Spontaneity of a Chemical Reaction

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1 Spontaneity of a Chemical Reaction We have learned that entropy is used to quantify the extent of disorder resulting from the dispersal of matter in a system. Also; entropy, like enthalpy and internal energy, is a state function and can be calculated for any isothermal changes using: qrev S = Suniv = Ssystem + Ssurroundings > 0 Third Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero, 0 K, is zero. Single microstate due to lack of all molecular motion. No: Translation motion = motion thru space Vibrational motion = bond bending and stretching Rotational motion = spinning in space on an axis T But, what about the entropy of a chemical reaction where the system is not isothermal? Standard Entropies Given that the entropy of a pure crystalline solid is zero at absolute zero, entropies of elements and compounds in their pure states can be calculated under some given environmental conditions (usually standard state, 1 atm, 298 K) These are called standard molar entropies. Standard molar entropies can be found in Appendix C Notice, standard molar entropies for pure substances are not equal to zero. Larger and more complex molecules have greater entropies. 1

2 Entropy Changes for Reactions Entropy changes for a reaction, or non-isothermal systems, can be estimated in a manner analogous to that by which H is estimated: 1. Calculate the entropy change entropy change of the system for the combustion of ethanol under standard state conditions. S sys = Σn S (products) - Σm S (reactants) where n and m are the coefficients in the balanced chemical equation. Our question now becomes, does the combustion of ethanol occur spontaneously? The 2 nd Law of Thermodynamics states that the entropy of the universe is always increased for spontaneous reactions. S univ = S system + S surroundings > 0 So, if we know something about the entropy of the universe, we can know something about the spontaneity of the reaction. To determine if the combustion of ethanol is a spontaneous reaction, we must determine the entropy change of the surroundings and compare to the entropy change of the system to see if the total entropy of the universe is increased. To do this, we should recognize that the surroundings serve essentially as a large constant-temperature heat source (or heat sink); therefore, we can think of the universe as an isothermal system. We know that for the surroundings, as an isothermal system: S surr = q rev / T From the first law of thermodynamics: q surr = -q sys From thermochemistry: q sys = H sys q surr = - H o sys Therefore: S surr = - H o sys / T So, the entropy of the surroundings is described by the heat transferred to the system per unit Kelvin. And H o sys can be calculated from: H sys = Σn H o f (prod) - Σn Ho f (react) 2

3 Again, using the 2 nd Law of Thermodynamics and the definition of change in entropy as: 2. Now determine if the combustion of ethanol is spontaneous. S univ = S sys + S surr If: S univ > 0 S univ < 0 S univ = 0 Then: spontaneous rxn Non spontaneous rxn Reversible Equilibrium 3. Prove that the dissolution of sodium chloride in aqueous solution is a spontaneous reaction under standard state conditions. Theretofore, we have been able to use the change in enthalpy of a system and the change of entropy of a system to attempt to predict the spontaneity of a reaction. Examine: 2C 4 H 10 (g) + 13O 2 (g) 9CO 2 (g) + 10H 2 O (g) H o = kj S o = J/K The data suggest that combustion of butane, like all combustion reactions, must be spontaneous. Energy is given off and the disorder increases. Therefore, for any process: H o sys S o sys Spontaneous - + Yes, always + - No, Never But, what about systems where enthalpy favors the reaction but entropy does not, or visa versa? Exothermic processes (favored by energy dispersal) are less likely to be spontaneous at high temperatures when entropy is not favored (nonfavored matter dispersal) Because: S surr = - H o sys / T; The higher the temperature, the smaller the value of S surr 3

4 Endothermic process (disfavored by energy dispersal) are more likely to be spontaneous at high temperatures when entropy is favored (favored matter dispersal) For example: Thermal Decomposition Endothermic Positive entropy Likely at high temperatures So, H o sys S o sys Spontaneous - + Yes, always + - No, Never - - Most likely at low T s + + Most likely at High T s 1. Identify the following reactions as spontaneous, non-spontaneous, more likely at low temperatures or more likely at high temperatures: H o sys S o sys A. CH 4 (g) 2O 2 (g) 2H 2 O (l) + CO B. Graphite + O 2 (g) CO 2 (g) C. 2Fe 2 O 3 (s) + C 4Fe (s) + 3CO 2 (g) A. N 2 (g) + 3 F 2 (g) 2NF 3 (g) What if there were a single thermodynamic function that could determine the spontaneity of a process as a function of both enthalpy and entropy? There is: G o sys = H o sys- T S o sys Gibbs Free Energy = energy of a system available to do useful work under standard state conditions. From: S univ = S surr + S sys And: S surr = - H o sys / T We get: S univ = - H o sys / T + S sys Rearranged: -T S univ = H o sys - T S sys Gibbs defined the change in free energy as: G sys = -T S univ Therefore: G sys = H o sys - T S sys Examine: C graphite + 2H 2 (g) CH 4 (g) H o sys = kj S o sys = J/K Energy is released, however, some energy is required to re-order the system since S is negative. Therefore, there is less free energy G o sys = kj 4

5 G o sys < 0, G o sys > 0, G o sys = 0, Work is available and therefore, the process is spontaneous No free work therefore, the process is not spontaneous Reversible process, at equilibrium 4. Calculate the standard free energy change for the formation of methane at 298 K. Use the enthalpy and entropy data in the appendices. G o rxn = Ho rxn - T So rxn 5. Is this reaction spontaneous? Standard Free Energies ( G o f ) = the free energy change that accompanies the formation of one mole of any compound from its free elements in their standard states (1atm, 298 K). Notice, (appendix C) the standard free energy of a pure substance in its standard state is zero. G o rxn = Σ G fo (products) - Σ G fo (reactants) Note that G f for an element = 0 6. Applying Hess Law, determine the enthalpy change, entropy change and free energy change for the oxidation of one mole of sulfur dioxide gas to form sulfur trioxide gas, then determine if the reaction is spontaneous and what drives the reaction. Remember, because H and S are state functions, G is also an additive state function. 7. Calculate the free energy change for the formation of carbon dioxide gas from the oxidation of both diamond and graphite forms of carbon. Under standard state conditions. Then, calculate the change in free energy for the conversion of diamond to graphite under standard state conditions. (Hess Law) 5

6 The definition of free energy, G = H TS, shows us that free energy is a function of temperature. G changes as the temperature changes. As a consequence, in certain instances, reactions can be spontaneous at one T and non-spontaneous at another. These instances arise when enthalpy and entropy are opposite as we have seen before. H o sys S o sys G o sys Spontaneous - + Always - Yes, always + - Always + No, Never - -? Most likely at low T s + +? Most likely at High T s Gibbs Free Energy can also be used to identify the temperature at which a system becomes spontaneous Remember, when G = 0, the system is reversible, meaning at equilibrium. Therefore if the change in free energy drops below zero, it is spontaneous. If the change in free energy rises above zero, it in nonspontaneous. 8. Use thermodynamics to estimate the boiling point of methanol. This is only an estimate because we are assuming that both enthalpy and entropy do not change with temp. 9. Calculate the boiling point of bromine using information in appendix C. 6

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