CEM 152 1
Reaction Spontaneity Can we learn anything about the probability of a reaction occurring based on reaction enthaplies? in general, a large, negative reaction enthalpy is indicative of a spontaneous reaction however, there is more to determining whether a reaction is spontaneous Example: Vaporization of water the phase change of water from a gas to liquid is exothermic H 2 O(g) H 2 O( ) DH = -44 kj and occurs spontaneously at low temperatures Example: Ammonium Chloride in water the reverse reaction is endothermic H 2 O( ) H 2 O(g) DH = +44 kj and occurs spontaneously above 100 o C Thermochemistry CEM 152
Reversible - Irreversible Reversible process System and surroundings can be restored to original states by exactly reversing the change. A reversible process reverses direction whenever an infinitesimal change is made to the system. Reversible processes produces the maximum amount of work on the surroundings. Irreversible process Reversing the change to the system does not restore it to its original state. Real processes can approximate reversible processes. CEM 152 3
Spontaneous Processes A spontaneous process is one which occurs without any external intervention snow water above 32 o C iron rust over time acid + base neutral system expansion of a gas to fill a volume Spontaneous processes have an inherent direction Spontaneous processes are irreversible In any spontaneous chemical process, the path between reactants and products is irreversible The spontaneity of a reaction will depend on the temperature of the system CEM 1524
Equilibrium Processes When we discussed the phase boundaries of a phase diagram, the boundary lines represented two phases in equilibrium For water at 0 o C and 1 atm pressure, the liquid and solid phases are in equilibrium to go from water to ice, a certain amount of heat (DH vap ) must be removed from the system to go from ice to water, a certain amount of heat (DH vap ) must be added to the system Equilibrium processes are reversible processes the system can go back and forth between states along exactly the same path CEM 1525
Entropy Entropy is a measure of the disorder of a system the more disordered a system, the larger its entropy Ludwig Boltzmann derived a quantitative relation for the entropy of any system S = entropy = k ln W k is the Boltzmann constant k = R/N A k = (8.314 J/molK)/(6.022 x 10 23 /mol) k = 1.38 x 10-23 J/K W is the number of possible arrangements of the system If W is large, the entropy will be large CEM 1526
Change in Entropy The change in entropy of a system is given by DS = S final - S initial both S and DS are state functions a positive DS indicates increased disorder a negative DS indicates decreased disorder For a reversible process at constant temperature DS = q rev /T For the isothermal expansion of an ideal gas DS= nr ln (V 2 /V 1 ) CEM 1527
Determination of Entropy I Example: Mercury freezes at -38.9 o C and its enthalpy of fusion is 2.33 kj/mol. Calculate the value of DS for the freezing of 1.00 mol Hg( ) at -38.9 o C. DS = q rev /T q ~ (DH fusion )(mol Hg( )) DH fusion = -2.33 kj/mol q ~ (-2.33 kj/mol)(1.00 mol) = -2.33 kj T = -38.9 o C + 273.15 = 234.25 K DS = (-2.33 kj)/(234.25 K) DS = -0.00994 kj/k DS = -9.94 J/K The fact that DS is negative reflects that the final system is more ordered CEM 1528
Determination of Entropy II Example: Calculate the entropy change when 1.5 mol of an ideal gas expands at constant temperature from 20.0 L to 90.0 L. DS= nr ln (V 2 /V 1 ) n = 1.5 mol R = 8.314 J/Kmol ln (90.0 L/20.0 L) = +1.504 DS = (1.5 mol)(8.314 J/Kmol)(+1.504) DS = +18.8 J/K The positive sign of DS reflects that the final system is more disordered. For the reverse process (which is not spontaneous) ln (20.0 L/90.0 L) = -1.504 DS = -18.8 J/K CEM 1529
Second Law of The total change in entropy of the universe DS univ can be expressed as DS univ = DS sys + DS surr DS sys is the entropy change of the system DS surr is the entropy change of the surroundings The second law of thermodynamics states in any reversible process DS univ = DS sys + DS surr = 0 in any irreversible process DS univ = DS sys + DS surr > 0 Unlike energy and mass, entropy is not conserved in a spontaneous chemical reaction CEM 15210
