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Thermodynamics vs. Kinetics

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Presentation on theme: "Thermodynamics vs. Kinetics"— Presentation transcript:

1

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

5 Spontaneous Physical and Chemical Processes
A waterfall runs downhill A lump of sugar dissolves in a cup of coffee At 1 atm, water freezes below 0 0C and ice melts above 0 0C Heat flows from a hotter object to a colder object A gas expands in an evacuated bulb Iron exposed to oxygen and water forms rust spontaneous nonspontaneous 18.2

6 Does a decrease in enthalpy mean a reaction proceeds spontaneously?
Spontaneous reactions CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH0 = kJ H+ (aq) + OH- (aq) H2O (l) DH0 = kJ H2O (s) H2O (l) DH0 = 6.01 kJ NH4NO3 (s) NH4+(aq) + NO3- (aq) DH0 = 25 kJ H2O 18.2

7 Entropy (S) is a measure of the randomness or disorder of a system.
DS = Sf - Si If the change from initial to final results in an increase in randomness Sf > Si DS > 0 For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than gas state Ssolid < Sliquid << Sgas H2O (s) H2O (l) DS > 0 18.3

8 Processes that lead to an increase in entropy (DS > 0)
18.2

9 How does the entropy of a system change for each of the following processes?
(a) Condensing water vapor Randomness decreases Entropy decreases (DS < 0) (b) Forming sucrose crystals from a supersaturated solution Randomness decreases Entropy decreases (DS < 0) (c) Heating hydrogen gas from 600C to 800C Randomness increases Entropy increases (DS > 0) (d) Subliming dry ice Randomness increases Entropy increases (DS > 0) 18.3

10 Entropy Changes in the System (DSsys)
When gases are produced (or consumed) If a reaction produces more gas molecules than it consumes, DS0 > 0. If the total number of gas molecules diminishes, DS0 < 0. If there is no net change in the total number of gas molecules, then DS0 may be positive or negative BUT DS0 will be a small number. What is the sign of the entropy change for the following reaction? 2Zn (s) + O2 (g) ZnO (s) The total number of gas molecules goes down, DS is negative. 18.4

11 Entropy State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. energy, enthalpy, pressure, volume, temperature , entropy Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. 18.3

12 18.4 First Law of Thermodynamics
Energy can be converted from one form to another but energy cannot be created or destroyed. Second Law of Thermodynamics The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process. Spontaneous process: DSuniv = DSsys + DSsurr > 0 Equilibrium process: DSuniv = DSsys + DSsurr = 0 18.4

13 Entropy Changes in the System (DSsys)
The standard entropy of reaction (DS0 ) is the entropy change for a reaction carried out at 1 atm and 250C. rxn aA + bB cC + dD DS0 rxn dS0(D) cS0(C) = [ + ] - bS0(B) aS0(A) DS0 rxn nS0(products) = S mS0(reactants) - What is the standard entropy change for the following reaction at 250C? 2CO (g) + O2 (g) CO2 (g) S0(CO) = J/K•mol S0(CO2) = J/K•mol S0(O2) = J/K•mol DS0 rxn = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)] DS0 rxn = – [ ] = J/K•mol 18.4

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

15 Copyright © Cengage Learning. All rights reserved
ΔSsurr The sign of ΔSsurr depends on the direction of the heat flow. The magnitude of ΔSsurr depends on the temperature. Copyright © Cengage Learning. All rights reserved

16 Copyright © Cengage Learning. All rights reserved
ΔSsurr Copyright © Cengage Learning. All rights reserved

17 Entropy Changes in the Surroundings (DSsurr)
Exothermic Process DSsurr > 0 Endothermic Process DSsurr < 0 18.4

18 Copyright © Cengage Learning. All rights reserved
ΔSsurr Copyright © Cengage Learning. All rights reserved

19 Copyright © Cengage Learning. All rights reserved
ΔSsurr Heat flow (constant P) = change in enthalpy = ΔH Copyright © Cengage Learning. All rights reserved

20 Copyright © Cengage Learning. All rights reserved

21 18.3 Third Law of Thermodynamics
The entropy of a perfect crystalline substance is zero at the absolute zero of temperature. S = k ln W W = 1 S = 0 18.3

22 For a constant-temperature process:
Gibbs Free Energy Spontaneous process: DSuniv = DSsys + DSsurr > 0 Equilibrium process: DSuniv = DSsys + DSsurr = 0 For a constant-temperature process: Gibbs free energy (G) DG = DHsys -TDSsys DG < The reaction is spontaneous in the forward direction. DG > The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction. DG = The reaction is at equilibrium. 18.5

23 The standard free-energy of reaction (DG0 ) is the free-energy change for a reaction when it occurs under standard-state conditions. rxn aA + bB cC + dD DG0 rxn dDG0 (D) f cDG0 (C) = [ + ] - bDG0 (B) aDG0 (A) DG0 rxn nDG0 (products) f = S mDG0 (reactants) - Standard free energy of formation (DG0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. f DG0 of any element in its stable form is zero. f 18.5

24 What is the standard free-energy change for the following reaction at 25 0C?
2C6H6 (l) + 15O2 (g) CO2 (g) + 6H2O (l) DG0 rxn nDG0 (products) f = S mDG0 (reactants) - DG0 rxn 6DG0 (H2O) f 12DG0 (CO2) = [ + ] - 2DG0 (C6H6) DG0 rxn = [ 12x– x–237.2 ] – [ 2x124.5 ] = kJ Is the reaction spontaneous at 25 0C? DG0 = kJ < 0 spontaneous 18.5

25 DG = DH - TDS 18.5

26 Gibbs Free Energy and Chemical Equilibrium
DG = DG0 + RT lnQ R is the gas constant (8.314 J/K•mol) T is the absolute temperature (K) Q is the reaction quotient At Equilibrium DG = 0 Q = K 0 = DG0 + RT lnK DG0 = - RT lnK 18.6

27 DG0 = - RT lnK 18.6

28 Copyright © Cengage Learning. All rights reserved
Maximum possible useful work obtainable from a process at constant temperature and pressure is equal to the change in free energy. wmax = ΔG Copyright © Cengage Learning. All rights reserved

29 Copyright © Cengage Learning. All rights reserved
Achieving the maximum work available from a spontaneous process can occur only via a hypothetical pathway. Any real pathway wastes energy. All real processes are irreversible. First law: You can’t win, you can only break even. Second law: You can’t break even. As we use energy, we degrade its usefulness. Copyright © Cengage Learning. All rights reserved

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