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Chemical Equilibrium Lab

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Background A chemical equation that is in equilibrium will naturally balance itself out should either the products are the reactants be increased, or other stresses such as temperature and pressure are placed on it. However, as with a lever in physics, the fulcrum or balancing point is not always in the middle. In most cases, one side of the equation will be favored more. Ergo, the equation will try to balance to this point of equilibrium.
Materials
We did several different examples of chemical equilibriums using different stresses and conditions. This includes adding or removing reactants, changing temperature, or adding additional compounds. Our materials included a large variety of chemicals and equipment used to conduct our labs. …show more content…

The first lab was testing the equilibrium of hydrogen ions within brewed tea. We put tea in three different wells. One was left alone for a control, one had acetic acid added to it, and the other had sodium hydroxide added to it. The solution was stirred after each substance was added to the tea to ensure even distribution. After finishing and recording results and observations, the trays were rinsed and dried. The second lab was testing the equilibrium of starch, iodine, and starch-iodine complex. For this lab we needed heated water and cold water. We created a diluted starch mixture in two wells and placed water of different temperatures of each one. The change in temperature would cause stress in the equilibrium, meaning it would have to rebalance itself. The wells were again cleaned and dried to prepare for the next …show more content…

When in bonded, it turned a deep hue of purple. However, when they were separate the mixture was clear. We could see the change in this scenario when we changed the temperature by placing either heated water or iced water under the different wells. When the temperature was raised, the color was much lighter and more clear, implying that the starch and iodine were not forming the bonded complex. However, when the temperature was lowered, it because a very dark shade of purple, almost appearing to be black. In the equation we could see that the formation of the complex produced heat as well, which represents why it responds the way it does. If you treat heat as a product, then by raising the temperature it will try to balance itself by shifting the balance to the other side, forming separated iodine and starch. The opposite is also true, and this is because the reaction is exothermic in one direction, but can be treated as endothermic in the other

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