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Part AThe reactant concentration in zero-order reaction was 100 M alter 195 and OOx10- M alter 395Whal is the rate constant for this reaction?Express your answer wi...

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Part AThe reactant concentration in zero-order reaction was 100 M alter 195 and OOx10- M alter 395Whal is the rate constant for this reaction?Express your answer with the appropriate units. Indicate the multiplication of units, as necessary; explicitly elther with multiplication dot or dash:View Available Hint(s)kothValueUnitsSubmitPrevious AnswersIncorrect; Try Agaln; attempts remalning The graph of concentration versus time yields straight line with slope of the Iine formed by connecting those

Part A The reactant concentration in zero-order reaction was 100 M alter 195 and OOx10- M alter 395 Whal is the rate constant for this reaction? Express your answer with the appropriate units. Indicate the multiplication of units, as necessary; explicitly elther with multiplication dot or dash: View Available Hint(s) koth Value Units Submit Previous Answers Incorrect; Try Agaln; attempts remalning The graph of concentration versus time yields straight line with slope of the Iine formed by connecting those points_ You may want t0 roviow Hint Calculato tho rise_ Therefore if you treat the given values as poit coordinates; you can calculate the slope of Part B Complete previous part(s) Part € The reaclant concentration in first-order reaction was 1Ox10 M alter 20.0 $ and 8.OOx10-: M alter 70.0 What is the rale constant for this reaction? Express your answer with the appropriate units. Indicate the multiplication of units; as necessary; explicitly either with multiplication dot or dash: View Available Hintl(s) E Mst Value Units Submit Part D The reactant concentration second-order eaction was 390 M after 200 and S0*10-? M after 895 What the rate constant for this reaction? Express your answer with the appropriate units. Indicate the multiplication of units; as necessary; explicitly either with multiplication dot or dash: View Available Hint(s) E] Value Units Submit kzad



Answers

The graphing calculator can run a program that can tell you the order of a chemical reaction, provided you indicate the reactant concentrations and reaction rates for two experiments involving the same reaction.
Go to Appendix C. If you are using a TI-83 Plus, you can download the program RXNORDER and run the application as directed. If you are using another calculator, your teacher will provide you with key-strokes and data sets to use. At the prompts, enter the reactant concentrations and reaction rates. Run the program as needed to find the order of the following reactions. (All rates are given in M/s.)
a. $2 \mathrm{N}_{2} \mathrm{O}_{5}(g) \rightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)$
$\mathrm{N}_{2} \mathrm{O}_{5} :$ conc. $1=0.025 \mathrm{M} ;$ conc. $2=0.040 \mathrm{M}$
rate $1=8.1 \times 10^{-5} ;$ rate $2=1.3 \times 10^{-4}$
b. $2 \mathrm{NO}_{2}(g) \rightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$
$\mathrm{NO}_{2} : \mathrm{conc.} 1=0.040 \mathrm{M} ; \mathrm{conc} .2=0.080 \mathrm{M}$
rate $1=0.0030 ;$ rate $2=0.012$
c. $2 \mathrm{H}_{2} \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{O}_{2}(g)$
$\mathrm{H}_{2} \mathrm{O}_{2} :$ conc. $1=0.522 \mathrm{M} ;$ conc. $2=0.887 \mathrm{M}$
rate $1=1.90 \times 10^{-4} ;$ rate $2=3.23 \times 10^{-4}$
d. $2 \mathrm{NOBr}(g) \rightarrow 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$
NOBr: conc. $1=1.27 \times 10^{-4} \mathrm{M} ;$ conc. $2=$ $4.04 \times 10^{-4} \mathrm{M}$ rate $1=6.26 \times 10^{-5} ;$ rate $2=6.33 \times 10^{-4}$
e. $2 \mathrm{HI}(g) \rightarrow \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g)$
HI: conc. $1=4.18 \times 10^{-4} \mathrm{M} ;$ conc. $2=$ $8.36 \times 10^{-4} \mathrm{M}$
rate $1=3.86 \times 10^{-5} ;$ rate $2=1.54 \times 10^{-4}$

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So, uh, we went to first draw a, uh, an energy profile for this reaction of hydrogen and flooring. And we know that there are a total of three steps, and so there should be an activation energy hump for each step. And we also know that the first step is thea rate determining step. Ah, or it has the highest activation energy, so that makes it three determining step. So the energy profile is going to look something like this with energy on the vertical and the reaction coordinate on the horizontal. So there will be reacting, set some level, and then a fairly high activation energy. And then, ah, we don't know anything about the activation energies for the other two. So we could ah, uh, come to a situation where the second step is very fast. And then the third step is not as fast. Uh, so something like this reactant and products. And then this is the transition state one for the first step in transition State two for the second transition state three intermediate one intermediate to and eso the big activation energy would be this first step. Then let's see, what else do we want to do here. Ah, reasonable mechanism. So we're looking at the reaction of ah h two and F two forming to H f. And so clearly, we've got a big change in the bonding here. The hydrogen atoms are bonded to each other, but they won't wind up that way so that H h Bond has to be broken. Ah, and likewise for the flooring Adams. They're bonded to each other and react. It's and bonded to the hydrogen send products, and so that FF bond has to be broken. So the bond energy data that they gave us here is of use because ah, the hydrogen. The hydrogen bond is very strong, but the flooring to flooring bond is very weak around 1/3 of the energy of the hydrogen bought. And so it's likely that the first thing that's gonna happen is the flooring molecule is going to separate into two flooring. Adams and then those may be fairly reactive toward the hydrogen eso that we get one HF there noticed the highest bond energy is the HF, so the tendency to form that bond is is great, is huge, is large. Ah, and that would leave us a free hydrogen. And then the flooring that's left over from step one would react with the hydrogen from step to make. The other H efforts of those would be our, uh, our steps. And this would be the slow step because, uh, the energy has got to be put in to break that bond. Part of the energy and breaking the age age bond is going to come from the flooring approaching one of those hydrogen atoms and forming that HF bond. And so that's why I wouldn't have as great an activation energy even though you're breaking, uh, those hydrogen hydrogen bonds. And then let's see, what else is there anything else we need to do or consider any other answer we need to give, um, of the reactant that was limiting in the experiments? Well, if you had a huge excess of flooring and you had small amounts of hydrogen ah, then even even at ah, that excess limiting relationship that we use mawr floor ain being involved in the slowest step, uh, would probably have and effect in increasing the reaction. But the same K was observed in ah, the same apparent great constant was observed in both runs and eso. Probably the hydrogen was the one that was in excess first and then in greater excess. And so it would have been the flooring. Ah, that would be limiting the one that we were actually watching, uh, reacting and decrease in concentration is three. Action went on.

So chemical kinetics is an area off physical chemistry on this allows us to understand the rates of chemical reactions where we study chemical process rates as well as transformations off reactant to products. So for a three step reaction where the first step is the limiting step, the diagram may look as follows. Well, we've got E, which is our relative energy against given time. We have, ah, reactant. And then we've got our rate limiting step here where it's the highest energy that we must achieve in order for the reaction to continue its three steps, as you can see, and then we achieve our products. So moving on to the next part, we have the FTO f is slow f at H 22 HF h is fast, and H ad F to H f is also fast. So our overall equation is f two at H two equals to eight. Jeff. So after two is our limiting re agent here. And the limiting re agent is the reactor that determines how much of the product is made. The other reactant are sometimes referred to as being in access in our equation.


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