Lab9 Report

Discussion The data obtained from this experiment indicates that the kinetic order of the iodide ion is first order within the iodine clock reaction. We can determine this by first taking the equation 𝑅𝑒???𝑖?? 𝑅??𝑒 = ?[?] ? [?] ? . We can then plug in the relative molarity of Iodide (A) and Hydrogen Peroxide (B) in trial 1. Next we look at trial 2, in which the molarity of the Iodide ion has been halved. We once again use the equation 𝑅𝑒???𝑖?? 𝑅??𝑒 = ?[?] ? [?] ? but this time plug in the values of A and B from trial 2 of the experiment. Now, since the ratio of reaction rates in trial one and two must be equal to the ratio of their ?[?] ? [?] ? counterparts, we can divide the equation derived in trial one by the equation derived in trial two: 2.97𝑒−4 1.32𝑒−4 = 𝑘[.111] ? [.196] ? 𝑘[.111] ? [.098] ? . Now if we eliminate like terms and simplify, we find that 2.25 = 2 ? . Here we can see that m is about equivalent to one, therefore iodide is first order. If we once again follow this same method but instead use trials one and three, we ’ re left with 2.97𝑒−4 1.66𝑒−4 = 𝑘[.111] ? [.196] ? 𝑘[.056] ? [.196] ? . Simplifying this yields the equation 1.789 = 2 ? . We can see here that n is also about equivalent to one, therefore hydrogen peroxide is also first order within the iodine clock reaction. In the last part of the experiment we measured the effect of temperature on reaction rate, and using this data and the Arrhenius plot/equation, we can find the activation energy of this reaction. First, the data was plotted as shown to produce the Arrhenius plot. Next, the Arrhenius equation: ? = ?𝑒 −𝐸𝑎 𝑅𝑇 was simplified by taking the natural logarithm of each side to find ln(?) = ln(?) − 𝐸𝑎 𝑅𝑇 . This was then further expanded to ln(?) = (− 𝐸𝑎 𝑅 ∗ 1 𝑡 ) + ln (?) this now resembles the linear equation of form ? = ?? + ? . Now, we can plug in the derived equation for our data, considering x is equivalent to 1 𝑡 and b is equivalent to ln (?) . Now, we can see that −4087.3 = − 𝐸𝑎 𝑅 where 𝑅 = 8.31 ? ? −1 ??? −1 . Now by solving for activation energy, we find that 𝐸? = 33,965.463 ????𝑒? .