Calorimetry 2: Hot and Cold Reactions Overview Getting Started Techniques Procedure FAQ Full Lab Manual Introduction & Goals Chemistry & Background Key Questions Prelab Problems Safety Procedure In Your Write-up Experiments Index ChemLab Home		Chemistry & Background The enthalpy changes of dissolution will determine whether a particular salt is appropriate as a hot or cold pack ingredient. By definition, exothermic processes release heat. That heat can then increase the temperature of the surroundings. If the surroundings include a strained or sore muscle, you have a useful hot pack. For a cold pack, an endothermic reaction that absorbs heat from the surroundings is required. Heat will flow from the surroundings, for example a twisted ankle, and its temperature will be reduced, with the hope that its swelling will also be reduced. A non-trivial part of selecting a salt is determining just how much solid should be used in your calorimeter to produce the desired temperature change. Each of the salts available for potential hot and cold packs will dissolve spontaneously in water. For spontaneous processes like this, the second law of thermodynamics tells us that the total entropy change must be greater than or equal to zero. The total entropy change is the sum of the entropy change of the surroundings and the system, where the surroundings are the rest of the universe, Stot = Ssurr + Ssys > or = 0 for spontaneous processes Heat flowing between the system and surroundings results in entropy change in the surroundings. For the case of constant temperature and pressure, we can write: -Hsys = Hsurr = T Ssurr The initial negative sign indicates that the heat into the surroundings equals the heat out of the system, and vice versa. We can rearrange this equation to yield, Ssurr = -Hsys / T Using this relationship, the total change in entropy can be expressed in terms of the properties of the system, rather than the entire universe, Stot = Ssurr + Ssys = -Hsys / T + Ssys > or = 0 for spontaneous processes Putting this together with the definition of Gibbs free energy, G = H - TS, gives Stot = -Hsys /T + Ssys = -Gsys / T > or = 0 for spontaneous processes Note that Stot and Gsys must have opposites signs, since absolute temperature is always greater than zero. Rearranging gives rise to new but equivalent criteria for spontaneous processes at constant temperature and pressure. Gsys must be less than or equal to zero: Gsys = Hsys - T Ssys < or = 0 for spontaneous processes at constant T and p Since we now have criteria for spontaneity that depend only on the system, we can drop the label, sys. For the spontaneous dissolving of ionic compounds, with the laboratory conditions used, the Gibbs free energy change, G, will be less than zero. Since G is a function of H and S, the spontaneity of a process involves a balance between its enthalpy and entropy changes. A negative value of H, an exothermic reaction, contributes to a negative G and to spontaneity. Since absolute temperature is always positive, the -TS term contributes to a negative G and a spontaneous reaction when S is positive, or the entropy of the system is increasing. As you can see, however, if H and S are both negative, H and -TS will oppose one another. In that case the balance, and the spontaneity of the reaction, will depend upon the temperature. We can predict the entropy change of the system for the dissolution of ionic compounds based on changes in spatial randomness. The initial state, the ionic crystal lattice, is more ordered than the final state, the ions dissolved in water. This results in a S that is greater than zero, since the entropy of the system is increasing. Consideration of spatial randomness can often, but not always, predict changes in entropy. Your textbook contains a more complete discussion of the statistical basis of entropy, which is more rigorous than spatial considerations alone. In order to tabulate the values of thermodynamic properties in a useful way, chemists have established standard conditions for reactions. As you read the procedure for this week's experiment, you should notice that the reactions will not be carried out under these standard conditions.
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