Asked by Mike
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)?
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)?
Answers
Answered by
DrBob222
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).
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).
Answered by
Mike
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
(8x-393.5)+9(-241.8)+x=5113.3
-5324.2+x=5113.3
x=0.96
Answered by
DrBob222
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
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
Answered by
Mike
yeah sorry hehe thanks!
Answered by
Joe
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
-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
Answered by
kassandra
I don't understand the math to this !!!
Answered by
Paige
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.
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.
Answered by
Ashli
Calculate the standard enthalpy change for the following reaction at 25 °C.
2CH3OH(g) + 3O2(g)> 2CO2(g) + 4H2O(g)
2CH3OH(g) + 3O2(g)> 2CO2(g) + 4H2O(g)
Answered by
milinag
the answer is -5070 kJ/mol
Answered by
Anonymous
its actually 210.9
Answered by
Jeidie
-210.9 is KJ/Mol is the correct answer.
I got that for my sapling
I got that for my sapling
Answered by
Jayce
The actual answer to this question would be -220 kj/mol . Wow youre welcome lol
Answered by
Eman
Jayce is correct, I just did this on sapling, the correct answer was -220kj/mol
Answered by
John
The answer is 210.9!
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