Calculate the(change in enthalpy) for the reaction:2C (s) + H2 (g) ---> C2H2 (g)
C2H2 (g) + 5/2 O2 (g) ---> 2CO2 (g) + H2O (l)    triangle(kJ) -1299.6
C(s) + O2 (g) ---> CO2 (g)                       (kJ) - 393.5
H2 (g) + 1/2 O2 (g) ---> H2O (l)                 (kJ) - 285.9

Hi teatea!!

We will use the Hess's law to solve this problem: if a process can be
written as the sum of several stepwise processes, the enthalpy change
of the total process equals the sum of the enthalpy changes of the
various steps.
Hess's Law says that heat of enthalpy are additive.
 ĢH(net) = Sum (ĢH(r)), equals the sum of enthalpies of reaction.
 Two rules:
 * If the chemical equation is reversed, the sign of ĢH changes.
 * If coefficients are multiplied, multiply ĢH by the same amount.


More info here: 
"Enthalpy":
http://www.chem.tamu.edu/class/majors/tutorialnotefiles/enthalpy.htm

2 C(s) + 2 O2(g)   ---> 2 CO2(g)                 ĢH1 = -787 kJ
        
H2(g) + 1/2 O2(g)  ---> H2O(l)                   ĢH2 = -285.9 kJ

2 CO2(g) + H2O(l)  ---> C2H2(g) + 5/2 O2(g)      ĢH3 = 1299.6 kJ 
______________________________________________________ 

Now adding the three equations together (putting all lefts on left,
and all rights on right) we have:
2C(s) + 5/2 O2(g) + H2(g) + 2CO2(g) + H2O(l) --> 2CO2(g) + H2O(l) +
C2H2(g) + 5/2 O2(g)
     
Cancelling the compounds that appear at both sides in equal quantities
we get the desired equation:
2C(s) + H2(g) ---> C2H2 (g)  

Applying the HessLs Law we must only add the enthalpy changes of the
used equations to find the desired enthalpy change:
ĢH = ĢH1 + ĢH2 + ĢH3 = 226.7 kJ


Hope this helps. If you need a clarification, please post a request
for it.

Regards.
livioflores-ga

---

Clarification of Answer by livioflores-ga on 20 Jul 2003 21:47 PDT

ooops!!

Something happens with the deltas (triangles that means variation of);
where appear the symbols Ģ replace them with Var (meaning variation
or change).
The answer is:
We will use the Hess's law to solve this problem: if a process can be
written as the sum of several stepwise processes, the enthalpy change
of the total process equals the sum of the enthalpy changes of the
various steps.
Hess's Law says that heat of enthalpy are additive. 
 VarH(net) = Sum (VarH(r)), equals the sum of enthalpies of reaction.
 Two rules: 
 * If the chemical equation is reversed, the sign of VarH changes. 
 * If coefficients are multiplied, multiply VarH by the same amount. 
 
 
More info here:  
"Enthalpy": 
http://www.chem.tamu.edu/class/majors/tutorialnotefiles/enthalpy.htm 
 
2 C(s) + 2 O2(g)   ---> 2 CO2(g)                 VarH1 = -787 kJ 
         
H2(g) + 1/2 O2(g)  ---> H2O(l)                   VarH2 = -285.9 kJ 
 
2 CO2(g) + H2O(l)  ---> C2H2(g) + 5/2 O2(g)      VarH3 = 1299.6 kJ  
______________________________________________________  
 
Now adding the three equations together (putting all lefts on left,
and all rights on right) we have:
2C(s) + 5/2 O2(g) + H2(g) + 2CO2(g) + H2O(l) --> 2CO2(g) + H2O(l) +
C2H2(g) + 5/2 O2(g)
      
Cancelling the compounds that appear at both sides in equal quantities
we get the desired equation:
2C(s) + H2(g) ---> C2H2 (g)   
 
Applying the Hess´s Law we must only add the enthalpy changes of the
used equations to find the desired enthalpy change:
VarH = VarH1 + VarH2 + VarH3 = 226.7 kJ

I hope that this clarifys the answer.

Sincerely.
livioflores-ga