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)?

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).

(8x-393.5)+9(-241.8)+x=5113.3

-5324.2+x=5113.3

x=0.96

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

-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

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.

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

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