐺𝑓𝑓𝑜𝑜 𝐶𝐶3𝐻𝐻8 =-23.47 kJ/mol 𝑜𝑜 ∆𝐺𝐺𝑟𝑟𝑟𝑟𝑟𝑟 = [3 ∗ ∆𝐺𝐺𝑓𝑓𝑜𝑜 𝐶𝐶𝐶𝐶2 + 4 ∗ ∆𝐺𝐺𝑓𝑓𝑜𝑜 𝐻𝐻2 ] – [∆𝐺𝐺𝑓𝑓𝑜𝑜 𝐶𝐶3𝐻𝐻8 + 5 ∗ ∆𝐺𝐺𝑓𝑓𝑜𝑜 𝑂𝑂2 ] 𝑜𝑜 ∆𝐺𝐺𝑟𝑟𝑟𝑟𝑟𝑟 = [3*(-394.4 kJ/mol)+ 4*(- 237.13 kJ/mol)] – [1*(-23.47 kJ/mol) + 1*(0 kJ/mol)] 𝑜𝑜 ∆𝐺𝐺𝑟𝑟𝑟𝑟𝑟𝑟 = -2108 kJ * ∆Go is less negative that ∆Ho since more ordered liquids have been produced 26  ∆G: Temperature Effects The Haber process for the production of ∆𝐺𝐺𝑓𝑓𝑜𝑜 𝑁𝑁𝑁𝑁3 =∆𝐻𝐻𝑓𝑓𝑜𝑜 𝑁𝑁𝑁𝑁3 - 𝑇𝑇∆𝑆𝑆𝑓𝑓𝑜𝑜 𝑁𝑁𝑁𝑁3 ammonia involves the equilibrium. Assume that ∆H° and ∆S° for this reaction do not change with temperature. (a) Predict the direction in which G for the reaction changes with increasing temperature. (b) Calculate At 25 °C = 298 K ∆G at 25 °C and at 500 °C. = -92.38 kJ/mol – (298K)*(-0.198.3 kJ/K.mol) N2(g) + 3 H2(g) ⇋ 2 NH3(g) = -92.38 kJ/mol + 59.1 kJ = -33.3 kJ ∆𝐻𝐻𝑓𝑓𝑜𝑜 𝑁𝑁𝑁𝑁3 = -92.38 kJ/mol ∆𝑆𝑆𝑓𝑓𝑜𝑜 𝑁𝑁𝑁𝑁3 =-198.3 J/K.mol At 500 oC = 773K = -92.38 kJ/mol – (773K)*(-0.198.3 kJ/K.mol) = -92.38 kJ/mol + 153 kJ = -33.3 kJ 27