6th Edition, 8.57

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Hai-Lin Yeh 1J
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6th Edition, 8.57

Postby Hai-Lin Yeh 1J » Tue Jan 29, 2019 4:21 pm

Determine the reaction enthalpy for the hydrogenation of ethyne to ethane, C2H2(g) + 2 H2(g) -> C2H6(g), from the following data: ΔHc(C2H2, g) = -1300. kJ·mol-1, ΔHc(C2H6, g) = -1560. kJ·mol-1, ΔHc(H2, g) = -286 kJ·mol-1. Can someone explain how to do this?

Andrea Zheng 1H
Posts: 61
Joined: Fri Sep 28, 2018 12:26 am

Re: 6th Edition, 8.57

Postby Andrea Zheng 1H » Wed Jan 30, 2019 5:58 pm

Hc is the standard enthalpy change of combustion. This means that the given values are the change in enthalpy when each substance reacts with O2, and thus the change in enthalpy when the substances are the reactants in the equation. Because C2H6 is a product in the equation given in the exercise, you would flip the sign of the Hc given to +1560. Because enthalpy is a state property, you can add up the Hc values. This means you would do (-1300)+(2*-286)+(1560) = -312 kJ/mol.

A way to visualize this would be to write out the combustion reaction for each of the substances and then manipulate the equations so that when added together, they cancel out to C2H2 + 2H2 -> C2H6. When manipulating the equation, if you reverse a reaction, the Hc would change signs (from positive to negative or vice versa).

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