q on Hess's law =/

[Tina324]

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

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

so originally i did H(products)-H(reactants) but didn't get the right answer which is B. i understand how to do hess's law..but why doesn't my way work? when can we NOT use hess's law?

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[heatwheat]

Are you sure you typed the question out right?

 

[funktion]

I just did it and got the right answer, you should set up the equations that are given with their enthalpy changes to arrive at the equation you need to calculate enthalpy change for. The way you tried to do it is incorrect because that is used to calculate enthalpy change from heat of formations for each reactant which are not provided in this question.

This is how I answered:

C2H5OH(l)+ 3O2(g) -->2CO2(g) + 3H2O(l)= –1367, reverse this equation and get:

2CO2(g) + 3H2O(l) --> C2H5OH(l)+ 3O2(g)= +1367 add this to:

2C(s) + 2O2(g)--> 2CO2(g) =(–394 * 2) need to multiply by 2 to be able to cancel out reactants, then add this to

2H2(g) + O2(g)--> 2 H2O(l)= (–286 * 2) if you add these equations together and cancel out same species on different side of arrow then you will arrive at equation of the question. just have to add up enthalpies and you get the answer. GL w/ your studies.

 

[bustedgrill]

Did you- by mistake- factor in the (s) into your calc?