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vibhav1991 (146)

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Pl help me in learning to find out enthalpy of formation (HESS'S LAW).
Solve the ques below in detail and explain it so that i can learn from it.
rate assured!!!!!!  thanx.
 
Q-Cal. enthalpy of formation of ethyl alcohol from following data==
 
1) C2H5OH + 3O2     2CO2(g)  + 3H2O   ( H= -1368 KJ)
 
2) C(s) + O2(g)    CO2(g)    ( H= -393.5 KJ) 
 
3) H2  +  1/2O2     H2O   ( H= -286.0 KJ)                 
 
 
 
 

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hitman47 (0)

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Hi Dude, this is fairly simple

equate the coeff of CO2 and H2O of eq. 1 with coeff of eq 2 and eq 3
than subtract add eq 2 and eq 3
now subtract them from eq 1

clearly speaking do this with the enthalpy values.

best of luck !!!!
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dev_22oct (1405)

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The (molar) enthalpy of formation is the heat released ( ) or absorbed ( ) in a chemical reaction at constant pressure when simple substances combine into a more complex substance. At standard conditions of pressure and temperature (1 atm and 298 K), it is denoted . For elements in their standard states, . For gases, the standard state is the atom or molecule existing at standard state conditions. For solids with different phases, the assignment is arbitrary. The hydrogen ion is also defined to have H+ = 0.
 
Now we want to calculate enthalpy of formation of ethyl alchohol.
therefore the reaction is
2C  + 3H2 +1/2O2------>C2H5OH   H = ?
now we have to calculate deltaH from the given data.
multiply eq no 2 with 2, eq no 3 with 3 and reverse eq 1. add all these changed reaction.

Devendra Kumar Shukla
M.Sc physical Chemistry
Currently working in Daswani classes
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dev_22oct (1405)

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2C + 2O2 -------> 2CO2                         -393.5 X2
 
2CO2 + 3H2O ------> C2H5OH + 3O2      1368
 
3H2 +3/2 O2 ------> 3H2O                      -286 X 3
--------------------------------------------------------------------------
2C + 3H2 +1/2O2----->C2H5OH              -277 KJ/mol
---------------------------------------------------------------------------

Devendra Kumar Shukla
M.Sc physical Chemistry
Currently working in Daswani classes
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