Chemistry and the material world Unit 4, Lecture 4 Matthias Lein

Transcription

1 Chemistry and the material world Unit 4, Lecture 4 Matthias Lein

2 Gibbs ree energy Gibbs ree energy to predict the direction o a chemical process. Exergonic and endergonic reactions. Temperature dependence o the Gibbs ree energy. The chemical equilibrium. The law o mass action. Reaction quotients. Gibbs ree energy and equilibrium constants. Equilibrium temperature.

3 Consider the reaction: N 2 (g) + 3 H 2 (g) 2 NH (g) 3 Let us calculate ΔH and ΔS or the ollowing reaction to decide in which direction each o these actors will drive the reaction. ΔH (N 2 ) = 0 kj/mol, ΔH (H 2 ) = 0 kj/mol, ΔH (NH 3 ) = kj/mol ΔS (N 2 ) = J/molK, ΔS (H 2 ) = J/molK, ΔS (NH 3 ) = J/molK ΔH = kj ΔS = J/K Enthalpy avours the reaction. It is exothermic, heat is released. Entropy dis-avours the reaction. Order is increased, the number o particles decreases. The questions remains: Will the reaction occur spontaneously?

4 The Gibbs ree energy G is deined as: G = H -TS a measure that connects the two previously treated thermodynamic unctions H, the enthalpy and S, the entropy. ΔG < 0 ΔG = 0 ΔG > 0 avored reaction (Spontaneous) Neither the orward nor the reverse reaction prevails (Equilibrium) disavored reaction (Non-spontaneous) Josiah Willard Gibbs United States,

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6 Just like reactions are called exothermic and endothermic depending on the heat produced ( ΔH), reactions are also called: ΔG < 0 ΔG > 0 exergonic (Spontaneous) endergonic (Non-spontaneous) Wether a reaction is exergonic or endergonic depends on the reaction enthalpy, the reaction entropy and the temperature at which the reaction is carried out.

7 Consider the reaction (at 25 C): N 2 (g) + 3 H 2 (g) 2 NH (g) 3 Let us calculate ΔH and ΔS or the ollowing reaction to decide in which direction each o these actors will drive the reaction. ΔH (N 2 ) = 0 kj/mol, ΔH (H 2 ) = 0 kj/mol, ΔH (NH 3 ) = kj/mol ΔS (N 2 ) = J/molK, ΔS (H 2 ) = J/molK, ΔS (NH 3 ) = J/molK ΔH = kj ΔS = J/K ΔG = ΔH -TΔS ΔG = -92,220 J - ( K J/K) ΔG = kJ Answer: at 25 C the reaction occurs spontaneously (it is exergonic)

8 How about the same reaction at 500 C? N 2 (g) + 3 H 2 (g) 2 NH (g) 3 ΔH = kj ΔS = J/K ΔG = ΔH -TΔS ΔG = -92,220 J - ( K J/K) ΔG = 61.4kJ Answer: at 500 C the reaction does not occur spontaneously (it is endergonic) The entropic term becomes more and more important as the temperature rises.

9 ΔG = ΔH -TΔS The sign o ΔG tells us the direction in which a reaction will proceed to reach an equilibrium and the magnitude o reaction still is ( ΔG = 0 means equilibrium). ΔG tells us how ar rom equilibrium the Thus, as the reaction progresses, the magnitude o ΔG will become smaller and smaller until an equilibrium is reached. The chemical reaction is then inished.

10 ΔG = ΔH -TΔS When we calculate ΔG values or reactions, we are comparing G values or the reactants and products. Thereore, the calculated reaction only represents the true ΔG value or the ΔG value or the very start o the reaction. Over the course o the reaction the magnitude o ΔG will go to zero.

11 Equilibrium: N 2 O 4 (g) 2 NO (g) 2 colourless brown

12 Equilibrium: N 2 O 4 (g) 2 NO (g) 2

13 Equilibrium: N 2 O 4 (g) 2 NO (g) 2

14 Equilibrium: N 2 O 4 (g) 2 NO (g) 2 Law o mass action (K c is the equilibrium constant) This equation only describes the equilibrium.

15 Equilibrium: N 2 O 4 (g) 2 NO (g) 2 Systems that are not at equilibrium can be described by a similar expression. Q c Q c is called the reaction quotient Remember that Q c and K c depend on temperature and pressure.

16 Equilibrium: N 2 O 4 (g) 2 NO (g) 2 The connection between the equilibrium constant and the Gibbs ree energy: ΔG = -RT ln K So, to calculate ΔG or a given reaction quotient: ΔG = ΔG +RT ln Q (note that the expression becomes zero or Q = K)

17 What does the magnitude o K c mean? K = 1 means that the reaction quotient is balanced. There is approximately the same amount o educt and product in the reaction mixture. K > 1 means that that the numerator in the reaction quotient is larger. There is more product in the reaction mixture. K < 1 means that the denominator in the reaction quotient is larger. There is more educt in the reaction mixture.

18 For example: 2 SO 2 (g) + O 2 (g) 2 SO (g) Δ G = kj/mol 3 What is the value o the equilibrium constant K? We know: So: Δ G = -RT ln K (R = J/molK, T = 298K) ln K = -ΔG / RT = -( kj/mol) / (8.314 J/molK 298K) = 56.5 K = e = 3 10

19 The equilibrium temperature We know that at equilibrium: ΔG = 0 which means that ΔH TΔS = 0 That means that i we know two quantities out o ΔH, ΔS and T, we can calculate the value o the third at the equilibrium.

20 For example: Br 2 (l) Br 2 (g) ΔH = 30.9 kj/mol, ΔS = 93.2 J/molK At what temperature is the two phases o liquid and gaseous Br 2 in equilibrium? At equilibrium: ΔG = ΔH TΔS = 0 so and ΔH = TΔS T = ΔH / ΔS T = J/mol / 93.2J/molK = 332 K 332 K = 59 C

21 Today we covered: Gibbs ree energy to predict the direction o a chemical process. Exergonic ( ΔG < 0) and endergonic ( ΔG > 0) reactions. Temperature dependence o the Gibbs ree energy. The chemical equilibrium. The law o mass action. Reaction quotients. Gibbs ree energy and equilibrium constants ( ΔG = -RT Equilibrium temperature ( ΔH = TΔ S). ln K).