UNIT ONE BOOKLET 6. Thermodynamic

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1 DUNCANRIG SECONDARY ADVANCED HIGHER CHEMISTRY UNIT ONE BOOKLET 6 Thermodynamic

2 Can we predict if a reaction will occur? What determines whether a reaction will be feasible or not? This is a question that can be answered by applying the principles of chemical thermodynamics, the study of the energy relationships associated with chemical reactions. The term used to label a reaction that proceeds without continual energy input is spontaneous. A spontaneous reaction is one that will occur all by itself, once the activation energy has been provided so that it can get started. These reactions may be fast or extremely slow. The burning of hydrogen, for example, is a spontaneous reaction. H 2 H 2 Once you add a little bit of energy, like the heat from a match, the hydrogen continues to burn without any outside help, until there is no more hydrogen to burn. It is possible to change water back into hydrogen and oxygen by passing electricity through the water - the process is definitely NOT spontaneous. This apparatus will stop producing hydrogen and oxygen if the electricity supply is switched off. Once the external energy supply is removed a non spontaneous reaction will stop even if all the reactants have not been used up.

3 Just like the skier, chemical reactions tend to be spontaneous if the natural flow of energy is DOWNHILL and so reach their lowest, most stable, energy state. This situation happens in exothermic reactions, where energy is released to the surrounds. In an exothermic reaction the products have less stored energy {ENTHALPY, H} than the reactants. The difference in energy is the energy released to the surroundings and is labelled Consider the combustion of methane CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) The change in enthalpy can be seen in the following reaction progress diagram H The diagram clearly shows that the products have less enthalpy than the reactants and are therefore much more stable - a favourable condition for a reaction to be spontaneous. The enthalpy change is found using H = H products H reactants In this case : H = (-969) (-75) = kj mol -1. The activation energy, E a, is also shown on this diagram.

4 In an endothermic the enthalpy of the products is greater than the reactants. The additional energy comes from the surroundings and so an endothermic reaction is accompanied by a fall in temperature. Consider the cracking of ethane C 2 H 6 (g) C 2 H 4 (g) + H 2 (g) H Endothermic reactions occur and so they must be feasible, even although the products are LESS STABLE than the reactants this is not a favourable condition for a spontaneous process. The enthalpy change for this reaction is given by H = H products H reactants H = (52) (-85) = 137 kj mol -1 Both these diagrams illustrate the First Law of Thermodynamics as the total energy of the chemicals plus the energy released or absorbed from the surroundings is constant.

5 If H for a reaction is NEGATIVE this is a good indication that a reaction will be feasible. However, as endothermic reactions { H positive} happen, this cannot be the only condition to determine whether a reaction will occur. In its very simplest sense entropy is related to how ordered something is. Systems that are highly ordered have LOW ENTROPY. HIGH ENTROPY is associated with lots of DISORDER. The drawings show examples of order and disorder the idea of entropy. The diagrams are illustrating that the natural order of a system is towards a situation where entropy increases. It is more probable that things become disordered rather than more ordered. Imagine two boxes with a gap between them. One box has a blue gas atom in it, the other has a red gas atom in it. What is the chance (probability) of both atoms ending up in the right hand box if they are left to diffuse? Do the same thing with 3 different atoms, 4 different atoms and finally with Avogadro s number of atoms. Hint: there is a formula

6 Common sense tells us that gases are much more disordered than liquids and liquids are more disordered than solids. It should be fairly obvious that the molecules in steam have a far more random nature than the molecules in water. Ice is a much more ordered substance than water. The change in entropy as ice is heated is shown in the following graph. Entropy has the symbol (S) The units for entropy are usually quoted as Joules per Kelvin per mole (J K -1 mol -1 ) The graph shows that, in general, entropy increase with increasing temperature disorder gets larger when substances are heated because they obtain more energy the particles will move around more. The graph also shows that entropy increases when the water changes state from solid to liquid to gas.

7 The table shows some standard entropy values, S o the values at 25 o C. Analysis of the table of entropy values reveals some of the trends in entropy. 1. Entropy increases from solids to liquids to gases. 2. The values for methane, ethane, propane and butane suggest that entropy increases with increasing molecular size. Why? The atoms joined together in a molecule move in a variety of ways. The diagram shows some of these movements. If there are more atoms, the there will be an increase in the number of stretches, vibrations and rotations the atoms can make. In other words more disorder. In general entropy will increase as molecular size increases