Entropy The synthesis of glucose from CO 2 and H 2 O and the synthesis of proteins directly from amino acids are both non-spontaneous processes under standard conditions. Yet it is necessary for these to exist, how is this possible? CEM 152 12
Enthalpies If the internal energy of a thermodynamic system is increased by 300 J while 75 J of expansion work is done, how much heat was transferred and in which direction? A 375 B 300 C 225 D 75 E none of the above Thermochemistry CEM 152
Molecules in Motion From Boltzmann s relation for the entropy of any system S = entropy = k ln W the entropy will depend on the number of arrangements a system may possess The entropy of a molecular system will be governed by the number of degrees of freedom associated with individual molecules The rotational, vibrational, and translational motions of a molecule provide a means for storing energy when the thermal energy of a system is increased, the energy stored by the system increases: entropy increases CEM 15214
Entropy Exercise Example: Identify which of the entropy expressions below are true. 1. 1 mol H 2 (g) at STP > 1 mol H 2 (g) at 100 o C and 1.0 atm? 2. 1 mol H 2 O(s) at 0 o C > 1 mol H 2 O( ) at 0 o C? 3. 1 mol H 2 (g) at STP > 1 mol SO 2 (g) at STP? 4. 1 mol of N 2 O 4 (g) at STP > 1 mol NO 2 (g) at STP A 2 B 4 C 1, 2 D 2, 3 E 3, 4 CEM 15215
Third Law of The third law of thermodynamics states the entropy of a pure crystalline substance at absolute zero is zero S(0K) = 0 J/K The system in this instance reaches the point of perfect order the solid has zero degrees of freedom the solid exhibits no disorder When the temperature is raised above absolute zero, the vibrational motion of the molecules will destroy the perfect order of the system the entropy of the system will increase from S = 0 with an increase in thermal energy CEM 15216
Calculation of Entropy Changes The absolute entropy for many substances at any temperature have been determined using experimental measurements of heat capacity as a function of temperature The molar entropy values for substances in their standard states are known as standard molar entropies, S o defined for a pure substance at 1 atm S o largest for gases S o not zero for elements in standard state The entropy change in a chemical reaction is given by o o DS ns (products) - ms o ( reactants) CEM 15217
Entropy Change Example Example: Calculate the standard entropy change for the reaction below at 298 K Al 2 O 3 (s) + 3H 2 (g) 2Al(s) + 3H 2 O(g) DS S o (298K), Al 2 O 3 (s) = 51.00 J/K S o (298K), H 2 (g) = 130.58 J/K S o (298K), Al(s) = 28.32 J/K S o (298K), H 2 O(g) = 188.83 J/K o o ns (products) - ms DS o = {2(28.43 J/K) + 3(188.83 J/K)} - {1(51.00 J/K) + 3(130.58 J/K)} DS o = {623.35 J/K} - {442.74 J/K} DS o = +180.62 J/K o ( reactants) CEM 15218
Entropy Which of the following reactions will have a positive entropy change? 1. 2H 2 S (g) + SO 2 (g) 3S (s) + 2H 2 O (g) 2. 2SO 3 (g) 2SO 2 (g) + O 2 (g) 3. H 2 (g) + ½ O 2 (g) H 2 O (l) 4. 2CH 3 OH (g) + 3O 2 (g) 2CO 2 (g) + 4H 2 O (g) A 2 B 4 C 1 D 2, 4 E 1, 3 CEM 152 19
Reaction Spontaneity In the discussion of entropy, we have observed that reactions which increase the disorder of the universe are spontaneous We also observed in our discussion of thermochemistry that reactions which have large, negative enthalpy values tend to be spontaneous The combination of DS and DH for a given reaction can be used to predict the spontaneity of any chemical reaction J. Willard Gibbs derived a new state function G, the free energy G = H - TS where H is enthalpy, S is entropy, and T is absolute temperature CEM 15220
Gibbs Free Energy For a process occurring at constant temperature DG = DH - TDS The change in the free energy of a system can provide valuable information of the spontaneity of a chemical reaction Under constant pressure and temperature conditions if DG is negative, reaction is spontaneous in the forward direction if DG is zero, the reaction is at equilibrium if DG is positive, the forward reaction is not spontaneous and work must be supplied to make it occur CEM 15221
