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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