Predicting Reactions with Gibbs Free Energy ```Name: Sebastian G. Status: student Grade: 9-12 Location: CT Date: October 2007 ``` Question: When you put values in for Gibbs free energy (enthalpy, entropy and temperature) and you want to find out if two substances mixed together will react, how do you put in the two values? Do you find the average? Replies: If you write the balanced equation for pure unmixed reactants and products, you can look up the Gibbs free energy in appropriate tables found in texts, handbooks, etc. These are expressed in (per mol) terms at 1 atm pressure, and the stated temperature (usually 298.15 K). If you do the calculation of "products" minus "reactants" taking the coefficients in the balanced equation into account and the resultant Gibbs free energy is negative, that reaction (as written) is spontaneous (in the thermodynamic sense). These cautions (in parentheses) are mentioned because some thermodynamically favored reactions can be prevented from "coming to equilibrium". A simple example is a balloon. Here there is a difference in pressure -- the higher pressure being inside the balloon -- and the pressures inside and outside the balloon should be the same. But the barrier of the balloon prevents that from occurring, at least very rapidly. Your perceptive question about what happens with mixed substances is much large to be derived concisely in a forum such as this. In fact the proper answer to your question makes up a significant part of a semester course in chemical thermodynamics!! But the result is: DGo = -RTln(Keq), where DGo is the standard free energy change for the balanced chemical reaction. R is the gas constant in consistent energy units, T is the absolute temperature, and ln(Keq) is the "natural log" of the equilibrium constant Keq. For a reaction of gases, say aA + bB = cC + eE, Keq is defined as: Keq = (PC)^c x (PE)^e / (PA)^a x (PB)^b. The pressures of the various gases being: (PC), ..., (PB) etc., c, e, a, b being the stoichiometric coefficients. Of course the pressures have to be be in consistent pressure units. In the case of solutions the P's become concentrations, that is, (MC), ..., (MB), etc. Pretty messy, but in a course, or a textbook, there are techniques presented to show you how to solve such equations systematically. What is "surprising" is that while the Gibbs free energy of pure solid and liquid reactants or products appears in the calculation of DGo, those numbers do not appear in Keq. This is not a mistake, that is how the derivation works out. Again, your question is perceptive, but the answer is too long to be answered in a short paragraph. Vince Calder Sebastian, Note that both enthalpy and entropy values have "delta" or change in front of the variable. While the 3rd law of thermodynamics states that we can measure the actual entropy of a system, there is no way to measure the actual enthalpy of the system. All we can do is measure the *change* in the enthalpy. In that sense, we are not using the single-event, one molecule reaction, in describing the change in enthalpy or entropy of the system. What we are reporting is the over-all enthalpy and entropy change of the whole system. In the case of enthalpy change - we measure how much heat is absorbed or released, in the case of entropy we can statistically obtain its entropy. Temperature, of course, is a measured value that is related to the average molecular speeds of the whole system. Greg (Roberto Gregorius) Hi Sebastian, No, this is not correct. You do not use averages. You need to also have Gibbs free energies for the products. Then you calculate delta G for the reaction, using the Gibbs free energies in the same way that you would handle enthalpies of formation to calculate the heat of reaction (see your textbook). Once this is done, if delta G < 0, products are favored at equilibrium and if delta G > 0, reactants are favored at equilibrium. However, even if delta G < 0 the reaction still might not be observed, or might be very slow, if there is a significant energy barrier between reactants and products (this is called the activation energy, Ea). You would need to know Ea for the particular reaction. If delta G < 0 and Ea is small or zero, the reaction will quickly proceed to products. However, if Ea is big it might be a very slow reaction. Hope this helps! Dr. Topper If you have values for the change in enthalpy and entropy for a reaction (there will be one value for the reaction of two substances together), you can plug it into the Gibbs free energy equation to see if the reaction will be spontaneous. Since DG = DH - TDS (where D should be the Greek symbol delta, symbolizing the change in one of these values), where DH is the enthalpy change and DS is the change in entropy. If the overall value of DG is negative the reaction will be spontaneous. For example, when water changes phases from liquid water to ice, this change has a negative DH value (since more hydrogen bonds are formed in the structure of ice this will release heat) and also a negative DS value since ice is more structured than water. You can see why at higher temperatures when -TDS is large and positive, DG will be positive and water will not freeze. At lower temperatures -TDS will be small, and the negative DH term will dominate leading to a negative DG value and spontaneous freezing. Keep in mind that just because a reaction is spontaneous however does not factor into how fast the reaction will occur. If there is a large activation barrier a spontaneous reaction may take a very long time to proceed. For example, iron rusting or (more morbidly) us reaching equilibrium (kicking the bucket) are spontaneous but slow processes. Ethan Greenblatt Stanford University Click here to return to the Chemistry Archives

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