# chemistry

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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)?

• chemistry -

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).

• chemistry -

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

• chemistry -

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
[(8*-393.5) + (9*-241.8)] - x = -5113.3

• chemistry -

yeah sorry hehe thanks!

• chemistry -

-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

• chemistry -

I don't understand the math to this !!!

• chemistry -

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

to get +230.2 = -x

move the negative to the other side
and you get -230 kj/mol

• chemistry -

Calculate the standard enthalpy change for the following reaction at 25 °C.

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

• chemistry -

• chemistry -

its actually 210.9

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