Post by: kenhung123 on June 22, 2010, 06:37:08 pm
E.g. we have
So what can we say about the reverse reaction? Like is the activation energy the same?
Post by: vexx on June 22, 2010, 07:15:33 pm
but anyway..
the reverse reaction will have different activation energy, if you look at the energy profile diagram for whatever is reacting, you will see that it has either absorbed or emitted energy, and so depending on the products having moved to a higher energy state or lower energy state- there will be a different required activation energy for the reverse reaction (if lower state, more energy required for reverse).. but they can have similar activation energies if the energy diagram shows they are on the same level...
also this reverse reaction will require the opposite amount of energy that it did initially to react (if it was a change of -200kj, then it'll be a change of 200 in reverse).
that answer your question? and also someone correct me if i'm wrong, i read through the first few pages so quickly that i may not have interpreted this section correctly!
Post by: kenhung123 on June 22, 2010, 08:19:29 pm
Post by: vexx on June 22, 2010, 09:31:24 pm
we have=200kJ mol-1, activation energy=150kJ mol-1 enthalpy of reactants 150kJ mol-1 and enthalpy of products 350kJ mol-1
uhh i haven't learnt this yet, but i think it's not too difficult..
since the change of energy is positive, it means energy was absorbed, 200kj mol-1, and since there was 150kj of activation energy, there would be 200-150=50kj of energy required as activation energy for the products to reverse the reaction..
(can anyone confirm please..?)
Post by: fady_22 on June 22, 2010, 10:50:31 pm
yes there iswe have
uhh i haven't learnt this yet, but i think it's not too difficult..
since the change of energy is positive, it means energy was absorbed, 200kj mol-1, and since there was 150kj of activation energy, there would be 200-150=50kj of energy required as activation energy for the products to reverse the reaction..
(can anyone confirm please..?)
For an endothermic reaction, I highly doubt that the activation energy is lower than the change in enthalpy.
Post by: vexx on June 22, 2010, 11:35:13 pm
yes there iswe have
uhh i haven't learnt this yet, but i think it's not too difficult..
since the change of energy is positive, it means energy was absorbed, 200kj mol-1, and since there was 150kj of activation energy, there would be 200-150=50kj of energy required as activation energy for the products to reverse the reaction..
(can anyone confirm please..?)
For an endothermic reaction, I highly doubt that the activation energy is lower than the change in enthalpy.
i need someone to confirm, i haven't learnt this yet, nor have i read much on it.. i wanted to have a stab, so someoneeee checkk :3
Post by: Whatlol on June 23, 2010, 01:26:43 pm
yes there iswe have
uhh i haven't learnt this yet, but i think it's not too difficult..
since the change of energy is positive, it means energy was absorbed, 200kj mol-1, and since there was 150kj of activation energy, there would be 200-150=50kj of energy required as activation energy for the products to reverse the reaction..
(can anyone confirm please..?)
For an endothermic reaction, I highly doubt that the activation energy is lower than the change in enthalpy.
Yea i think that is basically an impossible situation... you cant have activation energy as 150kjmol^-1 and change in chemical energy as 200kjmol^-1
Post by: kenhung123 on June 23, 2010, 02:00:36 pm
Post by: Martoman on June 23, 2010, 04:49:55 pm
Here is my botched effort at it.
To make things simple you zero the reactants at 0. The delta H is going to be whatever you stiupate it to be then, if its 50 you mark it at 50 if its -374 then you mark it at -374. Similarly, the activation energy is marked at whatever you call it to be. In this case it is 150.
http://img526.imageshack.us/i/diagraml.png/
You are right in saying that the backward reaction has an activation energy of 100. By looking at the diagram and starting from right to left you start at 50, and need to "hike up a hill" of 150. You're already 50 up. So treck a further 100.
Another obvious statement is that the new delta H is -50 as it releases heat.
Post by: kenhung123 on June 23, 2010, 06:37:47 pm
Post by: m@tty on June 23, 2010, 07:17:54 pm
Makes a lot of sense now martoman thanks
Yes, it does, now that you changed the information in the original post.
Post by: vexx on June 23, 2010, 07:20:04 pm
lol yeah i noticed that..Makes a lot of sense now martoman thanks
Yes, it does, now that you changed the information in the original post.
Post by: kenhung123 on June 23, 2010, 07:28:47 pm
Yea because it didn't make sense since the products had higher energy than the activation so i reduced it O.o? I was just wondering what I needed to do to find activation energy in reverse reactionMakes a lot of sense now martoman thanks
Yes, it does, now that you changed the information in the original post.
Post by: Mao on June 25, 2010, 08:08:37 am
Yea because it didn't make sense since the products had higher energy than the activation so i reduced it O.o? I was just wondering what I needed to do to find activation energy in reverse reactionMakes a lot of sense now martoman thanks
Yes, it does, now that you changed the information in the original post.
Be careful with that. Enthalpy of reactants and products are usually measured on a semi-absolute scale [i.e. measured with respect to some arbitrary zero point]. Change in enthalpy and activation energy are both relative quantities that measure change [relative to the enthalpy of reactants].
The restriction on activation energy is almost negligible: for endothermic reactions, it must be greater than
Post by: iffets12345 on June 25, 2010, 10:06:35 pm
I think the activation energy of reverse if like 150-50=100kJ mol-1?
I was thinking this too, is it correct? Since I thougth the reverse reaction would be the graph of the original flipped horizontally, so the energy released in the original becomes activation for the reverse reaction?