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Predicting and Calculating Entropy 15.3.1 15.3.2 15.3.3.

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Presentation on theme: "Predicting and Calculating Entropy 15.3.1 15.3.2 15.3.3."— Presentation transcript:

1 Predicting and Calculating Entropy 15.3.1 15.3.2 15.3.3

2 The Laws of Thermodynamics  The First Law The total energy of the universe (which hates you) is constant. ○ This is similar to the law of conservation of energy. It can be written as ○ ΔE universe = ΔE system + ΔE surroundings = 0 Energy can only be transferred ○ The flow of heat is considered one such transfer

3 The Laws of Thermodynamics  The Second Law All processes that occur spontaneously move in the direction of an increase in entropy of the universe (system + surroundings). ○ The probability of a state existing is known as its entropy (S) ○ In general terms the less order there is in a state, the greater the probability of the state and the greater its entropy. ○ Therefore this term describes the disorder present in a system ΔS universe = ΔS system + ΔS surroundings > 0

4 Entropy as a form of energy  Entropy (S) is a term coined by Rudolph Clausius in the 19th century.  Clausius was convinced of the significance of the ratio of heat delivered and the temperature at which it is delivered, qTqT

5 Entropy as a form of energy  Like total energy, E, and enthalpy, H, entropy is a state function.state function  Therefore,  S = S final  S initial

6 Entropy as a form of energy  For a process occurring at constant temperature (an isothermal process like the melting of ice): q rev = the heat that is transferred when the process is carried out reversibly at a constant temperature. T = temperature in Kelvin.

7 Linking S and  H: Phase changes A phase change is isothermal (no change in T). Entropy system For water:  H fusion = 6 kJ/mol  H vap = 41 kJ/mol If we do this reversibly:  S surr = –  S sys Q - What do you notice happening to S when T=O?

8 Third Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero is 0.

9 15.3.1- State and explain the factors that increase the entropy in a system.  These rules can be used to determine the net entropy change for a system. 1. Entropy increases when the number of molecules increases during a reaction. 2. Entropy increases with an increase in temperature. 3. Entropy increases when a gas is formed from a liquid or solid. 4. Entropy increases when a liquid is formed from a solid.

10 Is this system exhibiting an overall increase or decrease of entropy?

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13 How about for the universe?

14 Entropy practice problem 1  Would the change in entropy (ΔS) for the following process be increasing (+) or deceasing (-)? Why? Br 2 (l)  Br 2 (g)  Positive, because entropy increases as you move from a low energy state to a higher one. (# of moles of gas increased) 15.3.2 - Predict whether the entropy change (∆S) for a given reaction or process is positive or negative.

15 Entropy practice problem 2  Would the change in entropy (ΔS) for the following process be increasing (+) or deceasing (-)? Why? Ag + (aq) + Cl - (aq)  AgCl (s)  Negative, because entropy decreases as you move from a high energy state to a lower one. (# of moles of solid increased)

16 Entropy practice problem 3  Would the change in entropy (ΔS) for the following process be increasing (+) or deceasing (-)? Why? 2NO 2 (g)  N 2 O 4 (g)  Negative, because the number of moles of gas decreased

17 Entropy practice problem 4  Would the change in entropy (ΔS) for the following process be increasing (+) or deceasing (-)? Why? 2OH - (aq) + CO 2 (g)  H 2 O (l) + CO 3 2- (aq)  Negative, because the number of moles of gas decreased (1 to none) as well as the number of moles of total products decreased compared to the reactants (3 to 2)

18 Entropy practice problem 5  Would the change in entropy (ΔS) for the following process be increasing (+) or deceasing (-)? Why? H 2 (g) + Cl 2 (g)  2 HCl (g)  Negative, because the number of moles of gas decreased (2 to 1)

19 15.3.3 - Calculate the standard entropy change for a reaction ( ΔS o ) using standard entropy values ( S o ).  ΔS o = the change in standard molar entropy of an element This is the entropy associated with 1 mol of a substance in its standard state. Values can be found in Table 11 of the IB Data Booklet. Data Booklet ΔS o for any element in its standard state is zero!!

20 Standard Entropies  These are molar entropy values of substances in their standard states.  Standard entropies tend to increase with increasing molar mass.  Note that the units are different than those used for standard enthalpies!

21 Standard Entropies Larger and more complex molecules have greater entropies.

22 Entropy Changes Entropy changes for a reaction can be calculated the same way we used for  H: Note for pure elements:

23 Entropy practice problem 6  ΔS o = ΣS o products - ΣS o reactants ΔS o of H 2 O=188.7 J/mol K ΔS o of H 2 =131.0 J/mol K ΔS o of O 2 =205.0 J/mol K  Find the ΔS o for the reaction of Hydrogen and Oxygen Gasses to form Water. H 2 (g) + ½ O 2 (g)  H 2 O(g) Q – Should your answer be positive or negative? A – Negative!

24 Answer  ΔS o = ΣS o products - ΣS o reactants ΔS o of H 2 O=188.7 J/mol K ΔS o of H 2 =131.0 J/mol K ΔS o of O 2 =205.0 J/mol K  H 2 (g) + ½ O 2 (g)  H 2 O(g) 188.7 J/mol K – [131.0 + ½(205.0)] J/mol K = -44.8 J/mol K

25 HW  You should now have finished your calorimetry IA so…. Do the evens for the section 15.3 Exercises on page 156 Due Wednesday


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