| Home | | Table of Contents | | Thermodynamics Gateway Page | | In this module: | | Introduction | | Disorder in Atoms | | Disorder in Energy | | Measuring Entropy | | Entropy of Phase Changes | | Patterns in the Entropies of Substances | | Entropy in Thermochemical Equations | | The Second Law of Thermodynamics | | The Effect of Temperature | | Predicting How Reactions will Go | | Two Examples | | | Predicting How Reactions Will Go As was demonstrated on the previous page, in some reactions changing the temperature can change whether the reaction is product- or reactant-favored. In other reactions, temperature doesn't matter at all. How can you tell which is which? With both DSsyst and DHsyst affecting DSuniv, there are four possible combinations: 1. DH < 0, DS > 0 This is an exothermic reaction with a positive entropy change. DSsurr > 0, so DSuniv > 0 and the reaction is product-favored at all temperatures. | | Examples: | 2 CH4 (g) + 3 O2 (g) |  | 2 CO2 (g) + 2 H2O (g) | | CaCO3 (s) + 2 HCl (aq) | | CaCl2 (aq) + H2O (l) + CO2 (g) | | 2. DH > 0, DS < 0 This is an endothermic reaction with a negative entropy change. DSsurr < 0, so DSuniv < 0 and the reaction is reactant-favored at all temperatures. | | Example: | 6 CO2 (g) + 6 H2O (g) | | C6H12O6 (s) + 6 O2 (g) | | 3. DH > 0, DS > 0 This is an endothermic reaction with a positive entropy change. DSsyst > 0, but DSsurr < 0, so more information is needed to predict how this reaction will go. Since raising the temperature makes DSsurr smaller, this sort of reaction will be product-favored at high temperatures and reactant-favored at low temperatures. | | Example: | CaCO3 (s) | | CaO (s) + CO2 (g) | | 4. DH < 0, DS < 0 This is an exothermic reaction with a negative entropy change. DSsyst < 0, but DSsurr > 0, so more information is needed to decide. Since raising the temperature makes DSsurr smaller, this sort of reaction will be reactant-favored at high temperatures and product-favored at low temperatures. | | Example: | N2 (g) + 3 H2 (g) | | 2 NH3 (g) | | If you know DSº and DHº and they have the same sign (putting them into cases three or four above) you can use the Second Law of Thermodynamics to predict at what temperature the reaction will switch from reactant- to product-favored, or vice-versa. To a reasonable approximation, DSº and DHº are constant with temperature, so you can solve for the temperature where DSuniv is zero. This is the temperature where a reaction switches from being reactant-favored to product-favored, and vice versa.  Select positive or negative signs for DH and DS to see what sort of reaction results. Click on the mouse icon at left to clear the radio buttons and text. | | Sign of DH | Sign of DS | Product-favored? | | Positive |  exothermic endothermic | Positive | no yes no at low T, yes at high T yes at low T, no at high T | | Negative | Negative | | |  One of the steps in isolating zinc metal from its ore sphalerite (ZnS) is roasting it in air, producing zinc oxide and elemental sulfur: 8 ZnS (s) + 4 O2 (g) 8 ZnO (s) + S8 (s) This reaction is reactant-favored at 25 ºC. At what temperature does it become product-favored? | | | | Correct! Make sure your units are correct. The correct answer is 1263 K or 990 ºC. That is incorrect. Try again. | |
