For a particular isomer of C8H18, the following reaction produces 5113.3 kJ of heat

QuestionsCategory: ChemistryFor a particular isomer of C8H18, the following reaction produces 5113.3 kJ of heat
Lisa Howell asked 5 months ago
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)?

11 Answers
Lisa Howell answered 5 months ago
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).

Lisa Howell answered 5 months ago
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

Lisa Howell answered 5 months ago
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

Lisa Howell answered 5 months ago
yeah sorry hehe thanks!

Lisa Howell answered 5 months ago
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

Lisa Howell answered 5 months ago
I don’t understand the math to this !!!

Lisa Howell answered 5 months ago
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.

Lisa Howell answered 5 months ago
Calculate the standard enthalpy change for the following reaction at 25 °C.

2CH3OH(g) + 3O2(g)> 2CO2(g) + 4H2O(g)

Lisa Howell answered 5 months ago
the answer is -5070 kJ/mol

Lisa Howell answered 5 months ago
its actually 210.9

Lisa Howell answered 5 months ago
-210.9 is KJ/Mol is the correct answer.
I got that for my sapling

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