Chem 178 SIRate Laws and Arrhenius Equation - SolutionsWednesday, August 29, 2007(These are the last two problems that we didn’t finish in time on Wednesday)2. a) Graphically find the rate law for the following reaction:SO2Cl2(g) → SO2(g) + Cl2(g)Time (min) [SO2Cl2] (M)0 0.1000100 0.0876200 0.0768300 0.0673400 0.0590500 0.0517600 0.0453700 0.0397800 0.0348900 0.03051000 0.02671100 0.0234Upon graphing ln[SO2Cl2] (y-axis) vs. t (x-axis), you should see a straight line with a negative slope. In the SI session, we calculated that the negative of the slope of the line (and therefore k) was:k = 1.32 x 10-3 min-1b) If an experiment is started with a concentration of 0.80 M of SO2Cl2, how long will it take for the concentration to be 0.20 M?We know two things that are important:First, this is a first order reaction, so the half life is t1/2 = 0.693/k.Secondly, we know from part (a) that k = 1.32 x 10-3 min-1We started with 0.80 M of SO2Cl2. Half of 0.80 is 0.40, and half of that is 0.20. Thus, we’re looking for the time it takes for TWO half-lives to pass. So, we need to find 2t1/2.2t1/2 = 2 (0.693/k)2t1/2 = 2 (0.693/1.32 x 10-3 min-1)2t1/2 = 1050 min = 17.5 hChem 178 SIRate Laws and Arrhenius Equation - SolutionsWednesday, August 29, 20073. rate = k[N2O5]a) At 45 degrees C, the rate constant for the reaction with the rate law given above is6.22 x 10-4 s-1. If the initial concentration of the N2O5 in the solution is 0.100 M, how longwill it take for the concentration to drop to 0.0100 M?We know that this is a first order reaction. Thus, we can use the equation:ln[A]t = -kt + ln[A]0We’re solving for time t, so we can rearrange the equation to solve for t:ktt0]ln[]ln[ AA141022.6)100.0ln()0100.0ln(sMMtt = 3702 s = 61.7 min = 1.03 hb) If the initial concentration of N2O5 is 0.500 M, what will the concentration be after exactly 1 hour?We’ll use the same equation to start.ln[A]t = -kt + ln[A]0Now, we have to rearrange it to solve for [A]t by making each side an exponent of e:0ln[A] -kt ]ln[ eetA-kt 0][][ etAA [A]t = 0.500e-(6.22x10-4s-1)(3600s)[A]t = 0.0533
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