Mean Bond Enthalpy

Original post by liquid394

How do you work it out.
Here is a question:
Mean bond enhalpy values/KJMol-1:
H-H=436
C=C=612
C-C=348
Use the following equation and data above to calculate the value for the C-H bond enthalpy in ethane?

Thanks for any help in Advance :biggrin:

There appears not to be a following equation.

Reply 2

OP

Original post by joostan

There appears not to be a following equation.

Ofcourse:

CHCH + H2 ---> CH3CH3 DeltaH=-136KJmol-1

Reply 3

Original post by liquid394

Ofcourse:

CHCH + H2 ---> CH3CH3 DeltaH=-136KJmol-1

This equation is not true. . .

Reply 4

OP

Original post by joostan

This equation is not true. . .

Its what they gave in the book, I just want to work out what the Average enthalpy is for the C-H bond

Reply 5

Artymess

16

Original post by joostan

This equation is not true. . .

In what way isn't the equation true? :lolwut:

Unless I'm missing something painfully obvious, they're just reducing ethene :sly:

Edit: On closer inspection, you might be right. Either we're missing some hydrogens, or there's a triple bond between the carbons :erm:

(edited 11 years ago)

Reply 6

Original post by Artymess

In what way isn't the equation true? :lolwut:

Unless I'm missing something painfully obvious, they're just reducing ethene :sly:

Edit: On closer inspection, you might be right. Either we're missing some hydrogens, or there's a triple bond between the carbons :erm:

precisely

Reply 7

OP

Original post by Artymess

In what way isn't the equation true? :lolwut:

Unless I'm missing something painfully obvious, they're just reducing ethene :sly:

Edit: On closer inspection, you might be right. Either we're missing some hydrogens, or there's a triple bond between the carbons :erm:

Original post by joostan

precisely

If I drew the displayed formulas for CHCH + H2 ---> CH3CH3

It would be:

H-C---C-H (triple bond) + H-H (single) -----> H3-C-C-H3 (single)

Reply 8

Original post by liquid394

If I drew the displayed formulas for CHCH + H2 ---> CH3CH3

It would be:

H-C---C-H (triple bond) + H-H (single) -----> H3-C-C-H3 (single)

There aren't enough Hydrogens on the left :smile:

Reply 9

jadpan

you would need Delta H of a reaction involving the hydrogenation of ethene i believe to solve this not thought about it for long. you could then work out the bond breaking enthalpies reverse sign of foramation and the bonds formed and equate this to the overall delta H and do the maths, though i need pen and paper to show this and by the looks of it you dont have enough information.

Posted from TSR Mobile

Reply 10

Caribbeantwist

12

I think I've done this question!!!

This was posted from The Student Room's iPhone/iPad App

Reply 11

jadpan

c2h4 + h2 = c2h6

Posted from TSR Mobile

Reply 12

OP

Original post by joostan

There aren't enough Hydrogens on the left :smile:

Original post by alexandraa

I think I've done this question!!!

This was posted from The Student Room's iPhone/iPad App

Jadpan You're correct :biggrin:

Original post by jadpan

c2h4 + h2 = c2h6

Posted from TSR Mobile

The overall equation is :

H-C=C-H + H-H ------------> H3C-CH3
Ethene + Hydrogen ---------> Ethane

DeltaH=-136Kjmol-1

Use the equation above and the bond enthalpies bonded above on 1st post to calculate a value for C-H bond enthalpy in ethane.

Reply 13

Original post by liquid394

Jadpan You're correct :biggrin:

The overall equation is :

H-C=C-H + H-H ------------> H3C-CH3
Ethene + Hydrogen ---------> Ethane

DeltaH=-136Kjmol-1

Use the equation above and the bond enthalpies bonded above on 1st post to calculate a value for C-H bond enthalpy in ethane.

I'd suggest writing it like this:
C=C + 2C-H + H-H ==> C-C + 6C-H
As DeltaH=-136Kjmol^-1
You know that the difference between the bond enthalpies = delta H

Reply 14

jadpan