LeChateliers - Chemical Equilibrium-Le Chatelier's...

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Chemical Equilibrium—Le Chatelier’s Principle Background When working in the laboratory, one often makes observations that at first sight are surprising and hard to explain. One might add a reagent to a solution and obtain a precipitate. Addition of more of that reagent to the precipitate causes it to dissolve. A violet solution turns yellow on addition of a reagent. Subsequent addition of another reagent brings back first a green solution and then the original violet one. Clearly, chemical reactions are occurring, but how and why they behave as they do is not at once obvious. In this experiment we will examine and attempt to explain several observations of the sort we have mentioned. Central to our explanation will be recognition of the fact that chemical systems tend to exist in a state of equilibrium. If one disturbs the equilibrium in one way or another, the reaction may shift to the left or right, producing the kinds of effects we have mentioned. If one can understand the principles governing the equilibrium system, it is often possible to see how one might disturb the system, such as by adding a particular reagent or heat, and so cause it to change in a desirable way. Before proceeding to specific examples, let us examine the situation in a general way, noting the key principle that allows us to make a system in equilibrium behave as we wish. Consider the reaction A(aq) ↔ B(aq) + C(aq) (1) where A, B, and C are molecules or ions in solution. If we have a mixture of these species in equilibrium, it turns out that their concentrations are not completely unrelated. Rather, there is a condition that those concentrations must meet, namely that ([B] * [C])/ [A] = K c (2) where K c is a constant, called the equilibrium constant for the reaction. For a given reaction at any given temperature, K c has a particular value. When we say that has a particular value, we mean just that. For example, we might find that, for a given solution in which Reaction 1 can occur, when we substitute the equilibrium values for the molarities of A, B, and C into Equation 2, we get a value of 10 for K c . Now, suppose we add more of species A to that solution. What will happen? Remember, K c can’t change. If we substitute the new higher molarity of A into Equation 2 we get a value that is smaller than K c .. This means that the system is not in equilibrium, and must change in some way to get back to equilibrium. How can it do this? It can do this by shifting to the right, producing more B and C and using up some A. It must do this, and will, until the molarities of C, B, and A reach values that, on substitution into Equation 2, equal 10. At that point the system is once again in equilibrium. In the new equilibrium state, [B] and [C] are greater than they were initially, and [A] is larger than its initial value but smaller than if there had been no forced shift to the right.
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The conclusion you should reach on reading the last paragraph is that one can always cause a reaction to shift to
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This lab report was uploaded on 04/19/2008 for the course CHEMISTRY 142L taught by Professor Lab during the Spring '08 term at Bridgewater State University.

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LeChateliers - Chemical Equilibrium-Le Chatelier's...

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