AP CHEMISTRY NEED HELP PLEASE!
posted by Mandy on .
Calculate the enthalpy of combustion of C3H6:
C3H6 (g) + 9/2 O2 (g) yield 3CO2 + 3H2O
using the following data:
3C (s) + 3H2 (g) yield C3H6 (g) change in H= 53.3 kJ
C(s)+O2 (g) yield CO2 (g) change in H= -394kJ
H2 (g) + 1/2 O2 (g) yield H2O (l) change in heat= -286 kJ
(a) -1517 kJ
(b) 1304 kJ
(c) -626 kJ
I got answer b.
Can anyone please help me. Have a huge exam tomorrow. Teacher has been out sick and sub is no help. Thanks
You need to set these up so that if you add up the equations you get the equation in the question.
So we need to get to
C3H6 (g) + 9/2 O2 (g) -> 3CO2 + 3H2O
C(s)+O2 (g) -> CO2 (g) dH= -394kJ
and we need three of these so
3C(s)+3O2 (g) -> 3CO2 (g) dH= -1182kJ
H2 (g) + 1/2 O2 (g) -> H2O (l) dH= -286 kJ
again we need three of these
3H2 (g) + 3/2 O2 (g) -> 3H2O (l) dH= -858 kJ
The last equation
3C (s) + 3H2 (g) -> C3H6 (g) dH= 53.3 kJ
we need to reverse so
C3H6 (g) -> 3C (s) + 3H2 (g) dH= -53.3 kJ
(notice that I have revesed the sign on dH as I have reversed the equation)
If we now add equations A., B. and C. we will get the equation in the question (you should check that this is true). We can also add the dH values so
-1182kJ + -858 kJ + -53.3 kJ = -2093kJ