Chapter 8 8.57

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Nicole Anisgard Parra 2H
Posts: 39
Joined: Sat Jul 22, 2017 3:01 am

Chapter 8 8.57

Postby Nicole Anisgard Parra 2H » Tue Jan 16, 2018 6:41 pm

"Determine the reaction enthalpy for the hydrogenation of ethyne to ethane, C2H2(g) + H2(g) --> C2H6(g), given the following standard enthalpies of combustion: C2H2(g)= -1300 KJ/mol; C2H6(g)= -1560 KJ/mol; H2(g)= -286 KJ/mol."
While I was able to obtain the correct answer (-312 KJ) by making the enthalpy of combustion for C2H6 positive (1560 KJ), I am a little confused on how we would divide up the reaction into the reaction enthalpies of the steps we would use to obtain the final answer. Given that the standard enthalpies of combustion are given, I would assume you would write out the combustion reactions for C2H2, C2H6, and H2, but how does one write a combustion reaction for H2? For the C2H6 combustion reaction, would the equation be reversed in order to provide a positive enthalpy of combustion?

Rana YT 2L
Posts: 49
Joined: Thu Jul 27, 2017 3:01 am

Re: Chapter 8 8.57

Postby Rana YT 2L » Tue Jan 16, 2018 8:10 pm

you would write out the combustion analyses for each molecule and manipulate the equations and enthalpy values in a way so that you end up with the equation provided when all 3 combustion reactions are added together.
you would write the combustion reaction for H2 without it forming CO2.
yes, the equation would be reversed because it is being formed in the final equation

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