Standard Free Energies of Formation Similar to the case of standard enthalpies of formation and standard entropies, the standard free energies of formation, DG o f are tabulated for many substances (see Appendix C of BLB) The standard free energy change for a reaction can be determined using these standard free energies of formation DG o o f ng (products) - mg o f ( reactants) Such a calculation can reveal the spontaneity of a reaction under standard conditions (T = 298 K, P = 1atm) if DG o < 0, forward reaction spontaneous if DG o > 0, reverse reaction spontaneous CEM 15222
Free Energy Determination Example: Calculate the change in free energy for the following reaction at 298 K 2CH 3 OH( ) + 3O 2 (g) 2CO 2 (g) + 4H 2 O(g) DG DG o f(298k) CH 3 OH( ) = -166.23 kj/mol DG o f(298k) O 2 (g) = 0 kj/mol DG o f(298k) CO 2 (g) = -394.4 kj/mol DG o f(298k) H 2 O (g) = -228.57 kj/mol o o f ng (products) - mg o f ( reactants) DG o = {2(-394.4 kj/mol) + 4(-228.57 kj/mol)} - {2(-166.23 kj/mol) + 3(0 kj/mol)} DG o = {-1703.08 kj} - {-332.46 kj} DG o = -1370.62 kj CEM 15223
Free Energy and Temperature From the free energy relation DG = DH - TDS temperature has an influence on the free energy, and spontaneity, of any chemical process DH DS DG Reaction proceeds - + + - Always - Always + - - - if low T + + + if low T Reaction spontaneous at all temperatures Reaction nonspontaneous at all temperatures Reaction spontaneous at low temperatures Reaction spontaneous at high temperatures CEM 15224
Temperature and Reaction Spontaneity Example: Using standard enthalpies of formation and standard entropies calculate DH o and DS o at 298 K for the reaction 2SO 2 (g) + O 2 (g) 2SO 3 (g) and use these values to determine the DG o of this reaction at 298 K DH o f (298K) SO 2 (g) = -296.9 kj/mol DH o f (298K) O 2 (g) = 0 kj/mol DH o f (298K) SO 3 (g) = -395.2 kj/mol S o (298K) SO 2 (g) = 248.5 J/molK S o (298K) O 2 (g) = 205.0 J/molK S o (298K) SO 3 (g) = 256.2 J/molK A B C D E -253 kj -140 kj DG = DH - TDS DG = -196.6 140 kj kj - (298K)(-189.6 J/K) 253 kj 56304 kj DG = -140.0 kj CEM 15225
Vaporization of Water I Example: The phase change of water from a gas to liquid is exothermic H 2 O(g) H 2 O( ) DH = -44 kj and occurs spontaneously at low temperatures. The reverse reaction is endothermic H 2 O( ) H 2 O(g) DH = +44 kj and occurs spontaneously above 100 o C. Explain why this is so. DS S o (298K) H 2 O(g) = 188.83 J/molK S o (298K) H 2 O( ) = 69.91 J/molK o o ns (products) - ms DS o = {1(69.91 J/molK)} - {1(188.83 J/molK)} DS o = -118.92 J/K o ( reactants) CEM 15226
Boiling Point Estimate the boiling point of Br 2 (l). Assume that DH o and DS o do not vary as a function of temperature. A -330 K B -710 K C 0.33 K D 330 K E 710 K DH f o (kj/mol) DS o (J/mol K) Br 2 (l) 0 152.3 Br 2 (g) 30.71 245.3 CEM 152 28
Gibbs Free Energy: Example For the Haber process N 2 (g) + 3 H 2 (g) 2 NH 3 (g) What is DG o at 298 K? At what temperature is the reaction at equilibrium? Assume that DH o and DS o do not change with temperature. A 466 K B 117 K C 2.14 K D 0.466 K E 0.002 K DH o f (kj/mol) DS o (J/mol K) N 2 (g) 0 191.5 H 2 (g) 0 130.58 NH 3 (g) -46.19 192.5 CEM 152 29
Gibbs Free Energy and Equilibrium Most chemical reactions occur under non-standard conditions. To relate DG to DG o DG = DG o + RT ln Q For a system at equilibrium, DG = 0. At equilibrium Therefore 0 = DG o + RT ln K DG o = RT ln K CEM 152 30
Gibbs Free Energy and Equilibrium: Example For the Haber process N 2 (g) + 3 H 2 (g) 2 NH 3 (g) If a container at 400 K held an equilibrium mixture of 0.5 atm N 2, 0.75 atm H 2, and 2.0 atm NH 3, determine DG o. CEM 152 31
Gibbs Free Energy and Equilibrium: Example For the weak acid HNO 2, K a is 4.5*10-4. What is the value of DG o at equilibrium (298 K)? A -5.6 kj B -13.8 kj C -19.1 kj D 19.1 kj E 13.8 kj CEM 152 32
ln K Equilibrium constant Shown is a plot of ln(k) for a reaction at equilibrium as a function of temperature. What is the value of DS o? A 3.2 J/K B 31 J/K C huh? D 330 J/K E 3390 J/K 40 30 20 10 0 1 2 3 CEM 152 1000/ T (K) 33