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I) According collision theory ofchemical reactions; A) high energy collisions result few" successful reactions there isnt suflicient time for the products TacL...

Question

I) According collision theory ofchemical reactions; A) high energy collisions result few" successful reactions there isnt suflicient time for the products TacL B) low energy collisions result many successful reactions aS there sumlicient time: for the reactants fon products C)high cnergy collisions lead the successful fonation of products low energy collisions do not occur in the gas phase. E) all of the above2) Which ofthe following changes will increase reaction rate? An increase In Ihe c

I) According collision theory ofchemical reactions; A) high energy collisions result few" successful reactions there isnt suflicient time for the products TacL B) low energy collisions result many successful reactions aS there sumlicient time: for the reactants fon products C)high cnergy collisions lead the successful fonation of products low energy collisions do not occur in the gas phase. E) all of the above 2) Which ofthe following changes will increase reaction rate? An increase In Ihe concentrtion of mexclants An incrcase In temperature 3. Higher-energy collisions between reacting molecules and only and only and only DJ All 0f 4,2 and E) Neither 2.or } 3) Why does the rate of the reaction decrease Over time? A)Exelhermic reaclions lose heat #hich cook: Ihe reaction which decreases rcaction B) As the reaction proceeds. the concentration of the products results in fewer llisions: C) As the reaction procecds; decrease in the concentration of rcactants results in fewer successful collisions D) Not all molecules Will react and some choose stay their prescnc TOT nana of the above chemical equilibrium exists when: A) reacuants are completely changed t0 products B) there are equa) amounts of reactants and products C)the rate at which reactants lonu products becomes zero the sum of reactant and product concentrations cquals mol E) the rate which reactants form products is the same as the rale which products formn reactants 5) For the reaction 7(2) =25F(g)- '(g). the equilibrium expression is: Isaa A)S= CEFII B)6= USle C)k= Fwz Dat



Answers

- A chemical reaction with stoichiometry $\mathrm{A} \rightarrow$ products is said to follow an $n^{\text {th }}$ -order rate law if $\mathrm{A}$ is consumed at a rate proportional to the $n$ th power of its concentration in the reaction mixture. If $r_{\mathrm{A}}$ is the rate of consumption of A per unit reactor volume, then $$r_{\mathrm{A}}[\operatorname{mol} /(\mathbf{L} \cdot \mathbf{s})]=k C_{\mathrm{A}}^{n}$$ where $C_{A}(m o l / L)$ is the reactant concentration, and the constant of proportionality $k$ is the reaction rate constant. A reaction that follows this law is referred to as an $n^{\text {th }}$ order reaction. The rate constant is a strong function of temperature but is independent of the reactant concentration. (a) Suppose a first-order reaction ( $n=1$ ) is carried out in an isothermal batch reactor of constant volume $V$. Write a material balance on $A$ and integrate it to derive the expression $$C_{\mathrm{A}}=C_{\mathrm{A} 0} \exp (-k t)$$ where $C_{\mathrm{A} 0}$ is the concentration of $\mathrm{A}$ in the reactor at $t=0$ (b) The gas-phase decomposition of sulfuryl chloride $$\mathrm{SO}_{2} \mathrm{Cl}_{2} \rightarrow \mathrm{SO}_{2}+\mathrm{Cl}_{2}$$ is thought to follow a first-order rate law. The reaction is carried out in a constant-volume isothermal batch reactor and the concentration of $\mathrm{SO}_{2} \mathrm{Cl}_{2}$ is measured at several reaction times, with the following results: Verify the proposed rate law graphically [i.e., demonstrate that the expression given in Part (a) fits the data for $\left.C_{\mathrm{A}}(t)\right]$ and determine the rate constant $k,$ giving both its value and its units.

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.

So here we have some reactions to consider. Got a two negations form equilibrium K one k two The folding. The vast reaction to I in the case is full. We have to I know gaseous form at H two. That is an equilibrium with K three and K four to H. I guess this form So from Equation two, the rate of the forward reaction is equal to K one multiplied by I two concentration of right of reverse reaction. It's equal to K two. I squared. So at equilibrium the rate of the forward reaction is equal to the rate of the back reaction. It's my equilibrium. K one by two is equal to cater to. I squared so we can rearrange that For K one over K two is equal to concentration of Ise squared But by I two. So kay equilibrium is equal to I squared over I too where we have I two geishas form as an equilibrium with two I in the gaseous form where K equilibrium is equal to I squared over Bye two. So from the second equation we know that an equilibrium K three all right too H two is equal to K for H I, too. That's the forward and backwards reaction are equal. So K three K four equal to H I squared, divided by I squared. Okay, two. That's equal to K equilibrium. So therefore we can write out the overall reaction Page two and the geishas format. ITunes geishas formats include women to H I in the gaseous form where K C is equal to H. I squared over by two multiplied by age two. No.

