Entropy is a measure of the number of equivalent ways in which a system may exist.

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Paul Allen
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1 Chapter 17

2 1 ENTROPY (S) Entropy is a measure of the number of equivalent ways in which a system may exist. S = Units: -23 k = Boltzmann constant ( 1.38 x 10 J/K) Ù = number of equivalent ways for the system to exist Example 1: Consider coins being tossed: state H - H T - T H - T 4 H 4 T 3 H - T 3 T - H 2 H - 2 T ways of producing state entropy (J/K) The entropy of a system that has only one way of existing is. The state that is most likely to be produced is the one that has the ways of occurring. This event is the one with the entropy. Entropy is a measure of randomness. High entropy means ways for the system to exist and a high degree of randomness. Demonstration: Coin Toss Simulation

3 2 As the number of object increase, the chance that the system will be in either the highest entropy case (fraction of heads = ) or one very close to it is almost. The chances that 80 coins are tossed and they all end up heads is one chance in 2 x N a. IMPORTANT POINT: Changes occur in nature in such a way that the final state produced has ways of occurring than any other possible final state. SECOND LAW OF THERMODYNAMICS (A LAW OF PROPHECY) The entropy of the is always except at equilibrium when it is.

4 3 The universe can be thought of as a system and the surroundings: ÄS universe = The entropy of the system can decrease if. FACTORS THAT INFLUENCE ENTROPY OF THE SYSTEM 1. Temperature If I have one, $10 bill to give away, how many ways can the money be distributed within the class? If I have $1000 (in $10 bills) how many ways can the money be distributed? As the temperature increase, the entropy of the system 2. Number of Atoms 1 mole Mg (s) 2 mole Mg (s) As the number of atoms in the sample increases, the entropy. 3. Size or Mass of Atoms 1 mole O 2 1 mole of Cl 2 The larger or heavier the particle the the entropy. 4. Molecular Structure 2 moles NO 2(g) 1 mole N2O 4(g) For the same number of atoms, one larger molecule has entropy than several smaller ones. 5. Physical State 1 mole H2 O (l) 1 mole H2 O (g) 2 (s) A gas has entropy than a liquid. There is no value for H O listed in the table because. If it did the solid would have an entropy that is than the liquid.

5 4 Example: o Predict the signs of ÄS for these reactions. 6. Volume and Concentration The larger the volume of the system or the lower the concentration, the the entropy. Concentration plays a critical role in determining whether a chemical process will be possible or not. CALCULATING ENTROPY CHANGES ÄS univ is calculated from: The surroundings are usually defined such that the only entropy changes occurring in the surroundings are due to the transfer of energy. If the temperature of the surroundings are constant, the change in entropy in the surroundings can be calculated from: During an exothermic reaction the entropy in the surroundings. This is the reason why most observed reactions are exothermic. When the same amount of energy is transferred to the surroundings it will produce a larger entropy change if the temperature of the surroundings is. ÄS sys is a state function and can be calculated by a variety of ways. Whenever the symbol ÄS is used it is assumed to be ÄS FROM TABLE OF ENTROPY OF FORMATION

6 5 Example 1: Calculate the ÄS for the following reaction at 298 K. ADDING REACTIONS 1. If equations are added the ÄS values are 2. If an equation is reversed, the sign of ÄS. 3. If an equation is multiplied by a constant, the value of ÄS. FREE ENERGY Calculating entropy changes in the universe to predict whether a process is possible sounds a bit arrogant. So a new function is defined to do the same thing but sounds more humble. ÄS univ = ÄS sys + ÄS surr -TÄS univ = ÄH - TÄS (constant P,T) Define -TÄS univ to be the change in free energy ( Ä ), the energy that is free ( ) to do or transfer to other reactions to allow them to occur. Under standard conditions: ÄH and ÄS change much with T, ÄG change with T. Very important to remember.

7 6 THERMODYNAMIC PROPERTIES AT 298 K o o o f f substance ÄH (kj/mol) S (J/mol*K) ÄG (kj/mol) C graphite C diamond CO 2(g) CH 4(g) C2H5 OH (l) Cl 2(g) Cl (aq) HCl (g) H (g) H 2(g) H (aq) OH (aq) H2 O (l) H2 O (g) KClO 3(s) Mg (s) Mg (aq) MgO (s) MgCl 2(s) NO 2(g) N2O 4(g) NaHCO 3(s) Na2CO 3(s) O (g) O 2(g) O 3(g) P (white) P (red)

8 7 SIGN CONVENTIONS type of change means ÄS universe ÄG spontaneous non-spontaneous equilibrium As time goes on, the entropy of the universe is and the amount of free energy in the universe is. The energy is still in the universe, it is just. SPONTANEOUS Spontaneous means it does not imply. Thermodynamics tells us if the reaction is. Kinetics tells us if the spontaneous reaction is fast or slow. Why would a spontaneous reaction be slow?. The magnitude of ÄG has no bearing on the rate of a spontaneous reaction. What does it mean when ÄG becomes more negative? reaction relative rate Fast Slow The values of Ä G for the reactions are above are kj/mol and -822 kj/mol. Which o 298 K reaction has the value kj/mol?

