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Case Study_background - Topic 8 1 Topic 8 Case Study...

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Unformatted text preview: Topic 8 1 Topic 8 Case Study Chemical Kinetics and Catalysis Introduction The formation of products from reactants is the essence of chemistry. As you will find out in Chapter 19 of your textbook, thermodynamics can be used to identify reactions that are spontaneous (favorable), as well as the equilibrium state of the reaction. However, on a practical level the rate with which reactants are formed is as important as the overall favorability. If the reaction is too slow, products may never form. If the reaction is too fast, the reaction can destroy the system (i.e. an explosion). There are several strategies for implementing control over the rate of a reaction: the concentration of the reactants and products on which the rate depends can be varied, the temperature at which the reaction takes place can be adjusted, a catalyst can be used to speed up a reaction, or an inhibitor can be used to slow down the reaction. In this topic, you will identify several of the key variables the impact the rate of a chemical reaction, and you will design a chemical experiment to investigate a variable of your choice. In case study, you will see examples of some relevant chemical reactions, and you should expect to contribute to the discussion of how to design and implement an experiment to quantitatively measure the kinetics of a reaction and to gain insight into molecular level factors that control the rate of the reaction. Chemical Foundations Rate of Reaction For the reaction: aA + bB → cC + dD the rate of the reaction can be expressed as, rate = k [A] x [B] y where k is the rate constant and x and y are the orders of reaction (powers) with respect to the concentrations of A and B respectively. The concentrations that appear in this equation are not limited only to reactants but can include any chemical species that impacts the rate, such as a catalyst or inhibitor. To make this concrete, consider the reaction that you will focus on in the laboratory portion of this topic, the decomposition of hydrogen peroxide: 2 H 2 O 2 (aq) & O 2 (g) + 2 H 2 O (l). In this reaction, as in most reactions, reactants must overcome a potential energy barrier to become products. This barrier is called the activation energy. In this case, the barrier might be imagined to be a state in which the oxygen-hydrogen bonds of one hydrogen peroxide molecule are being stretched (broken) as oxygen-hydrogen bonds on an adjacent molecule are being formed. The course of the reaction might be pictured as, H H O O H H...
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This note was uploaded on 10/11/2011 for the course CHEM 012 taught by Professor Mounaamaalouf,reneescole, during the Spring '11 term at Iowa State.

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Case Study_background - Topic 8 1 Topic 8 Case Study...

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