1. Title: Lesson 9 Entropy and Gibb’s Free Energy Learning Objectives: – Calculate Gibb’s Free Energy from ∆G = ∆H - T∆S – Calculate Gibb’s Free Energy from Hess Cycles – Predict and explain the effect of temperature changes on Gibb’s Free Energy

2. Main Menu Refresh 2CH 3 OH(g) + H 2 (g) → C 2 H 6 (g) + 2H 2 O(g)  The standard entropy for CH 3 OH(g) at 298 K is 238 J K –1 mol –1, for H 2 (g) is 131 J K –1 mol –1 and for H 2 O(g) is 189 J K –1 mol –1.  Using information from Table 12 of the Data Booklet, determine the entropy change for this reaction.

3. Main Menu To Do:  Categorise the following as either spontaneous or non-spontaneous:  Cooking an egg, acid reacting with metal, the reaction in a battery, electrolysis, respiration, photosynthesis  Think of an example of each of the following:  A spontaneous exothermic reaction  A non-spontaneous endothermic reaction  A spontaneous endothermic reaction  A non-spontaneous exothermic reaction (difficult!)

4. Main Menu Gibbs free energy is a useful accounting tool  For chemical reactions neither ∆H(system) nor ∆S(system) alone can reliably be used to predict the feasibility of a reaction.  To calculate feasibility we use this expression:  Tidying up and moving T to the other side:  Multiplying by -1 and reversing the inequality:  This gives the function known as Gibbs free energy (∆G(system)):  ∆G(system) must be negative for a spontaneous process. Units: kJ or J mol -1  Whereas ∆H(system) is a measure of the quantity of heat change, ∆G(system) gives a measure of the quality of the energy available. (Ability of energy free to do useful work rather than leave the system as heat)

5. Main Menu Using ∆G(system) to predict the feasibility of a change  We generally assume that both the enthalpy and entropy changes of the system do not change with temperature.  Temperature, T, is alike a tap which adjusts the significance of the term ∆S(system) in determining the value of ∆G(system).  At low temperature:  This shows that all exothermic reactions can occur at low temperatures.  At high temperature:  This means all reactions which have a positive value of ∆S(system) can be feasible at high temperatures even if they are endothermic.

6. Main Menu 1. If G is negative, the reaction is spontaneous in the forward direction. 2. If G is equal to zero, the reaction is at equilibrium. 3. If G is positive, then the reaction is non-spontaneous in the forward direction, but the reverse reaction will be spontaneous. 4. for elements at standard state (pure elements at 25ºC and 1 atm are assigned a value of zero). The Gibb’s free energy equation can be used to calculate the phase change temperature of a substance. During a phase change, equilibrium exists between phases, so if the G is zero, we know that the reaction is in equilibrium. What does the Gibb’s free energy value tell us about a reaction?

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11. Summarising the Effect Of ∆H, ∆S and T on Spontaneity of Reaction ∆G = ∆H - T∆S  Task: Use the equation to help yourself reason these through

12. Main Menu Spontaneity  We can calculate Gibbs free energy in two ways:  Method 1:  Directly from (appropriate) data using ∆G = ∆H - T∆S < 0  Method 2:  Indirectly from standard ∆G f o values in the data booklet, using a Hess Cycle  Note: Similar to ∆H f o values, ∆G f o for any element in its standard state is zero.

13. Main Menu Method 1: From ∆G = ∆H - T∆S  Is the reaction of ethene with hydrogen (to form ethane), spontaneous at room temperature (298K)?  C 2 H 4 (g) +H 2 (g)  C 2 H 6 (g)  ∆H o = -137 kJ mol -1  ∆S o = -121 J K -1 mol -1  Calculate Gibbs Free Energy  ∆G = ∆H - T∆S  ∆G = -137 – (298 x -121)/1000Divide by 1000 to convert to kJ  ∆G = -149 – (-36) = -101 kJ mol -1  Evaluate the answer  The answer is negative, which means the reaction is spontaneous at room temperature

14. Main Menu Method 2: Using standard ∆G f o values  Is the reaction of ethene with hydrogen (to form ethane), spontaneous at room temperature (298K)?  C 2 H 4 (g) +H 2 (g)  C 2 H 6 (g)  ∆G f o (C 2 H 4 ) = 68 kJ mol -1, ∆G f o (H 2 ) = 0 kJ mol -1, ∆G f o (C 2 H 6 ) = -33 kJ mol -1  Construct a Hess Cycle C 2 H 4 (g) + H 2 (g)C 2 H 6 (g) 2C(s) + 3H 2 (g) ∆G ∆G 1 = 68 + 0 = 68 ∆G o = ∑ ∆G f (products) - ∑ ∆G f (reactants) = -68 + -33 = -101 kJ mol -1 ∆G 2 = -33  Evaluate Answer  The value is negative so the reaction is spontaneous  Note this is the same as the previous answer, which is to be expected.

