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Overview All chemical reactions have a specific rate defining the progress of reactants being converted to products

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Overview All chemical reactions have a specific rate defining the progress of reactants being converted to products. This rate can be influenced by temperature, concentration, presence of a catalyst and surface area of the reactants. The general form of a rate equation is shown below: ???? = ?[?]?[?]? where A and B are concentrations of different species, m and n are reaction orders, and k is the rate constant. The rate of chemical reactions change over time as reactants are depleted. The rate constant, however, is fixed for any single reaction at a given temperature. In this experiment, we will explore two methods for determining the rate law Determining a Rate Law There are two common methods for determining reaction orders. In the method of initial rates, a reaction is run several times varying the starting concentration of a reactant (keeping all other concentrations constant) and the initial reaction rate determined for each experiment. The reaction order for the reactant can then be determined using the relationship. ???? 1 ] ???? 2 ???????? ????? = [?] ??? [[?]1 ] 2 ??? [ Where 1 and 2 refer to the data obtained in each of two experiments. An alternative approach for reactions involving one reactant is to graphically determine which integrated rate law agrees with the experimental data. In this method data is collected for concentration of the reactant and time as the reaction progresses and then how the two are related is observed. In particular, if a plot of ln[Reactant] versus time is linear the reaction is 1 st order, if a plot of 1/[Reactant] versus time is linear the reaction is 2nd order and if a plot of [Reactant] versus time is linear the reaction is 0 th order. This is summarized in the table below. Order Rate Law Integrated Rate Law Linear Plot Slope Units of k 0th rate = k [A]t=[A]0 -kt [A] versus t -k M/s 1st rate = k[A] ln[A]t = ln[A]0 - kt ln[A] versus t -k s-1 2nd rate = k[A]2 1/[A]t = 1/[A]o + kt 1/[A] versus t k M-1s-1 Catalytic decomposition of H2O2 In this experiment, the catalytic decomposition of hydrogen peroxide over a platinum catalyst is explored. Since the platinum is a catalyst, only one species is involved which decomposes into two products according to the reaction below: 2H2O2(aq) → O2(g) + 2H2O(l) Because one of the products, O2, is a gas, the increase in pressure of the system over time can be measured and the Ideal Gas Law (PV = nRT) used to relate pressure to moles. Once that is done for several different concentrations of the reactant, the reaction order and rate constant can be determined. Experimental Procedure 50.0 mL of 0.882 M H2O2 and a platinum catalyst disc with a volume of 1.0 cm3 was placed in a test tube that had a total volume of 150.0 mL. The test tube was then immediately fitted with a pressure sensor and the increase in pressure due to the oxygen produced was monitored twice per minute for 3 minutes. This procedure was then repeated with 50.0 mL of 0.706 M H2O2. Both experiments were conducted at a temperature of 25.0 oC. Experimental Data The following experimental data was collected. T = 25.00 oC [H2O2]0 = 0.882 M Time (s) Poxygen (torr) 0 0.00 0 30.00 90 30.00 72 60.00 178 60.00 142 90.00 263 90.00 211 120.00 348 120.00 278 150.00 430 150.00 344 180.00 510 180.00 408 Time (s) Poxygen (torr) 0.00 [H2O2]0 = 0.706 M Questions: 1. Are the differences in the two sets of data what you would expect? Explain your answer. 2. What volume in liters will the oxygen produced occupy? 3. Complete the following tables. Give an example calculation for each column (attach additional sheets of paper). Give your answers in scientific notation to 4 significant figures. [H2O2]0 = 0.882 M [H2O2]0 = 0.706 M Time (s) Poxygen (torr) 0.00 0.00 30.00 90.0 60.00 178.0 90.00 263.0 120.00 348.0 150.00 430.0 180.00 510.0 Time (s) Poxygen (torr) 0.00 0.00 30.00 72.0 60.00 142.0 90.00 211.0 120.00 278.0 150.00 344.0 180.00 408.0 Moles of O2 produced Moles of H2O2 [H2O2] (M) consumed 0.882 Moles of O2 produced Moles of H2O2 [H2O2] (M) consumed 0.706 4. Calculate the average rate of decomposition of H2O2 between 0 and 3 minutes for each experiment. Clearly show your work. Give your answers in scientific notation to 4 significant figures. 5. Assuming the average rate of reaction over the first three minutes is a good approximation to the initial rate, use the method of initial rates to calculate the rate law and rate constant for this reaction. Clearly show your work. Give your answer for the rate constant to 4 significant figures. 6. Create a plot of ln[H2O2] versus time for each experiment and determine the slope of the best fit lines. Attach your graphs to this report. 7. Are your graphs consistent with the rate law and rate constant determined in 4? Explain your answer.