8 As already stated entropy will increase with increasing temperature. This is due to increasing motion of the particles in the substance. As a substance cools this motion will slow down and if the temperature is low enough it will stop altogether. The lowest temperature possible is known as ABSOLUTE ZERO. It has a value of -273 o c or zero Kelvin, 0 K. At zero Kelvin the solid on the right will have no thermal energy and the atoms will have no motion, not even the smallest vibration. A crystal structure like this will be perfectly ordered and therefore have ZERO ENTROPY. This is summarised in the 3 rd law of thermodynamics. Use the table of entropy values on page 6 to help with these questions. 1. Which substance in the table is most ordered? 2. Why does the increasing trend in entropy from methane to butane NOT continue with pentane? 3. Calculate the entropy value for the following. a. 3.5 moles of sodium chloride b mole of butane gas c. 60 g of diamond 4. Calculate the increase in entropy if 2 moles of water are boiled. 5. Would you expect graphite to have a higher or lower entropy than diamond. Explain your answer.

9 In order for a chemical reaction to be feasible both enthalpy and entropy have to be taken into consideration. If a reaction is exothermic, this is an indication that the reaction will be spontaneous. If the entropy change in a reaction increases when reactants become products this is also a favourable condition to make a reaction spontaneous. Is the entropy change positive or negative in the following processes? 1. Mg(s) + ½O 2 (g) MgO(s) 2. N 2 (g) + 3H 2 (g) 2NH 3 (g) 3. H 2 O(g) H 2 O(l) 4. Na + Cl - (s) Na + (aq) + Cl - (aq) 5. NH 4 NO 3 (s) N 2 O(g) + 2H 2 O(g) The fact that both entropy and enthalpy are involved in reaction feasibility is summarised in the Second Law of Thermodynamics. Consider the burning of magnesium! This means for a reaction to be spontaneous there MUST be an increase in the overall entropy of the reaction system. As it produces a solid product form a gas the entropy will decrease this appears to contradict the 2 nd Law. However, the reaction produces heat and light which increase the entropy of the material surrounding the burning magnesium. When considered together the overall entropy increases this reaction is therefore spontaneous.

10 The American Mathematical Physicist J.Willard Gibb s derived an equation which brought together the idea that enthalpy and entropy were the driving forces in chemical reactions. His equation, called Gibb s Free Energy is as important to chemists as E=mc 2 is to physicists. This equation decides the feasibility of a chemical reaction. Very simply it indicates if a reaction will proceed at a particular temperature. We know that if a reaction is exothermic that this is a favourable sign the reaction will be spontaneous. If the entropy value of the reaction itself also increases this is another favourable sign. Including these facts in the equation means that a NEGATIVE H value and a POSITIVE S value will give a NEGATIVE value for G. It is this value which decides if a chemical reaction is FEASIBLE. Now - three calculations. 1. Calculate the three components of the Free Energy equation. 2. Use the Free Energy equation to determine reaction feasibility at a particular temperature 3. Use the Free energy equation to find the temperature at which a reaction becomes feasible.

11 The feasibility of a reaction can be predicted from a consideration of the signs of H and S. The table shows how free energy is affected by four possible conditions of enthalpy and entropy. Consider reaction A the combustion of methanol. The free energy value of this reaction is always negative no matter the temperature the reaction is always spontaneous. If the combustion of methanol is always spontaneous, why doesn t it burn at room temperature? 1. All reactions require their activation energy. At room temperature there is insufficient energy to start the reaction. 2. Spontaneous means that once started the reaction will continue on its own. 3. Another view is that it is burning just very slowly thermodynamics tell us nothing about the speed of the change just that the change is possible.

12 Before attempting Free Energy calculations we need to consider three important facts regarding thermodynamics. In data tables the values for free energy enthalpy and entropy are always quoted at what is known as STANDARD CONDITIONS. This is because the values can vary with temperature, pressure and concentration. Standard conditions are Temperature of 298 K (25 o C) Pressure of 1 atmosphere Concentration of 1 mol l -1 The symbols denoting standard conditions are G o, H o and S o o The standard enthalpy of formation, H f is often used to find the overall enthalpy change for a reaction. The enthalpy of formation is defined as the enthalpy change when 1 MOLE of a compound is formed from its elements under standard conditions. Equations can be written for the enthalpy of formation E.g Enthalpy of formation of methane is C(s) + 2H 2(g) CH 4(g) Enthalpy of formation of water is H 2(g) + ½ O 2(g) H 2O(l) Notice that state symbols must be used to show the equations complies with the definition and that the substance formed must be 1 mole. The enthalpy of formation of any ELEMENT is defined as ZERO. Given G o, H o or S o values for the chemicals in a reaction it is possible to calculate the G, H or S for the reaction in which they are involved. This is done by using the formula; Where is the sum of and X o is the free energy, enthalpy of entropy of the chemicals.

