Ch. 8 #57

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Michelle Dong 1F
Posts: 110
Joined: Fri Sep 29, 2017 7:04 am

Ch. 8 #57

Postby Michelle Dong 1F » Mon Jan 15, 2018 9:33 am

Determine the reaction enthalpy for the hydrogenation of ethyne to ethane, C2H2 (g) + H2 (g) --> C2H6 (g), from the following data: delta Hc (C2H2,g)= -1300 kj/mol, delta Hc (C2H6,g) = -1560 kj/mol, delta Hc (H2,g)= -286 kj/mol.

Can someone explain how to do this? Why is it that delta Hc(C2H6,g) becomes positive? Why does delta Hc (C2H2,g) and delta Hc (H2,g) stay negative?

Curtis Tam 1J
Posts: 105
Joined: Thu Jul 13, 2017 3:00 am

Re: Ch. 8 #57

Postby Curtis Tam 1J » Mon Jan 15, 2018 10:33 am

Here, they give you the enthalpy of combustion, so first you would have to write combustion equations for each of the molecules they give you. But do so for one mole of each combustible.

C2H2 + 5/2O2 ----> 2CO2 + H2O
C2H6 + 7/2O2 ----> 2CO2 + 3H2O
H2 + 1/2O2 ----> 2 H2-=O

Rearrange the equations for each equation above so that the net equation is:

C2H2 +2H2 ----> C2H6

You are going to have to reverse some equations or multiply them in order to get the net equation above. Then just add the adjusted enthalpies of combustion to get the total enthalpy of combustion.


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