The Third Law of Thermodynamics Predicting S for Physical and Chemical Changes It is often a relatively simple matter to predict whether a particular change in a reaction will cause the energy of the reactants to become more spread out (have greater entropy) or less spread out (have lesser entropy).  One of the things to look for is the state of the reactants and products. Gases have more entropy than liquids which have more than solids. During chemical reactions, the freedom of movement of the atoms often changes because of changes in the complexity of the molecules. For example, consider the reaction 2 NO2(g) ----- N2O4(g) Among the reactants are six atoms combined into two molecules of NO2. Among the products, these same six atoms are confined to one molecule. In the reaction vessel, dividing the energy of the six atoms between two molecules allows their energy to be  spread out more in their free movement than when the atoms are combined into just one molecule. Therefore, we can conclude that as this reaction occurs, there is an entropy decrease. Two general rules for predicting entropy changes. 1.   Look at the states first. (gases liquids solids) 2.   If  both states are the same then look at the number of moles of reactants and products and decide if there has been an increase in the number of moles or a decrease. The entropy of a pure crystal is zero at absolute zero. The entropy of a substance (i.e; the extent of its dispersal of energy at a given temperature) varies with the temperature of the substance. The lower the temperature, the lower the entropy. For example, at a pressure of 1 atm and a temperature above 100oC, water exists as a highly disordered gas with the energy of its molecules widely spread out: a very high entropy. If confined, the molecules of water vapour will be spread evenly throughout their container, and they will be in constant motion. When the system is cooled, the water vapour eventually condenses to form a liquid. Although the molecules can still move somewhat freely, their lesser energy is less widely dispersed; they are now confined to the bottom of the container. Their energy is not as great nor as widely spread out as in the gas and thus the entropy of the liquid is lower. Further cooling decreases the entropy even more, and below 0oC, the water molecules join together to form ice, a crystalline solid. The molecules are now not at all free to move, particularly in comparison to that of the gas, and the entropy of the system is very low. Yet even in the crystalline form, the water molecules still have some entropy. There is enough thermal energy left to cause them to vibrate within the general area of their lattice sites. Thus at any particular instance, we would find the molecules near, but probably not exactly at, their lattice positions. If we cool the solid further, we decrease the thermal energy and the molecules spend less time away from their lattice positions. The amount of energy and its spreading out decreasese and the entropy decreases also. Finally, at absolute zero, the ice will be in a state of absolutely minimal energy of molecular movement and its entropy will be zero. This leads us to the statement of the Third Law of Thermodynamics. At absolute zero, the entropy of a pure crystal is also zero. i.e.     S = 0   at     T = 0K Because we know the point at which entropy has a value of zero, it is possible by measurement and calculation to determine the actual amount of entropy that a substance possesses at temperatures above 0 K. If the entropy of one mole of a substance is determined at a temperature of 298 K (25oC) and a pressure of 1 atm, we call it the standard entropy, So. In your databook there is a listing of standard entropies next to the standard enthalpies. ********* IMPORTANT *********                       Entropy has units of J/K degree J/K not kJ/K!!!!! Once we know the entropies of a variety of substances we can calculate the standard entropy change, ΔSo, for chemical reactions in much the same way that we calculated Ho. ΔSo = (sum of So of products) - (sum of So of reactants) The values of ΔSfo have not been calculated for you. If you need them, then you must calculate then from the values of So. Sample problem Urea (from urine) hydrolyzes slowly in the presence of water to produce ammonia and carbon dioxide. CO(NH2)2(aq) + H2O(l) ---- CO2(g) + 2 NH3(g) What is the standard entropy change, in J/K, for this reaction when 1 mole of urea reacts with water? ΔSo = [CO2(g) + (2)NH3(g)] - [CO(NH2)2(aq) + H2O(l)]           = [213.6 J/K + (2)192.5 J/K] - [173.8 J/K + 96.96 J/K]           = (598.6 J/K)-(243.8 J/K)           = 354.8 J/K
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