15. Main Menu The Effect of Changing the Temperature  In the previous example:  ∆G = -101 kJ mol -1, ∆H o = -137 kJ mol -1, ∆S o = -121 J K -1 mol -1, T = 298 K ∆G = ∆H - T∆S  What happens if we increase the temperature?  Since ∆H and both ∆S are both negative, ∆ G can be either positive or negative depending on the temperature...  A high temperature (in this case 1132 K) will make the ‘T∆S’ term large enough to counter-balance the ∆H term. This will mean that the reaction is not spontaneous.  At any temperature below 1132 K, the reaction will be spontaneous  This does not mean that the reaction will be fast, just that it will (in theory) happen

16. Main Menu Why 1132K?  This slide just explains the maths behind working out when the reaction becomes spontaneous.  This is interesting but not needed!  The transition from spontaneous to non-spontaneous happens at ∆G = 0, so if we use this we can determine the temperature necessary for this as follows:  ∆G = ∆H - T∆SThe initial equation  0 = ∆H - T∆SSet ∆G = 0  T∆S = ∆HRearrange to make T subject  T = ∆H/∆S  T = -137 / (-121/1000) Sub in values  T = 1132 KEvaluate ∆G = -101 kJ mol -1, ∆H o = -137 kJ mol -1, ∆S o = -121 J K -1 mol -1, T = 298 K

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24. Practice Questions  Using data from the data booklet, calculate ∆G o for the following reactions using both methods, and comment on whether the reaction is spontaneous. Assume 298 K.  Note: in some cases you will first need to calculate ∆H o and ∆S o.  Extension: for non-spontaneous reactions, find the minimum temperature necessary to make them spontaneous, and for spontaneous reactions, find the maximum temperature where they are non- spontaneous 1. S(s) + O 2 (g)  SO 2 (g) – see next slide for example 2. C 5 H 12 (l)  CH 4 (g) + 2C 2 H 4 (g) 3. CH 4 (g) + 3Cl 2 (g)  CHCl 3 (g) + 3HCl(g) Substance S o J K -1 mol -1 SO 2 (g)248 O 2 (g)205 HCl(g)187 Cl 2 (g)223 S(g)32 Substance ∆H o f kJ mol -1 HCl(g)-92 Substance ∆G o f kJ mol -1 SO 2 (g)-300

25. Main Menu S(s) + O 2 (g)  SO 2 (g)  Method 1:   H = -297 kJ mol-1 This is in the data booklet as  H c for sulfur   S = 248 – (32 + 205) = 11 J K -1 mol -1 Data is on previous slide  ∆G = ∆H - T∆S = -297 – (298 x 11)/1000 = -300 kJ mol -1  Method 2: S(s) + O 2 (g)SO 2 (g) S(s) + O 2 (g) ∆G ∆G 1 = 0 + 0 = 0 ∆G o = -0 + -300 = -300 kJ mol -1 ∆G 2 = -300 At what temperature does it become non spontaneous? This reaction is always spontaneous, since  H is negative and  S is positive.

26. Main Menu Gibbs free energy and equilibrium  So far we have considered reactions in which it is assumed that all reactants are converted into products.  Many reactions do not go to completion, but instead reach equilibrium.  The extent of the reaction can be quantified by the ratio of concentrations: [products]/[reactants].  As ∆G becomes more negative, the reaction will favour the products. (More spontaneous reaction)

27. Main Menu Summary of ∆G(reaction) and the extent of the reaction  Above 30 kJmol -1 = No reaction  Less than 30 kJmol -1 but more than -30 kJmol -1 = Partial reaction  Below -30 kJmol -1  = Complete reaction

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31. Summarising  ∆G = ∆H - T∆S  The mathematics means that:  Some reactions are spontaneous at all temperatures  Some reactions are only spontaneous above certain temperatures  Some reactions are only spontaneous below certain temperatures  ∆S can be calculated using Hess cycles in much the same way as ∆G and ∆H