9 8 CALCULATION OF ÄG ÄG is a state function so its value depends only on the initial and final state. 1. From Table of ÄGf ÄGf for elements in standard states are defined as. WARNING: If tabulated values of ÄGf are used then the temperature must be Example : Calculate the ÄG for the reaction below at 298 K. C2H5 OH (l) + 3 O 2 (g) > 2 CO 2 (g) + 3 H2 O(l) 2. FROM ÄH, ÄS WARNING: The most common mistake in this method is to forget. The main advantage of this method is that it allows the calculation of ÄG at. o o 298 K Example 1: Calculate the ÄG at 50 C for the reaction above. ÄH = kj and o 298 K ÄS is on page ADDING REACTIONS If equation is reversed the sign of ÄG is. If the coefficients in the equation are doubled, ÄG is. When equations are added, the ÄG 's.

10 9 o 298 Example: Calculate the value of Ä G for this reaction given the information below. TEMPERATURE - THE GREAT MEDIATOR ÄH ÄS ÄG spontaneous ( )@ Example 1: conditions? At what temperature does the following process become possible under standard H2 O (l) > H2 O (g) The point at which the reaction becomes spontaneous occurs when ÄG changes from to, or, in other words, is. This would be called a state of and it is unlikely that this would be at 298 K. ÄH = ÄS = = kj

11 10 COUPLED REACTIONS Demonstration: apparatus. Test a mixture of magnesium oxide and water for ions with a conductivity MgO (s) + H2 O (l) > Mg (aq) + 2 OH (aq) ÄG = 36.2 kj Observations: meaning Are results consistent with ÄG? + Add acid (H ). Observations: MgO (s) + H2 O (l) < Mg (aq) + 2 OH (aq) ÄG = 36.2 kj H (aq) + 2OH (aq) > 2 H2 O (l) ÄG = kj 298 ÄG = IMPORTANT POINTS: 1. Non-spontaneous reactions can be made to occur by coupling them to a one. 2. The of the spontaneous reaction is used by the non-spontaneous one in order to occur. 3. Coupling occurs whenever two equations have a chemical. ÄG vs ÄG The symbol means. ÄG is the change in free energy for WHEN THE CONCENTRATIONS OF REACTANTS AND PRODUCTS ARE M. If ÄG is (-) then the reaction is spontaneous; ÄG is (+) the reaction is spontaneous IF THE CONCENTRATIONS OF REACTANTS AND PRODUCTS ARE 1 M.

12 11 At equilibrium = 0. ÄG is the change in free energy when the concentrations are. ÄG and ÄG are related by: At equilibrium, Q = and ÄG = so: Which equilibrium constant ( K or K p ) is it? Since ÄG is for standard states and the standard state concentration of a gas is measured as, it is if gases are present or if there are no gases present in the chemical equation. Example: Calculate the equilibrium constant for this reaction at 298 K. Which equilibrium constant did you calculate? o298k Example : Calculate the equilibrium constant for a reaction whose ÄG = -50 kj Conclusion: o Reactions that have large negative ÄG have K values and proceed in basically just one direction. Such reactions are often called irreversible since the equilibrium strongly favors the formation of the products. 298K -8 Example : Calculate ÄG if the concentration of the ions are each 1 x 10 M. The o298k ÄG for the reaction is 79.8 kj/mol H2 O (l) H3 O (aq) + OH (aq) Concentration plays a critical role in determining whether or not a process is possible and how much free energy can be released or will be required to make the process possible.

13 o Example: Look on page 9. Can any water evaporate at 25 C? If the pressure of the gaseous water is 5 mm, will more liquid water evaporate? What is the maximum o pressure of the gaseous water at 25 C? 12 EFFECT OF TEMPERATURE ON K slope = intercept = o298 K, o 298 K o298 K Example: Calculate ÄG, ÄH, and ÄS.

Lecture #13. Chapter 17 Enthalpy and Entropy

Lecture #13. Chapter 17 Enthalpy and Entropy Lecture #13 Chapter 17 Enthalpy and Entropy First Law of Thermodynamics Energy cannot be created or destroyed The total energy of the universe cannot change Energy can be transferred from one place to