Question

what is the change in enthalpy for the following reaction?
2H2(g) + 2C(s) + O2(g) d C2H4OH(l)?

C2H5OH(l) + 2O2(g) -> 2CO2(g) + 2H2O(l),  
DH = -875 kJ
C(s) + O2(g) -> CO2(g),  DH = -394.51 kJ
H2(g) + ½O2(g) -> H2O(l),  DH = -285.8 kJ

is it -194.7 kJ...?

Answer 1

nooo it will be -477 KJ

Answer 2

could you explain how?

Answer 3

reverse the equation
2CO2(g) + 2H2O(l)===> C2H5OH(l) + 2O2(g) DH = +875 kJ -1

multiply next equations by 2
2C(s) + 2O2(g) -> 2CO2(g), DH = -394.51*2 kJ -2
2H2(g) + O2(g) -> 2H2O(l), DH = -285.8 8*2 kJ -3

add all the three you get

2H2(g) + 2C(s) + O2(g) d C2H4OH(l) =-789-571+875 = -477 KJ
hope u got it

Answer 4

but wait, would that be -486 kJ instead?

Answer 5

wouldn't*

Answer 6

yaa ur right its 485.8 something :)

Answer 7

okay!(: Thank you.
I truly appreciate your help.

Answer 8

you are welcome :)

Answer 9

could you help me with another one?
If so, do I need to post in a different thingy or can I just ask in here?

Answer 10

@chmvijay

Answer 11

you just post in another :)

Answer 12

do I have to 'close' this one?