Essay, Pages 7 (1546 words)
Copper (II) sulfate pentahydrate crystals are light blue and granular. After being heated, the crystals turned into a white powder and condensation formed in the upper part of test tube. It is represented by the equation CuSO4•5(H2O)(s) > CuSO4(s) + 5(H2O)(g) When 5 drops of water were added to the white powder, bubbling occurred and the powder turned into a blue liquid, and then into the original blue coloured crystals (copper (II) sulfate pentahydrate). It is represented by the equation CuSO4(s) + 5(H2O)(l) > CuSO4•5(H2O)(s) B Sodium sulfate solution is a colourless, transparent liquid.
Barium chloride is a colourless, transparent liquid. After adding a medicine dropper of barium chloride solution, the solution immediately became cloudy and milky looking. A precipitate was formed during the reaction. It is represented by the equation BaCl2(aq) + Na2SO4(aq) > BaSO4(s) + 2NaCl(aq) D Solid magnesium is a slightly shiny and malleable metal. Hydrochloric acid is a colourless, transparent liquid. After adding hydrochloric acid, magnesium began to fizz, releasing bubbles. The bottom of the test tube got warmer. Condensation formed in the upper part of test tube and the magnesium looked white and got much smaller.
The reaction is represented by the equation Mg(s) + 2HCl(aq) > MgCl2(aq) + H2(g) E Potassium iodide is a white granular solid. Hydrogen peroxide is a cooler than air, colourless, transparent liquid. After adding hydrogen peroxide, potassium iodide solids disappeared, followed by bubbling and colour change of solution to yellow, but still transparent.
Temperature of solution is still cooler than air. After inserting a glowing splint into the mouth of the test tube, a pop sound is made and the flint is re-ignited. But the flame quickly died. The reaction is represented by the equation 2H2O2(aq) ––KI> 2H2O(l) + O2(g)
F The iron nail is a hard, shiny, grey, metal solid. The copper (II) sulfate solution is a light blue, translucent liquid. After the iron nail has been immersed in the copper (II) sulfate solution for a while, it had chunks of reddish brown rust-like substances on it. The reaction is represented by this equation CuSO4(aq) + Fe(s) > FeSO4(aq) + Cu(s) Conclusion Experiment A demonstrated a decomposition reaction and a synthesis reaction. The decomposition reaction occurred when copper (II) sulfate pentahydrate was heated. The compound decomposed to copper (II) sulfate and water. The water is the condensation.
Exposure to heat is what caused this product’s decomposition, so it is an endothermic reaction. The reason for this is that for chemical reactions to happen, at the smallest scale, the individual chemical molecules have to bump into one another. As molecules warm up they begin to vibrate more and the chances of them bumping in to one another increases – the more bumps that happen, the faster the reactions go. It is represented by the equation AB > A + B or (in the case of the lab) CuSO4•5(H2O)(s) > CuSO4(s) + 5(H2O)(g) The synthesis reaction occurred when water was added to copper (II) sulfate.
The compound synthesized to copper (II) sulfate pentahydrate. Synthesis can occur when one compound or molecule is introduced to another. It is represented by the equation A + B > AB or (in the case of the lab) CuSO4(s) + 5(H2O)(l) > CuSO4•5(H2O)(s) Experiment B demonstrated a double displacement reaction/precipitate reaction. The reaction occurred when aqueous sodium sulfate and aqueous barium chloride were mixed together. The milky precipitate is BaSO4, which was a product of the reaction. The compound is a precipitate because it is insoluble (says so on the solubility table).
In double displacement reactions two ionic compounds switch cations. In order to switch cations, the ions must first be separated, in a solution for example, so that they may react with other ions present. A precipitate forms because the combination of a positive and a negative ion in solution forms a compound that is insoluble in water and precipitates out of the solution. The precipitate ions cannot re-dissolve in the mixture so they are rapidly removed from the solution. Double displacement is represented by the equation AB + CD > AD + CB or (in the case of the lab) BaCl2(aq) + Na2SO4(aq) > BaSO4(s) + 2NaCl(aq).
