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6 and 527. 9 Approved by DOJ Civil Harassment Prevention CH110 Page 1 of 5 To the Person in 2 The court has granted the temporary orders checked as granted below. CH110 Clerk stamps date here when form is filed. Temporary Restraining Order To keep other people from seeing what you entered on your form please press the Clear This Form button at the end of the form when finished. Person in 1 must complete items 1...
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Welcome to determining the form of the rate law so let's look at an example reaction so we have ammonium reacting with nitrite no.2 minus ammonium plus like right react to give us nitrogen gas and 2 and water to h2o and so now that's right our radar we can say that rate is equal to negative the change in concentration of ammonium divided by the change in time and we also know that that equals k our rate constant times the concentration of ammonium to some power pen that we don't know times the concentration of nitrite to some power animal that we also don't know so we need to determine the order of the reaction with respect to ammonium and to the night right that is we needed to determine what end at em are so remember that we need to determine these experimental mean we can't do this just by looking at the reaction so we need some data from some experiment so let's pretend that I've done an experiment and let's say I've done three trials where I very the initial ammonium concentration and the initial nitrite concentration and I measure the initial rate of the reaction now if we compare the ratios for the rate laws or trials two and one so again I'll go back and try to what's this one on the middle and try one is that one so you can see that the rate here doubled in going from one to two we've doubled from 1.3 5 times 10 972 2.70 x 10 divided 7 and let's see what we changed we kept the initial ammonium concentration of the same but we doubled the night right concentration so doubling the nitrate concentration causes the initial rate two double so if we take rate to / rape one we're going to calculate this fraction so right to / 81 what we know that the rate was equal to ka times the ammonium concentration to some power n times the night right concentration to some power M so now note that in putting in the subscript 2 for the rate I'm going to put in subscript 2 for the concentration of ammonia and a subscript 2 of conservation of nitrite so we have the equation for the rate law with rate tube with trial 2 divided by the rate and the concentrations for trial one again make sure you put in this N and M there so we know that the rate for the second trial was to pluck 70 times 10 to the minus 7 moles per liter per second or over per second and we know that we're dividing that by 1.3 5 times 10 of 97 molar per second so our units there will cancel heater and this fraction of course simplifies to just two so we have 2 is equal to K times 0.100 over to the N power times zero point zero zero five zero mower ready to the nth power I've made a mistake here let's go back and look it looks like I've taken the concentration from trial one and match that with the rate from trial to so these two need to be flipped because this fraction gives us two and when we flip this not least this side of the equation will also give us too because of course what we put these two then those will cancel and the K here cancels as well that's a now we're left with again once we flip it 0.010 or / 0.00 50 or so we're left with 2 to the M on this side so again I can't stop a bunch of stuff there and we left with 2 is equal to 2 again this should be flipped so we have 2 100 is equal to 2 10 to the mpower well the only way that that can be true is if m is equal to 1 so this means that now our reaction order with respect to no.2 minus the nitrite is 1 so we say this is the reaction is first order with respect to nitrite so if we then compare the ratios of the rate laws for trial three and try out too so again we're putting great three with concentrations from trial three divided by rate from trial to and the concentrations from trial to well again we've doubled the concentration I started we've done with the rate in our units are going to cancel their our k's are going to cancel now in this case the ammonium concentration will cancel but the nitrate concentration will cancel along with empower there so now we can figure out what this end is and in this case I've actually plugged in the concentrations in the correct places and so now i have to point zero zero because again we've doubled our rate from 2.72 5.45 107 molar for a second and then here our molar cancels and this fraction as a value of tube it's still raised to the nth power and now the only way that we can have two equal to 2 to the N is if n is equal 1 so again our reaction order is with respect to ammonium is one and so we say that the reaction is first order with respect to ammonia so the overall reaction order is the sum of N and M & P and Q and so on if you have further reactants involved so n plus M is 1 plus 1 which is too so the overall reaction is second order but its first order with respect to each of the reactants so now we can calculate the value of our rate constant k again remember we have great sequel k times concentration of ammonium to the ends power Titus concentration of nitrite empower and now we know that N and M are each one so if we want we can rearrange this equation solving for K so we put K over our left we get rate is it to about in concentration to end times nitrate concentration to the end and now we can plug in some values we know that n is 1 we know that n is 1 here I've picked the values for trial one the rate for trial one is 1.3 5 times 10 to minus 7 lower per second and the concentrations are 0 100 m / n 0.005 04 and then when we work that out we get a value for K of 2.7 talks to the minus 4 note here that our units then become liters per mole per second that was determining the form of the rate law
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