9 - Stress and Other Equilibrium topics

9 - Stress and Other Equilibrium topics - Equilibrium...

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Equilibrium Lecture Three: A bit more detail and some additional kinds of problems. Lecture Overview: We get even more involved in the details by equilibria by relating Δ G to K relating K c to K p comparing Q to K to figure out the direction of a reaction developing LeChatelier’s principle to explain system stress But first, a reminder of how to calculate Q and K from data. Why do I seem to start every lecture this way? Because it is the essence of how you should view chemical equilbria. What is the ratio of materials in a system and at what point do their rates stop changing so that Q = K. Calculate Q at: Note that Q=K during this interval Time 0 minutes Time 2 minutes Time 4 minutes Time 6 minutes Q= B A = 0 4 =0 Q= B A = 2 2 =1 Q= B A = 3 1 =3 = K Q= B A = 3 1 =3 = K 4 3 2 1 Q=K at equilibrium 1 2 3 4 5 6 7 Time(minutes) 0 Q<K
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Time u out u for a few u interesting thoughts on K Interesting thought 1. Solids and pure liquids are not included in equilibrium expression calculations When you construct an equilibrium expression from the chemical reaction, you put the materials on the right in the numerator and the materials on the left on the numerator. H 2 O h H + + OH - K = [H + ][OH - ]/[H 2 O] But these values are concentrations, and while the concentration of a gas can change (with pressure) and the concentration of a solute in a solvent can change, what about the concentration of pure liquids and solids? They don’t change—they are constant which means there is no reason to include these constant values in the equilibrium expression. So as a rule, for all solids and pure liquids, assume [ ] = 1. Here with [H 2 O], which has a constant concentration of 55.4 M (do the calculation yourself), we set [H 2 O] = 1 and the equilibrium expression for water dissociation is actually K = [H + ][OH - ] Interesting thought 2. K can have very large and very small values—get used to using the exponent button on your calculator Although I emphasized the point that reactions never go to completion and that there are always both product and reactant in a closed system at equilibrium, this doesn’t mean that there aren’t a lot of reactions that go nearly to completion and produce very large equilibrium constants. Consider the reaction for the dissociation of N2 to make atomic nitrogen: Example: N 2 h 2N Δ G=+456kJ K=8.6 x 10 -81 or the solubility product for silver sulfide Example: Ag 2 S h 2Ag + + S = K=8 x 10 -51 What bout reactions that produce really large values of K, are there any of those? Sure, what about the reverse of these reactions: 2N h N 2 Δ G = -456kJ K=8.6 x 10 81
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Interesting thought 3: In the RICE calculation, how do you know which signs to put in the Δ row of the array Consider the RICE expression. When you are doing calculations, either the stuff on the left is getting smaller and the stuff on
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9 - Stress and Other Equilibrium topics - Equilibrium...

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