When not all bond enthalpies are given

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Shelbyhamilton 3H
Posts: 37
Joined: Fri Sep 26, 2014 2:02 pm

When not all bond enthalpies are given

Postby Shelbyhamilton 3H » Tue Jan 27, 2015 11:09 pm

Hi- this question is concerning the winter 2014 midterm question 1 & 2 C. So it asks for the enthalpy change for the reaction using the molar bond enthalpies given above.

so the equation is CH3OH (g) + NH3 (g) -----> CH3NH2 (g) + H20 (g)

but it only gives us the bond enthalpies for C-N, C-O, N-H, And O-H.
Isn't there a C-H bond??
How can we calculate the delta h of the reaction if we don't take into account the C-H bond?

thank you so much!

Niharika Reddy 1D
Posts: 127
Joined: Fri Sep 26, 2014 2:02 pm

Re: When not all bond enthalpies are given

Postby Niharika Reddy 1D » Tue Jan 27, 2015 11:20 pm

There are C-H bonds in the reaction, but the same number of C-H bonds are present on both sides of the equation. If you were to account for them by saying 3 C-H bonds were broken, you would have to say 3 C-H bonds were formed, which cancels the bond enthalpy of the C-H bonds out when calculating the molar enthalpy change for the reaction. (You would add 3x(bond enthalpy of C-H) for the bonds broken and then subtract 3x(bond enthalpy of C-H) for bonds formed).

When calculating the enthalpy of a reaction using bond enthalpies, check to see which bonds are unique to either side. If the same number of the same bond is present on both sides, you can ignore that particular bond in the calculations, such as C-H in this case. The C-H bonds were neither broken not formed, so we don't need to know their bond enthalpies because they don't affect the calculation.


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