8.57

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RenuChepuru1L
Posts: 58
Joined: Thu Jul 27, 2017 3:00 am

8.57

Postby RenuChepuru1L » Sat Jan 20, 2018 1:48 pm

Can someone do this question step by step? should I just know what hydrogenation means or is there another way to figure out initial reactants and products?

Zane Mills 1E
Posts: 80
Joined: Fri Sep 25, 2015 3:00 am
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Re: 8.57

Postby Zane Mills 1E » Sat Jan 20, 2018 1:59 pm

Hydrogenation is just to react with hydrogen (H2) which usually takes away double bonds in the process. For this you should just be able to use the given enthalpy changes in the balanced equations (reverse the seccond, add the first, and double the third eqn) then when you add up the entalpies in the end it will represent your desired eqn and dH will be -312kJ/mol

Diane Bui 2J
Posts: 61
Joined: Sat Jul 22, 2017 3:00 am

Re: 8.57

Postby Diane Bui 2J » Sat Jan 20, 2018 2:26 pm

In this problem, the standard enthalpy of combustion for C2H2, C2H6, and H2 is given. Hence, the combustion equations for these reactions should be determined:
Balanced combustion equation for 1 mole of C2H2:
C2H2 + 2.5O2 --> 2CO2 + H2O reaction enthalpy: -1300 kJ

Balanced combustion equation for 1 mole of C2H6:
C2H6 + 3.5O2 --> 2CO2 + 3H2O reaction enthalpy: -1560 kJ

Balanced combustion equation for 1 mole of H2:
H2 + 0.5O2 --> H2O reaction enthalpy: -286 kJ

Given these equations and enthalpies of combustion, manipulate the equations using Hess's Law to result in the final balanced equation of:
C2H2 + 2H2 --> C2H6

Adding up the enthalpies, you should result in a reaction enthalpy of -312 kJ/mol.

Danah Albaaj 1I
Posts: 50
Joined: Fri Sep 29, 2017 7:07 am

Re: 8.57

Postby Danah Albaaj 1I » Sat Jan 20, 2018 2:29 pm

The values of each standard enthalpy of combustion they provided can just be plugged in to find the reaction enthalpy. So your equation to find the enthalpy will look like this
deltaH = [1 mol(-1560 kJ/mol)] - [1 mol(-1300 kJ/mol) + 2 mol(-286 kJ/mol)]
The final answer will then be -312 kJ. If you're still confused, it might help to write out the combustion equations and look at solving from that angle.


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