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A reaction is known have AG" of 3,000 kJ mol- What can be predicted about the kinetics of this reaction? a) it will exhibit very rapid kinetics b) it will exhi...

Question

A reaction is known have AG" of 3,000 kJ mol- What can be predicted about the kinetics of this reaction? a) it will exhibit very rapid kinetics b) it will exhibit very slow kinetics c) the kinetics can be predicted only after the reactants and the products in the reaction are known d) the kinetics of the reaction cannot be predicted from this information As catalysts _ enzymes are significantly less effective than nonenzymatic catalysts b) slightly less effective than nonenzymatic catalysts

A reaction is known have AG" of 3,000 kJ mol- What can be predicted about the kinetics of this reaction? a) it will exhibit very rapid kinetics b) it will exhibit very slow kinetics c) the kinetics can be predicted only after the reactants and the products in the reaction are known d) the kinetics of the reaction cannot be predicted from this information As catalysts _ enzymes are significantly less effective than nonenzymatic catalysts b) slightly less effective than nonenzymatic catalysts significantly more effective than nonenzymatic catalysts d) slightly more effective than nonenzymatic catalysts The rate of reaction depends on 2) the free energy change b) the activation energy c) the enthalpy change d) the entropy change The main difference between catalyzed and an uncatalyzed reaction that a) the activation energy of the catalyzed reaction is lower b) the catalyzed reaction has more favorable free energy change c) the catalyzed reaction has more favorable enthalpy change Id) the catalyzed reaction has more favorable entropy change The transition state of reaction iS a) the arrangement of atoms that can easily g0 from reactants t0 products b) the one in which reactants are placed in the reaction vessel c) the final stage of the reaction when the products are just about to appear d) none of the above The kinetic order of reaction can be determined by inspection from the coefficients of the balanced equation must be determined experimentally always depends on the concentration of enzyme never depends on concentrations of reactants



Answers

Is each of these statements true? If not, explain why.
(a) At a given T, all molecules have the same kinetic energy.
(b) Halving the P of a gaseous reaction doubles the rate.
(c) A higher activation energy gives a lower reaction rate
(d) A temperature rise of $10^{\circ} \mathrm{C}$ doubles the rate of any reaction.
(e) If reactant molecules collide with greater energy than the activation energy, they change into product molecules.
(f) The activation energy of a reaction depends on temperature.
(g) The rate of a reaction increases as the reaction proceeds.
(h) Activation energy depends on collision frequency.
(i) A catalyst increases the rate by increasing collision frequency.
(j) Exothermic reactions are faster than endothermic reactions.
(k) Temperature has no effect on the frequency factor (A).
(l) The activation energy of a reaction is lowered by a catalyst.
(m) For most reactions, $\Delta H_{\text { ran }}$ is lowered by a catalyst.
(o) The initial rate of a reaction is its maximum rate.
(p) A bimolecular reaction is generally twice as fast as a unimolecular reaction.
(q) The molecularity of an elementary reaction is proportional to the molecular complexity of the reactant(s).

Mhm. For this conceptual question, they give us information associated with three different reactions. Reaction. A has a half life that's independent of concentration. This only occurs for a first order reaction where the first order half life equation is here independent of concentration. For reaction be they state that the half life doubles when the concentration doubles. This would only occur for a zero order reaction where the concentration is in the numerator of the half life equation, so doubling the concentration doubles the half life And for the last one it tells us that the half life doubles when the concentration is cut in half. This would only occur for a second order reaction where the half life equation has the concentration in the denominator. Mhm. So of all of the statements reaction is first order reaction be a second order and so forth statements A through D. None of them correctly identify reaction A His first order reaction be a zero order and reactions see a second order. So none of the answers are correct.

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're giving an example for reaction mechanism with certain steps. So for our first element, these are both all elementary reactions. So we can derive our rate equation directly from our elementary reaction. We have to a. and equilibrium with a. two. And this is a fast step in part two, we have a two plus E. Yield B plus city. And this is a slow step. So the rate of reaction and reality is only going to depend on the rate determining step. Since whenever you have a change in concentration in one of your rate determining re agents, it's going to lead to this fast reaction to immediately replenish the concentration of everything you lost. So the reaction only really depends on the rate of the slow step. So since this is an elementary reaction we can write a rate law directly from the rate equation, which is equivalent to K. Times the concentration of a two times E. However, this isn't useful since we have the concentration of an intermediate. Since intermediates are you uh produced and consumed in the reaction and they aren't readily measured. So, first we're essentially going to write rate law for the Let's just call this rate constant K1 And the reverse rate constant K -1. Then this rate constant K two. So for the forward reaction in our first step, the rate is going to be a second order reaction as a result of basically this being an elementary reaction. So the rate would be equal. We have to choose the correct rate constant. Here would be equivalent to K1 times the concentration of a squared. We can directly get this from an elementary reaction. Reverse reactions dependent on the other rate constant K. And nurse one. So part B. The rate is equivalent to the concentration of K in verse one times the concentration of A two. Mhm. So, since the reaction is in equilibrium for for our first step, the rate of the forward reaction would be equivalent to the rate of the reverse reaction. So we can set these two equal to one another. It's okay. One times the concentration of a squared Over equals K -1 concentration of a. two. So we can rearrange this for the concentration of the intermediate and now we can substitute everything into our equation. Here, I should write this is K two. So for the rate determining step, the rate is equivalent two. You can write this as K prime times the concentration of a squared times the concentration of E. So Kay prime in this case equals KK one. K 2 over K minus one. Yeah. Edward. So our final, basically our final part is the overall reaction equation. So we can see that the A. Two cancels out. So this results in we're just going to add up the steps mm and this gives the equation for our final overall chemical reaction, and this is our final answer.


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