Posted by Mike on Sunday, January 29, 2012 at 11:38pm.
delta Hf = dHf
dHf rxn = (n*dHf products) - (n*dHf reactants).
Substitute 5113.3 kJ for dHf rxn and solve for the only unknown in the equation which is dHf C8H18. You can look up dHf for CO2 and H2O in your text or notes. Be sure and use H2O(g) and not H2O(l).
so it would be 0.96 ?is that right?
(8x-393.5)+9(-241.8)+x=5113.3
-5324.2+x=5113.3
x=0.96
I don't get 0.96 and if your math is right (I don't think it is) you should have
x = 5113.3 + 5324.2 which isn't close to 0.96
My answer is something like -211.2 kJ. Check your algebra.
[(8*-393.5) + (9*-241.8)] - x = -5113.3
yeah sorry hehe thanks!
correct answer is:
-5133.3-[(8x-393.5)+9(-241.8)] = 190.9
since it is an enthalpy of formation, the value has to be negative,
thus the correct answer is -190.9
I don't understand the math to this !!!
C8H18(g) + 25/2(O2)(g) -> 8CO2(g) + 9H2O(g)
STANDARD ENTHALPY VALUES (H):
02(g) = 0 kj/mol
CO2 (g) = -393.5 kj/mol
H20(g) = -241.8 kj/mol
H total = -5094 kJ
H of C8H18(g) is your unknown, so call that x.
use: H of reaction = H of products - H of reactants
so, -5094 kJ = [8(CO2) + 9(H2O)] - [x +12.5(O2)]
plug in your standard enthalpy values.
-5094kJ = [8(-393.5) + 9(-241.8)] - [X + 12.5(0)]
-5094 kJ = [-3148 + (-2176.2)] - [x + 0]
-5094 kJ = -5324.2 - x
add -5324.2 to -5094
to get +230.2 = -x
move the negative to the other side
and you get -230 kj/mol
that is the correct answer.
Calculate the standard enthalpy change for the following reaction at 25 °C.
2CH3OH(g) + 3O2(g)> 2CO2(g) + 4H2O(g)
the answer is -5070 kJ/mol
its actually 210.9