Assign.# 6.5 – 2nd and 3rd Law of Thermodynamics

Constraints of the 1st Law 1st Law and ΔH allows us to clearly study energy, but it has limitations. ΔH does not provide a good description of whether a reaction is likely to occur There are examples of both positive and negative ΔH values leading to reactions that are favorable.

Spontaneous Process Spontaneous A spontaneous process is one that proceeds without any outside assistance. A non-spontaneous process is one that requires assistance in order to occur Spontaneity does not mean a reaction occurs “fast”. Only tells us the energy of a reaction Spontaneous Provide examples of how water freezing can be both spontaneous and non-spontaneous based on the temperature. Additionally, a gas filling a room is a spontaneous process. Non – Spontaneous

Endothermic and Spontaneous Why are reactions/processes that have a +ΔH able to occur (e.g. reactions that feel cold)? Only reactions that have –ΔH should be favored by the universe The answer is due to: Entropy!!!

Reversible vs. Irreversible Processes Reversible Process is any process that can be restored to its original state without no overall net change. Irreversible Process is a process that cannot be simply reversed to return system to its original state Irreversible Process Changes in temperature are irreversible processes Hot Cold Not Possible!

Isothermal Process

Major Idea All real processes are irreversible Therefore, all spontaneous processes are irreversible

ΔS and Spontaneity Need to be able to characterize a reaction that we are unfamiliar with as spontaneous or non-spontaneous Entropy is a state function so: ΔS = Sfinal – Sinitial For a process under constant pressure, ΔS can be calculated by: qrev represents the heat required to make the process reversible

ΔS and Phase Changes Phase changes are isothermal processes. Imagine examining the melting of a substance. Therefore, we can say: qrev = ΔHfusion This means that for phase changes:

Example-1 Elemental mercury is a silver liquid at room temperature. Its normal freezing point is -38.9 oC, and its molar enthalpy of fusion is ΔHfusion = 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?

Example-2 The normal boiling point of ethanol, C2H5OH, is 78.3 oC, and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the change in entropy in the system when 68.3 g of C2H5OH (g) at 1 atm condenses to liquid at the normal boiling point?

ΔS and the 2nd Law of Thermodynamics Entropy is not conserved (unlike energy in the 1st law) Example: 1 mole of water freezes at T < 0 oC H2O (l)  H2O (s) What is ΔSsys? Assume the surroundings are around 310 K. We know that Sliquid > Ssolid Therefore ΔSsys < 0 because solid has less disorder than liquid This also means that ΔHsys < 0 because heat was released to the surroundings This ΔHsys leads to an increase in the overall entropy of the surroundings. If you were to calculate the ΔS for both system and surrounding, then you would discover that: ΔSsys + ΔSsurr = ΔSuniversie > 0 for all spontaneous processes THE UNIVERSE FAVORS DISORDER!!!!

ΔSsurroundings and ΔHsystem As we lose energy, entropy increases. So, we can say that: Furthermore, we know that entropy increases as T increases, so: Therefore, it is possible to conclude that:

Two Important Equations When does S equal 0? When does ΔSsurr equal 0? There is no entropy when there is only one microstate that exists ΔS = 0 when the temperature reaches 0 K. This means that at 0 K there is no entropy because there is only one microstate that exists.

3rd Law of Thermodynamics The 3rd Law of Thermodynamics claims that the entropy of a pure crystalline substance at absolute zero is 0. There is only one possible microstate because all motion has ceased, which means there is 0 entropy in the system. This is one of the proofs that absolute zero is a quantity that must exist.

