## 8.57

Essly Mendoza 1J
Posts: 25
Joined: Fri Sep 29, 2017 7:04 am

### 8.57

Hi guys I'm having trouble on this problem right here: Determine the reaction enthalpy for the hydrogenation of ethyne to ethane,
C2H2(g)+ 2 H2(g) --> C2H6(g), from the following data: Delta Hc (C2H2, g)= -1300. kJ·mol^-1, Delta Hc (C2H6, g)= -1560. kJ·mol^-1, Delta Hc (H2, g)=- 286 kJ·mol^-1. How do we use this information to get started? thank you!

Shawn Patel 1I
Posts: 54
Joined: Thu Jul 27, 2017 3:01 am

### Re: 8.57

Hey,
The question is referring to enthalpy of combustion, so you would create a combustion reaction for each of the reactants and products. Then with the combustion reactions, you would balance them and order them so that if you would add them, you would be left with the original equation. This means that you would flip C2H6 combustion reaction and make the change in enthalpy positive 1560. Then you would multiply the enthalpy of the hydrogen reaction by two, and add all the enthalpies together.

Joyce Lee 1C
Posts: 52
Joined: Fri Sep 29, 2017 7:03 am
Been upvoted: 1 time

### Re: 8.57

I think of it as a little puzzle. Basically, you either reverse the given reactions and enthalpies or multiply them by an integer in a way that makes their sum equal to the reaction for the hydrogenation of ethyne to ethane. The given ΔHc implies combustion.
C2H2(g)+ 2 H2(g) --> C2H6(g) This is the outcome we want.
The given reactions are:
1) ΔHc (C2H2, g)= -1300. kJ/mol
C2H2(g) + 5/2 O2 (g) --> 2CO2 (g) + H2O (l)
2) ΔHc (C2H6, g)= -1560. kJ/mol
C2H6 (g) + 7/2 O2 (g) --> 2CO2 (g) + 3H2O (l)
3) ΔHc (H2, g)= -286 kJ/mol
H2 (g) + 1/2 O2 (g) --> H20 (l)

You reverse the second reaction and get a new enthalpy of +1560 kJ/mol
Then add it to the first reaction, the 2CO2 cancels out on both sides and the sum of the enthalpies is 260 kJ.
Then multiply the third reaction by 2 and add it to the result from above.
You should get C2H2 (g) + 2H2 (g) --> C2H6 (g) and ΔH = -312kJ

Essly Mendoza 1J
Posts: 25
Joined: Fri Sep 29, 2017 7:04 am