For a particular isomer of C8H18, the following reaction produces 5113.3 kJ of heat per mole of C8H18(g) consumed, under standard conditions.
C8H18(g) + 25/2(O2)(g) -> 8CO2(g) + 9H2O(g)
DeltaH= -5113.3 kJ
What is the standard enthalpy of formation of this isomer of C8H18(g)?
-210.9 is KJ/Mol is the correct answer.
I got that for my sapling
I don't understand the math to this !!!
The actual answer to this question would be -220 kj/mol . Wow youre welcome lol
yeah sorry hehe thanks!
Jayce is correct, I just did this on sapling, the correct answer was -220kj/mol
The answer is 210.9!
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).
Calculate the standard enthalpy change for the following reaction at 25 °C.
2CH3OH(g) + 3O2(g)> 2CO2(g) + 4H2O(g)
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
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.