TSR Chemistry Society

imtired

i've just looked up lymph system, i had no idea what it was but i take it the fatty acids are in an environment which allows them to be transported as liquids.

Pretty much yeah. If you look up "chyle" instead of lymph you'll find better information. It's basically lymph of the digestive system (though there is more to it) which is very rich in fatty acids and triglycerides.

Reply 783

OP

quick question pleeease:

1/2N2 + 1 1/2 H2---> NH3
(is this equation correct btw-this isn't the question though!)

Basically if the above equation is correct, how would I got about calculating the enthalpy of formation from these mean bond enthalpies:

H-H = 436
C-C = 348
C=C = 612
N=N = 944 (triple bond btw)
N-H = 388

I ask because I have half moles and I'm confused as to how to interpret the info with the half moles I have on the reactants

Thanks

Reply 784

Malsi101

quick question pleeease:

1/2N2 + 1 1/2 H2---> NH3
(is this equation correct btw-this isn't the question though!)

Basically if the above equation is correct, how would I got about calculating the enthalpy of formation from these mean bond enthalpies:

H-H = 436
C-C = 348
C=C = 612
N=N = 944 (triple bond btw)
N-H = 388

I ask because I have half moles and I'm confused as to how to interpret the info with the half moles I have on the reactants

Thanks

You'd just do it as normal using the ("bonds broken - bonds formed" equation), but doing this:
0.5(944)
1.5(436)
3(388)

Do you understand that? I will explain further if not :smile:

Reply 785

OP

yes thanks

it's just sometimes I get confused as to why for example NH3 would be times 3 but 2NH3 is times 6. Is there any rules for like determining how many times to multiply things by e.g. with H20 it's times by 2 and with 2H20 it's times by 4. I guess it's not that difficult to see why but I just get confused easily :redface:

Thanks for that though I was hoping you could help me with this:

use the following equation and the data from the table above(it's the data I gave in my first question) to calculate a value for the enthalpy of ethane:

C2H4 + H2 -----> C2H6

DeltaH = -136 kJmol^-1

It has the bonds of the equation drawn out btw and if you like I can draw them if you need to see

Reply 786

OP

Like how the beeezers would I work out the bond enthalpy of C2H4 ..... I don't know what to times by given the above data!

Reply 787

Malsi101

Like how the beeezers would I work out the bond enthalpy of C2H4 ..... I don't know what to times by given the above data!

Right.

In a molecule of C₂H₄, the bonds will be as follows:
1 C=C bond (612)
4 C-H bonds (You'll beed the mean bond enthalpy for the C-H bond)

C₂H₄ + H₂ -> C₂H₆

So:

[612 + 4(C-H) + 436] - [348 + 6(C-H)]

Reply 788

http://store.aqa.org.uk/qual/gceasa/qp-ms/AQA-CHM2-W-QP-JAN06.PDF
According to question 1)(b) on that exam paper, the mean bond enthalpy for a C-H bond is 412. As all the other data in the question is identical to the data you've been given (C=C and C-C), I think it's safe to use that.

Reply 789

OP

Ah yes I don't have the C-H it's told me I need to calculate it-find it out!

Seeing as they've given me the enthalpy change I think they want me to work out something I don't know (the C-H bond enthalpy in Ethane) from things I do know: the above data and the enthalpy change which is -136

But I have no idea how to !!! Because it doesn't want me to find the bonds formed it wants just one value in ethane - the C-H so does that mean I just divide by 6 once I've find out the bonds made in ethane?

If this doesn't make sense....I'm sorry and I'll repeat the question :smile:

Reply 790

Malsi101

Ah yes I don't have the C-H it's told me I need to calculate it-find it out!

Seeing as they've given me the enthalpy change I think they want me to work out something I don't know (the C-H bond enthalpy in Ethane) from things I do know: the above data and the enthalpy change which is -136

But I have no idea how to !!! Because it doesn't want me to find the bonds formed it wants just one value in ethane - the C-H so does that mean I just divide by 6 once I've find out the bonds made in ethane?

If this doesn't make sense....I'm sorry and I'll repeat the question :smile:

I'll post a (potentially helpful :tongue:

) reply in a few minutes :smile:

Reply 791

OP

it's confusing because i can easily rearrange the equation however I can;t find out the bonds broken as there's no C-H value for even that :eyeball:

Reply 792

Malsi101

it's confusing because i can easily rearrange the equation however I can;t find out the bonds broken as there's no C-H value for even that :eyeball:

What's making me have to think is that there are 4(C-H) on the left, and 6(C-H) on the right.

[612 + 4(C-H) + 436] - [348 + 6(C-H)] = -136
[4(C-H) + 1048] - [6(C-H) + 348] = -136

Is currently my status :tongue:

Reply 793

OP

TBH I think it's just as simple as -136/6

Reply 794

OP