Okay. So to start with two given information here is that when one mole of nitro Glazer and decomposes, we end up with a collection of gases that at one atmosphere and 1950 degrees Celsius take up 1323 leaders. And in the first part of the question is, if we have 0.4 moles of nitro glycerin, how many moles of gas are we going to create eso to start off? Let's figure out how many moles of gas we got from one mole of nitro glycerin. So one atmosphere 13 23 leaders 1950 Celsius. Let's go ahead and use PV equals NRT are ideal gas law we're solving for end So P V over r t equals end. And so what that tells us is our pressure is one atmosphere are volume is 1323 leaders are our value is 13230.8 to 1. And of course, as always, we need to convert into Kelvin S O. That would be 22 23. Uh, kill them. So let's go ahead and do that math out. So 13. 23 divided by 230.8 to 1 divide by to 2 to 3. That gives a 7.25 moles of gas created for every one mole of nitro glycerine. So if we take our 7.25 and multiply it by 0.4 we get 0.29 moles of hot gas being created. So that is part a of the question. Uh, part B is telling us that we're going to Ah, I would do this in a 500 milliliters flask on. Cool it to negative 10 degrees Celsius. Ah, and then the pressure inside the flask is 623 after product, A solidifies. So, uh, what we're gonna do here is we're going to look at product a solidifying some. First of all, it is how many old moles of a were present. And what is its likely identity? Well, we know that whenever we have hydrocarbons, combusting or exploding, usually we get co two and H 20 as products. And so h 20 is a very likely product that would solidify by the time we got to negative 10 degrees Celsius. But what we do need to figure out then is how many moles of gas we have left. We're going to assume that the ice, if it is ice, has no volume. It's pretty common to assume solids have no volume on and we're going to at a new page. We're going to use P V over r T equals and to figure out how many moles of gas we have left from our original 0.2 nights. So the pressure now is 623 millimeters of mercury. The volume is 0.5 leaders. The our value since now we're in millimeters of mercury is going to be 62.4 and the temperature is negative. 10 Celsius, which is 263 killed in so are in value now is zero point zero one nine. So originally it was 0.29 Now it's 0.19 which tells us that we lost zero point 01 moles of H 20 Uh uh make sure that's what the question is. Actually asked how many moles of a were present? Ah yes. So we have 0.1 moles of a were present. Uh, part C is asking us eso when when these air passed through, when what's left is passed through a tube of powdered ally to oh, gas be reacts to form Ally to CEO. Three. The remaining gas to see India collected another 500 milliliters flask, a found of a pressure of 260 millimeters of mercury at 25 degrees Celsius. How many moles of B were present and what is its likely identity? OK, well, uh, let's start with the identity cause that's pretty easy If we have alli to oh, we have to add something to it to get Ally to CEO three. Then clearly what we are adding to it is CEO, too. So for adding CO two, that's that's the identity of be there. And the question is, ah, when. So the question is how many moles of of that co two did we have. So we now have a 500 milliliters flask, 260 mil millimeters of mercury and 25 degrees Celsius. We can use the same equation using our new pressure of 260 millimeters of mercury. Another 500 millimeter flask are. Our value is still is still 62.4 and our temperature is now 25 degrees Celsius, which is 298 Kelvin. And so what we have left then to 60 times 600.5. Uh, divide 62.4. Divide to 98 gives us 0.7 is what's left. Which means that what we lost is 0.3 Moles of co two. Okay, so that's part C. Finally, we have part D on and then I guess we have party afterward. But let's move on to part D, Sephardi says, When gases cnd were passed through a hot tube of powdered copper gas, see reacted to form copper two oxide, the remaining gas D was collected in 1/3 500 militar flask found to have a mass of 5000.168 grams and a pressure of 223 millimeters of mercury, a 25 degrees Celsius. How many moles of CND were present and what are their likely identities? Okay, so moving on to a new page s So we know that when we added copper to something, we ended up with copper two oxide. So obviously this gas had to be 02 That's what gas see is and what we are going to remember here is that the remaining gas. So the nitrogen gas that's left in a 500 military flask Ah, a lot of pressure of 223 millimeters of mercury, it 25 Celsius. So going back and using our same equation. 223 millimeters millimeters of mercury millimeters of mercury s times are 0.5 leaders divide by our 62.4 and by our 298 gives us to 23 times 0.5 divide 62.4 divide to 98 point 006 moles of whatever D is, which means that since we had 0.7 before, we have point 00 one moles of oh too. And 0.6 moles of D that 0.6 moles of d we also know takes up a mass of 0.168 grams 0.168 grams. Divide by 0.6 Moles gives us a Moeller mass of 28 grams per mole and based on the ah, compounds that we have, the elements that we have in our original that tells us that are 28 grams from all is probably in two. So then the last thing we have to do is we have to add it all together. We have to get our, uh, our full balanced equation here. So we know that we end up with 0.6 moles. Pretty much everything as we've been doing this has has come out very close to the to the thousands, starting with the 0.4 moles of nitro glycerin. So if we just multiply everything through by 1000 we should be able to get our equation. So we have our four moles of the Nitro glycerin, which is C three h five in 309 And what do we get out of it? We get ah, first the water vapor and how much water vapour did we get? We got 0.10 moles, so multiply that by 1000 and we have 10 h 20 and then be waas points Here is there are three moles of CO two So that's three co two, see, was our one. Oh, too. I'm not gonna write the one. I'm just gonna right 02 And then finally six into and nice and convenient that we finished right with that end to right there. And now, let's double check. Always good to double check. Make sure that we have proper balancing here. So our carbon is not balanced properly. Why not? Did we end up with the wrong amount of carbon there? Yes, I see what I did. Okay, I subtracted this 0.0 Zahra seven from the amount of water instead of from the amount of Ah, the amount that was left the 0.19 So I should have got 0.0. Let me go ahead and fix that right here on this part. Here. It's not 0.3 It is actually 0.12 moles of carbon dioxide. And so what that's going to do here is it's gonna do same thing instead of three carbon dioxides. Of course, it is going to be 12 carbon dioxides eso Now we can see that we've got 12 carbons and we've got 12 carbons right here. We've got 20 hydrogen ins, and here they all are 20. Hydrogen is right there. We've got 12 nitrogen ins and here's 12. Nitrogen is right here and then we've got 36 oxygen's and out of those 36 Oxygen's there's 10 of them here, plus another 24 that's 34 plus another two here. That's 36. So this balances out. That is the correct equation for the explosion of nitro glycerine.


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