[NL-bucky]

∆Hreaction = (enthalpy of bonds broken) - (enthalpy of bonds formed) (from your own answer to problem 1)

So propagation A: +7 kJ/mol
propagation B: -108 kJ/mol

Adding those together gives an overall reaction enthalpy of -101 kJ/mol, which is your answer to problem 1 (what a surprise! see how you can check the consistency of your answers to see whether you are on the right track?)

This value is the difference in enthalpy between the beginning and the and of the reaction, so the reaction curve would indeed look like a parabole (it always does) with its maximum more or less in the middle.

[NL-bucky]

Now, for the radical fluorination, the mechanism is identical to the chlorination, except for the F instead of the Cl of course. The values are different of course.

The problem states that you do not need to calculate the energy for the first step of the radical fluorination per se, but maybe it is illustrative that you do. You will see that the value is much different than for the chlorination, also changing the reaction enthalpy curve.

[theanonymous]

∆Hreaction = (enthalpy of bonds broken) - (enthalpy of bonds formed) (from your own answer to problem 1)

So propagation A: +7 kJ/mol
propagation B: -108 kJ/mol

Adding those together gives an overall reaction enthalpy of -101 kJ/mol, which is your answer to problem 1 (what a surprise! see how you can check the consistency of your answers to see whether you are on the right track?)

This value is the difference in enthalpy between the beginning and the and of the reaction, so the reaction curve would indeed look like a parabole (it always does) with its maximum more or less in the middle.

Ohhhh I get it!!!!
Ok so now the potential energy diagram should look like this:



Sorry for taking so long to respond...I have to draw it on my paper first, take a picture with my macbook, edit the photo, and upload on photobucket xD

Ok, now for the radical fluorination...hold on

[theanonymous]

Now, for the radical fluorination, the mechanism is identical to the chlorination, except for the F instead of the Cl of course. The values are different of course.

The problem states that you do not need to calculate the energy for the first step of the radical fluorination per se, but maybe it is illustrative that you do. You will see that the value is much different than for the chlorination, also changing the reaction enthalpy curve.

Yeah I did that in 2B after you corrected me!



Ok so the potential energy diagram for fluorination of CH4 would look like this:

[NL-bucky]

It doesn't does matter for the answer of the question, but please be aware:

The +sign in front of the value means the enthalpy after the reaction is higher than before the reaction. This means that you still have the graph reversed.

Implications (not necessary to answer the question, but gives you more insight):  You probably learned that this reaction step (the first propagation of the chlorination) is thus infavorable, and it in fact is! That does not mean that the reaction will not occur, just that the first step is endothermic. The total reaction is still exothermic, as the sign for the total reaction enthalpy is negative.

The graph for the fluorination is also reversed. The reactants are here higher in energy than the products.

We are almost there!

[NL-bucky]

What feature of the energy diagram represents the speed of the reaction?

[theanonymous]

Wait so....

the graph for chlorination - I have to reverse it?

Like this?



Fluorination:

[theanonymous]

What feature of the energy diagram represents the speed of the reaction?

The transition state!

The transition state comes much earlier in the graph for fluorination (in the reaction coordinate) than in chlorination, which means that it is faster. Is that the answer to Number 5?

[NL-bucky]

The graphs are correct!

For the answer to the final question: the transition state is important, but it is not the x-coordinate of the TS that determines the speed of the reaction...

[NL-bucky]

I am sorry, I should be off, so I will post the answers with explanations now.

Indeed the y-coordinate of the transition state determines the speed of the reaction! It is called the activation energy of the reaction, and you can view it as the barrier the molecules have to overcome to become products.

A catalyst speeds up a reaction not by changing anything about the reactants or the products, but by lowering this barrier. If you would add a catalyst and then draw the graph, the reactants and products would be at the same position, yet the peak would be much lower.

Now we compare the two graphs with the Hammond postulate in hand. The chlorination has the TS more or less in the middle, so it resembles product nor reactant. The fluorination has the TS more towards the reactants. We cannot compare the two barriers quantitatively, but intuitively, which barrier should be higher?

[theanonymous]

I am sorry, I should be off, so I will post the answers with explanations now.

Indeed the y-coordinate of the transition state determines the speed of the reaction! It is called the activation energy of the reaction, and you can view it as the barrier the molecules have to overcome to become products.

A catalyst speeds up a reaction not by changing anything about the reactants or the products, but by lowering this barrier. If you would add a catalyst and then draw the graph, the reactants and products would be at the same position, yet the peak would be much lower.

Now we compare the two graphs with the Hammond postulate in hand. The chlorination has the TS more or less in the middle, so it resembles product nor reactant. The fluorination has the TS more towards the reactants. We cannot compare the two barriers quantitatively, but intuitively, which barrier should be higher?

Chlorinations activation barrier will be higher (i mean)

[NL-bucky]

You can view the x-coordinate of your graph like a trajectory the molecules have to follow to get to the products. This implies that the molecules should only change a little to reach the TS for the fluorination, and a lot more for the chlorination. The graphs show you that the barrier for the fluorination is therefore lower, just because the change the molecules have to make to reach the TS is a lot smaller.

Keep in mind that the fluorination is not faster because the reaction enthalpy for the fluorination is bigger, but because the barrier is lower. This time they coincide, and they do this more often, but it is not necessarily the case. It is however very useful to use it as a rule of thumb when comparing two similar reactions, as we did here.

Hope I helped you! And if there is more questions, that someone else can help you further.

Now addressing the more senior members of this forum: did I do well explaining? It seemed fun to help people, and I was thinking of helping more often, so tips are more than welcome

[theanonymous]

You can view the x-coordinate of your graph like a trajectory the molecules have to follow to get to the products. This implies that the molecules should only change a little to reach the TS for the fluorination, and a lot more for the chlorination. The graphs show you that the barrier for the fluorination is therefore lower, just because the change the molecules have to make to reach the TS is a lot smaller.

Keep in mind that the fluorination is not faster because the reaction enthalpy for the fluorination is bigger, but because the barrier is lower. This time they coincide, and they do this more often, but it is not necessarily the case. It is however very useful to use it as a rule of thumb when comparing two similar reactions, as we did here.

Hope I helped you! And if there is more questions, that someone else can help you further.

Now addressing the more senior members of this forum: did I do well explaining? It seemed fun to help people, and I was thinking of helping more often, so tips are more than welcome

Ohhh that makes sense!!!

Thanks so much!!!!!