Question

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

CHoices are
(a) -1517 kJ
(b) 1304 kJ
(c) -626 kJ
(d) -2093

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

Answer 1

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

A.
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

B.
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

C.
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