Entropy Change for Chemical Reactions We use calorimetry to measure ΔH We do not have a nice way to measure ΔS in a reaction We do know that every substance must have a point where S = 0 due to the 3rd Law As temperature increases from 0 K, the value of S also increases. We therefore can determine standard entropies (So) Standard Entropies is the entropy gained by taking 1 mol of perfect crystalline substance at 0 K and bringing it to standard conditions Measured in J/(mol x K)

Important Points on So Unlike ΔHf, So is not zero for elements So is greatest in gas, than liquids, and least in solids So increases as substances have larger molar masses So increases as the number of atoms in a chemical formula increases Increased number of possible microstates

Calculating ΔSo We can use So values to calculate ΔSo through the equation:

Example-3 For the reaction: N2 (g) + 3 H2 (g)  2 NH3 (g) Calculate ΔSo for the reaction. You know that So (N2) = 191.5 J/(mol x K), that So (H2) = 130.6 J/(mol x K), and that So (NH3) = 192.5 J/(mol x K).

Al2O3 (s) + 3 H2 (g)  2 Al (s) + 3 H2O (g) Example-4 Calculate ΔSo for the following reaction: Al2O3 (s) + 3 H2 (g)  2 Al (s) + 3 H2O (g) You know that So (Al2O3) = 51.00 J/ (mol x K), So (H2) = 130.6 J/(mol x K), So (Al) = 28.32 J/(mol x K), and So (H2O ) = 188.8 J/(mol x K).

Entropy Changes and Spontaneity We now have a tool to calculate ΔSsystem, but how do we combine this with ΔSsurroundings and ΔSuniverse? We know that: We then can add surroundings and system to determine if the reaction is spontaneous.

Example-5 For the reaction: N2 (g) + 3 H2 (g)  2 NH3 (g) Determine whether or not the reaction is spontaneous under standard conditions. You know that ΔHf for NH3 is -46.19 kJ. Additionally you know that So (N2) = 191.5 J/(mol x K) So (H2) = 130.6 J/(mol x K) So (NH3) = 192.5 J/(mol x K).

Al2O3 (s) + 3 H2 (g)  2 Al (s) + 3 H2O (g) Example-6 For the following reaction: Al2O3 (s) + 3 H2 (g)  2 Al (s) + 3 H2O (g) Determine whether or not the reaction is spontaneous under standard conditions. You know that ΔHf (Al2O3) = -1669.8 kJ/mol and ΔHf (H2O) = -241.82 kJ/mol You know that: So (Al2O3) = 51.00 J/ (mol x K) So (H2) = 130.6 J/(mol x K) So (Al) = 28.32 J/(mol x K) So (H2O ) = 188.8 J/(mol x K).

Question 1 Which molecule should have a higher So? Why? C2H6 or C2H2

Question 2 Calculate ΔSo for the following chemical reaction: C2H4 (g) + H2 (g)  C2H6 (g) So (C2H4) = 219 J/ (mol x K) So (H2) = 114.60 J/ (mol x K) So (C2H6) = 229.5 J/ (mol x K)

Question 3 Calculate ΔSo for the following chemical reaction: N2O4 (g)  2NO2 So (N2O4) = 304.3 J/ (mol x K) So (NO2) = 240.45 J/ (mol x K)

Be(OH)2 (s)  BeO (s) + H2O (g) Question 4 Determine if the following reaction is spontaneous at 298 K: Be(OH)2 (s)  BeO (s) + H2O (g) So (Be(OH)2) = 50.21 J/ (mol x K) So (BeO) = 13.77 J/ (mol x K) So (H2O) = 188.83 J/ (mol x K) ΔHf (Be(OH)2) = -905.8 kJ/mol ΔHf (BeO) = -608.5 kJ/mol ΔHf (H2O) = -241.82 kJ/mol

3 CH3OH (g) + O2 (g)  2 CO2 (g) + 4 H2O (g) Question 5 Determine if the following reaction is spontaneous at 298 K: 3 CH3OH (g) + O2 (g)  2 CO2 (g) + 4 H2O (g) So (CH3OH) = 237.6 J/ (mol x K) So (O2) = 205.0 J/ (mol x K) So (H2O) =188.83 J/ (mol x K) So (CO2) = 213.6 J/ (mol x K) ΔHf (CH3OH) = -201.2 kJ/mol ΔHf (CO2) = -393.5 kJ/mol ΔHf (H2O) = -241.82 kJ/mol

Practice 6.5 Page 285 #59, 60, 61, 62