Hess's Law question...

[beastly115]

H2O(g) → H2O(l) ∆H (kJ/mole) = –44
C(s) + O2(g) → CO2(g) –394
H2(g) + 1/2O2(g) → H2O(l) –286
C2H5OH(l)+ 3O2(g) → 2CO2(g) + 3H2O(l) –1367

Calculate the enthalpy change for the reaction:
2C(s) + 2H2(g) + H2O(l) → C2H5OH(l)
A. –226 kJ/mole
B. +7 kJ/mole
C. +109 kJ/mole
D. +344 kJ/mole
E. +687 kJ/mole

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The answer is B, but in the answer the first rxn where H2O(g) goes to H2O(l) is not involved in the calculation to arrive at choice B. Why not?

Thanks.

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[armorshell]

Because that value is used in the next question

 

[checkamundo]

You only have to use the rxns that will give you the final rxn with the appropriate coefficients. You don't need to use the first rxn to get the final rxn given. If you had used the first rxn, there's no other reaction that has H2O(g) so you wouldn't be able to cancel it out to get the final reaction.

 

[beastly115]

You only have to use the rxns that will give you the final rxn with the appropriate coefficients. You don't need to use the first rxn to get the final rxn given. If you had used the first rxn, there's no other reaction that has H2O(g) so you wouldn't be able to cancel it out to get the final reaction.

Ok thanks.

 

[asckwan]

Could someone give a detailed calculation please?

 

[beastly115]

2 C(s) + 2 O2(g) → 2 CO2(g) 2(–394)
H2(g) + O2(g) → 2 H2O(l) 2(–286)
2 CO2(g) + 3 H2O(l) → C2H5OH(l)+ 3 O2(g) –(–1367)
________________________________________________________________________________
2 C(s) + 2 H2(g) + H2O(l) → C2H5OH(l) ∆H= 2(–394) + 2(–286) – (–1367)

 

[Rock23]

on this question the kaplan book says to that you have to use the enthalpy of the products minus the enthalpy of the reactants. if we did that though you wouldnt come out to the same answer.

 

[Chga]

yeah you would

 

[OSUDDS]

yeah you would

No, one way you get -7, the other way +7....

 

[tom_servo_dds]

No, one way you get -7, the other way +7....

Product = 1367
Reactants = 2(-394) + 2(-286)

1367 - 1360 = 7+