## question 8.57

ERIKTORRESDisc3C
Posts: 18
Joined: Wed Sep 21, 2016 2:58 pm

### question 8.57

I had trouble starting this problem: "Determine the reaction enthalpy for the hydrogenation of ethyne to ethane,from the following data: $C_{2}H_{2 }(g)+ 2H_{2}(g)\rightarrow$$\Delta H_{c}^{\circ}( C_{2}H_{2 }(g))=-1300 kJ.mol^{-1}, \Delta H_{c}^{\circ}( C_{2}H_{6 }(g))= -1560 kJ.mol^{-1}, \Delta H_{c}^{\circ}( H_{2 }(g))=-268 kJ.mol^{-1}$$$ "

Eljie_2F
Posts: 19
Joined: Fri Jun 17, 2016 11:28 am

### Re: question 8.57

Since delta H sub c is H of combustion, to use the given values for compounds that are created (products), you must reverse the sign of the given value and then add it to the value of the reactants. This will give you the reaction enthalpy for hydrogenation.

Anthony_Imperial_1G
Posts: 22
Joined: Wed Sep 21, 2016 2:59 pm

### Re: question 8.57

Because this is a combustion reaction (subscript of C), C2H2, C2H6, and H2 will be ignited with O2 to form H20 and CO2. You should have three different reactions that show this. Since your final equation is show as C2H2+2H2 ---> C2H6, you have to rearrange and balance equations in a way that will equal your final equation. Don't worry about having fractions for coefficients on certain molecules because it's the only way to reach 1 mole of C2H2 and C2H6.

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