13 1. The equation for the reaction of ethyne and hydrogen chloride is C 2 H 2 + 2HCl CH 2 ClCH 2 Cl The thermochemical data for the substances involved in the reaction is shown in the table below. Compound S o /J K -1 mol -1 Hf o /kj mol -1 C2H HCl CH2ClCH2Cl Use the information to calculate both the enthalpy change and the entropy change for this reaction. We can use to calculate the enthalpy change H o reaction = H f [CH 2 ClCH 2 Cl] {2 H f [HCl] + H f [C 2 H 2 ]} = [-166] [2(-92.3) + (227)] = kj mol -1 The entropy change is calculated in exactly the same way... S o reaction = S [CH 2 ClCH 2 Cl] {2 S [HCl] + S [C 2 H 2 ]} = [208] [2(187) + (201)] = -367 J K -1 mol -1 Note that as the values are given per mole of substance the stoichiometric coefficients (balancing numbers) must be used in these calculations. These results indicate that this reaction is exothermic ( H is negative) and that there has been a decrease in entropy ( S is negative).

14 2. The equation for the decomposition of zinc carbonate is ZnCO 3 ZnO + CO 2 Calculate the free energy, G o, for this reaction given that H o is 71 kj mol -1 and S o is J K -1 mol -1 Use G o = (175.1/1000) = = 19 kj mol -1 In the free energy equation the value for temperature must be in Kelvin. In this example it appears that no value for the temperature has been given. However, as the standard state symbols have been used, we can assume a temperature of 298 K. Notice that the entropy value has been divided by This is done to match up the units. The units for enthalpy are given in kilojoules and the units for entropy are given in Joules. Both must be the same before being used in the free energy equation. Not doing this is a very common mistake in exams. As the free energy value is positive this reaction is not spontaneous at 298 K. For this reaction to occur the temperature would need to be much higher. This would increase the entropy value which would eventually overcome the unfavourable positive enthalpy value. Anyone trying to change zinc carbonate to zinc oxide at 298 K (room temperature) would probably have to wait a very long time indeed.

15 3. Barium carbonate decomposes on heating. BaCO 3 (s) BaO(s) + CO 2 (g) ΔH = +266 kj mol 1 a. Using the data from the table below, calculate the standard entropy change, ΔS, in J K 1 mol 1, for the reaction. S o reaction = {S [CO 2 ] + S [BaO]} - S [BaCO 3 ] = [ ] [112] = J K -1 mol -1 b. Calculate the temperature at which the decomposition of barium carbonate just becomes feasible. To answer this question we use a variation of the free energy formula... This formula relies on the fact that, at equilibrium, the value of G o is ZERO. Substituting the values into this expression gives This temperature provides the activation energy to make this reaction proceed at a reasonable rate.

16 1. Use the information in the table to calculate the free energy change in the following reactions a. 2Mg + CO2 2MgO + C b. 2CuO + C CO2 + 2Cu Explain wether both these reactions are feasible at 298 K 2. The equation for the decomposition of magnesium carbonate is shown below a. Use the thermodynamic data to calculate the free energy change, in kj mol -1 at 400K for the reaction. b. Is the reaction feasible at this temperature? 3. Consider the thermodynamic data shown for the Haber process. Use the data given above, along with data book values to calculate the temperature at which the Haber process becomes feasible. 4. Chloroform was one of the first anaesthetics used in surgery. Use the thermodynamic data to calculate a boiling point for chloroform. CHCl3(l) CHCl3(g) S = 94.2 J K -1 mol -1 H = 31.3 kj mol -1

17 5. Consider the reactions and their thermodynamic data. a. Why does the reaction between aluminium and oxygen have the largest difference between G o and H o? b. The reaction between hydrogen and chlorine is spontaneous at 298 K. Explain why there is no observable reaction between hydrogen and chlorine until the mixture is exposed to ultraviolet light. c. Why does the data confirm that ammonium chloride dissolves spontaneously in water and that the temperature of the water will decrease as the ammonium chloride dissolves?

18 10. This table contains some thermodynamic data for hydrogen, oxygen and water. a. Calculate the temperature above which the reaction between hydrogen and oxygen to form gaseous water is not feasible. b. State what would happen to a sample of gaseous water that was heated to a temperature higher than that of your answer to part a. Give a reason for your answer. 11. The oxides nitrogen monoxide (NO) and nitrogen dioxide (NO 2 ) both contribute to atmospheric pollution. The table gives some data for these oxides and for oxygen. Nitrogen monoxide is formed in internal combustion engines. When nitrogen monoxide comes into contact with air, it reacts with oxygen to form nitrogen dioxide. a. Calculate the enthalpy change for this reaction. b. Calculate the entropy change for this reaction. c. Calculate the temperature that this reaction becomes thermodynamically feasible.

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