Lab4 Report

Discussion In the beginning of this experiment, it was necessary to calculate the calorimeter constant in order to complete further experiments. To find this 𝐶 ?𝑎? we start by finding the initial temperature of cold water and hot water, and measure the equilibrium temperature they settle to when mixed inside the calorimeter. The water should settle at the average of the two temperatures if measured in equal amounts, but the calorimeter walls also absorb heat, allowing us to determine the calorimeter constant. Because all of the heat given off by the hot water is absorbed by either the calorimeter or the cold water, we can set up the equation ? ℎ?𝑡 = −(? ???? + ? ?𝑎??𝑟𝑖??𝑡?𝑟 ) . We can define this equation further using the equation for heat transfer: (? ∗ ? ∗ ∆𝑇) ℎ?𝑡 = −(? ∗ ? ∗ ∆𝑇) ???? − (? ∗ ∆𝑇) ?𝑎??𝑟𝑖??𝑡?𝑟 . This can be further simplified to 𝐶 ?𝑎? = − ? ℎ2𝑜 ? ℎ2𝑜 (∆𝑇 𝐻 +∆𝑇 𝐶 ) ∆𝑇 𝑐𝑎𝑙 . In trial one, where ∆𝑇 𝐻 = -26 degrees Celsius, ∆𝑇 ? = 25 degrees Celsius, ∆𝑇 ?𝑎? = 25 degrees Celsius, ? ℎ2? =30 grams, and 𝐶 ℎ2? =4.184 J/g degree Celsius, we can find 𝐶 ?𝑎? =5.021 J/ degree Celsius. Once we have an average 𝐶 ?𝑎? , we can use this value to calculate the enthalpy of a decomposition reaction of hydrogen peroxide. We know that heat released due to the decomposition is transferred into the water and calorimeter, therefore ? 𝑟𝑥? = −(? ℎ2? + ? ?𝑎??𝑟𝑖??𝑡?𝑟 ) , we also know that ? 𝑟𝑥? is equal to the extrinsic enthalpy of this reaction because pressure and volume are constant. Expanding out this relationship gives us ∆𝐻 = −(? ℎ2? 𝐶 ℎ2? ∆𝑇 + 𝐶 ?𝑎? ∆𝑇) . In trial one of the experiment, ? ℎ2? = 32.3 𝑔?𝑎?? , both heat capacities are the same as earlier, and ∆𝑇 = 19.2 ??𝑔???? 𝐶?????? . Plugging in these numbers to the equation, we ’ re given ∆𝐻 = −2.69 ???𝑜?𝑜???? . If we divide this value by the number of hydrogen peroxide present in the sample, we will get ∆𝐻 ??? . Since there are 0.025 mols of hydrogen peroxide present (molarity of the solution times the volume added) ∆𝐻 ??? = -107.55 kJ/mol. With this enthalpy of reaction for the decomposition of hydrogen peroxide, we then determined the enthalpy of oxidation of copper due to hydrogen peroxide using Hess ’ s Law. We want to find the enthalpy of reaction for: 𝐻 2 ? 2 (𝑎?) + 𝐶?(?) → 𝐶??(?) + 𝐻 2 ? (𝐿) , and we know that 𝐻 2 ? 2 (𝑎?) → 𝐻 2 ? (𝐿) + 0.5? 2 , with ∆𝐻 ??? = -107.55 kJ/mol, so we simply need an equation that has copper and oxygen on the reactant side, and copper oxide on the product side. This reaction is defined by 𝐶?(?) + 0.5? 2 (𝑔) → 𝐶??(?) , and we can look up that it has an enthalpy of formation of ∆𝐻 ? = -157.3 kJ/mol. Now we can simply combine these two equations, and their enthalpies to form the oxidation reaction due to hydrogen peroxide. The added enthalpies tell us that our experimental enthalpy of reaction is ∆𝐻 𝑅𝑥? = -264.9 kJ/mol. A second way to derive this value (the theoretical way) is to determine the ∆𝐻 𝑅𝑥? using other defined enthalpies of formation. Using Hess ’ s Law, we know that ? 𝑅𝑥? = ∆𝐻 𝑅𝑥? = ∑ ∆𝐻 𝐹 (??𝑜?????) − ∑ ∆𝐻 𝐹 (𝑅?𝑎??𝑎???) . We can look up standard enthalpies of formation for copper( ∆𝐻 𝐹 = 0) , hydrogen peroxide( ∆𝐻 𝐹 = −187.8?𝐽/?𝑜?) , copper oxide ( ∆𝐻 𝐹 = −157.3 ?𝐽/?𝑜?) , and water ( ∆𝐻 𝐹 = −241.8 ?𝐽. ?𝑜?) online or in the textbook. Summing the enthalpies of the products, and subtracting the reactants, gives us ∆𝐻 𝑅𝑥? = −211.3 ?𝐽/?𝑜? .