Experiment D demonstrated a single displacement reaction/exothermic reaction. The reaction occurred when hydrochloric acid was added to solid magnesium. Since Mg is higher than H on the activity series (more reactive), therefore Cl ions are more attracted to Mg ions and will leave H to bond with Mg. Mg is more reactive than H because it has two electrons in its valence shell, these two electrons are easily lost because they are far from the nucleus (compared to that of H) so there is less attraction, resulting in low ionization energy (energy needed to remove electrons) . Magnesium must lose its outer electrons in order to react.
Hydrogen has one valence electron and it is closer to the nucleus (compared to that of Mg) so there is more attraction, resulting in higher ionization energy. Therefore it is easier for Mg to react (lose electron) compared to H. The bubbles are H2 gas forming. Heat was released so it is an exothermic reaction. An exothermic reaction is a chemical reaction that is accompanied by the release of heat. In other words, the energy needed for the reaction to occur is less than the total energy released. As a result of this, the extra energy is released, usually in the form of heat. Single displacement is represented by the equation.
A + BC > AC + B or (in the case of the lab) Mg(s) + 2HCl(aq) > MgCl2(aq) + H2(g) Experiment E is a catalytic decomposition reaction. Hydrogen peroxide is an unstable compound that will decompose with time, just on exposure to sunlight or heat (energy), to make water and oxygen gas. It should be kept in an opaque container in a cool environment (hence the cool temperature). The rate of decomposition of hydrogen peroxide can be increased with a catalyst, like potassium iodide, which is a substance that speeds up a chemical reaction but remains chemically unchanged at the end of the reaction.
The mechanism of catalysis involves the negative iodine ions only. The mechanism is a multi sequence reaction H2O2(aq) + I-(aq)> IO-(aq) + H2O(l) H2O2(aq) + IO-(aq) > I-(aq) + H2O(l) + O2(g) *When the IO ion was made, it would react with another H2O2 molecule until there were no more H2O2 molecules left. The second step regenerated the negative iodine ion (thus acting as a catalyst). The iodine ions were not affected by the reaction, meaning it did not bond with any other substances other than it had originally (potassium) by the end of the reaction.
The iodine acted as a catalyst, it sped up the reaction without bonding in the end. The potassium did not play any role in the reaction, it was only bonded with iodine in the beginning. In fact, another substance with similar chemical properties as potassium, sodium for example, could have been bonded with iodine before the reaction, and the reaction would have been the same. Decomposition is represented by the equation AB > A + B or (in the case of the lab) 2H2O2(aq) ––KI> 2H2O(l) + O2(g) The glowing splint popped because it ignited hydrogen gas. But hydrogen gas was not made from the reaction so I assume it was in the air before.
The hydrogen gas was oxidized, it combined with oxygen gas violently and quickly to form water. The oxidization caused vibrations which is the popping sound. The splint re lit because oxygen gas is also present. The fire died eventually because the oxygen had been used up in the test tube. The glowing splint test proved that the above reaction occurred, because when a glowing splint is exposed to O2 gas, it re-ignites and that is what had happened, it is a complete combustion because there was an excess of O2 gas. Experiment F is a single displacement reaction/precipitate reaction.
The reaction occurred when iron and copper (II) sulfate came into contact. Since Fe is higher than Cu on the activity series, therefore SO4 molecules are more attracted to Fe ions and will leave Cu to bond with Fe. The precipitate created is copper, which collects on the nail. Single displacement reaction is represented by the equation A + BC > AC + B or (in the case of the lab) CuSO4(aq) + Fe(s) > FeSO4(aq) + Cu(s) Errors 1. After heating copper (II) pentahydrate, we did not allow the test tube to cool down fully before adding water. When water was added it began to boil and evaporate.
This could have changed our results because we do not know if the evaporation took away substances from the test tube, so some substances that should be there may be missing. This will give us inaccurate results. 2. We may have handled the magnesium strip and the iron nail for too long. Contaminants on our skin such as oils or other chemicals could become residue on the magnesium and iron and react in the experiment, which would give us inaccurate results (shouldn’t have happened). 3. There were people walking by the bunsen burner. The wind that their movement created may have affected the temperature of flame and thus the results.