Unit 4 questions

[Collin Li]

As temperature increases, the system favours the position of equilibrium that would counteract the increase in energy. It would follow the endothermic reaction.

For an exothermic reaction, this means that as T increases, K will decrease, and approach 0 (position of equilibrium lies towards more reactants and less products).

For an endothermic reaction, this means that as T increases, K will increase, and approach infinite (position of equilibrium lies towards more products and less reactants).

[bec]

But why does the gradient of the endothermic reaction graph get progressively steeper as temperature increases? I understand why the K-value itself increases, but not why the rate of increase becomes higher with higher temperatures.

[Mao]

But why does the gradient of the endothermic reaction graph get progressively steeper as temperature increases? I understand why the K-value itself increases, but not why the rate of increase becomes higher with higher temperatures.

increase in temperature increases kinetic energy of molecules, that means there are more collisions, as well as the collisions being of higher energy
when you take into account the multitudes of factors that are involved, the graph wouldn't be linear. as for why dK/dt increases, that is definitely not on the course

[bec]

thanks mao, i won't worry about understanding it then!

[Collin Li]

Well, it should make sense that for when K approaches zero as T approaches infinite, you can't represent that linearly. You'd have to have it tapering off (asymptotically)

It'd then be a bit farfetched to say that in the other case (K approaches infinite, T approaches infinite), that it could be linear, because the reverse still utilises the same underlying principle, so hence it should have a similar curvature (and not a different type of relationship).

Hand waving and flip-floppy argument, but it works for me and VCE.

[Mao]

Well, it should make sense that for when K approaches zero as T approaches infinite, you can't represent that linearly. You'd have to have it tapering off (asymptotically)

It'd then be a bit farfetched to say that in the other case (K approaches infinite, T approaches infinite), that it could be linear, because the reverse still utilises the same underlying principle, so hence it should have a similar curvature (and not a different type of relationship).

Hand waving and flip-floppy argument, but it works for me and VCE.

I did not understand one bit of that

[Collin Li]

Yeah, I just realised it doesn't make any sense. Intuitively it works for me though, haha.

I've never seen those K-T graphs, but I think they'd be the logarithmic and the exponential function.

[bec]

Well, it should make sense that for when K approaches zero as T approaches infinite, you can't represent that linearly. You'd have to have it tapering off (asymptotically)

It'd then be a bit farfetched to say that in the other case (K approaches infinite, T approaches infinite), that it could be linear, because the reverse still utilises the same underlying principle, so hence it should have a similar curvature (and not a different type of relationship).

Hand waving and flip-floppy argument, but it works for me and VCE.

I did not understand one bit of that

um, neither did I...I'm just going to leave it there and not worry about understanding it!

Anyway, moving on: how do I work out the value of K_w for water at 50oC given that the pH is 6.6?
Is this covered in the Heinemann textbook at all? I couldn't find it.

Thanks!

[Mao]

given that water self ionises with this equation:



i.e. there are equal amounts of and

also since

[bec]

Thanks

[lanvins]

Another question

The equilibrium constant for the reaction given by the equation 2HI(g)------->H2(g) + I2(g) is 48.8 at 445oC. An equilibrium mixture in a 2L vessel at this temperature contains .22mol of H2 and .11mol of I2


1. Another mixture was prepared by placing 4mol of HI in a 2L vessel at 330oC. At equilibrium .44 mol of H2 and .44mol of I2 were present. calculate the value of the equilibrium at this temperature.

i got .0121 but the books says it's .020
 

[orangez]

(no units here as they cancel out)

EDIT: fixed something up

[lanvins]

why did you do this?

[orangez]

Sorry, it's just meant to be n(HI) used = 0.88 mol -- you get this from the mole ratio. Hence, the the amount of HI present at equilibrium = 4 - 0.88 = 3.12 mol.

[lanvins]

oic.....thankz