Hess's law

HESS"S LAW: says that when there is more that one reaction, the total enthalphy change is the sum of the enthalpy changes of each reaction.


\Delta H^\theta = \Sigma(\Delta H_{f~products}^\theta ) - \Sigma(\Delta H_{f~reactants}^\theta)

 
Calculate the DHf of CH3OH(g) from its elements in their standard states at 25oC and 1 atm.


CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2H2O(l) : DH= -726.4kJ

C(graphite) + O2(g) --> CO2(g) DH=-394kJ

H2(g) + 1/2O2(g) --> H2O DH= -285.8 kJ

Overall

C(graphite) + 2H2(g) + 1/2O2(g) -->CH3OH(g)

 

  

C(graphite) + O2(g) --> CO2(g) 

 DH=          -394kJ

2(H2(g) + 1/2O2(g) --> H2O) 

 DH= 2(-285.8 kJ)

Flipped CO2(g) + 2H2O(l) -->CH3OH(g) + 3/2O2(g)

 DH=      +726.4kJ
   
     

DH=      -239.2kJ 

 

Determine The DH for this reaction

2B(s) + (3/2)O2(g) → B2O3(s)

 Given

    
B2O3(s) + 3H2O(g) → 3O2(g) + B2H6(g)   ΔH =  +2035 kJ Flipped 3O2(g) + B2H6(g) → B2O3(s) + 3H2O(g) ΔH =  -(+2035 kJ)  
H2O(l) → H2O(g)                                     ΔH =  +44 kJ    Flipped  3( H2O(g)→H2O(l) ) ΔH = 3 x -(+44)=  -132 kJ  

H2(g) + (1/2)O2(g) → H2O(l)                    ΔH =  -286 kJ

Flipped  3(H2O(l) →H2(g) + (1/2)O2(g) ) ΔH =3 x -(-286) = +858 kJ  
2B(s) + 3H2(g) → B2H6(g)                        ΔH =   +36 kJ 2B(s) + 3H2(g) → B2H6(g) ΔH = +36 kJ  
   
 

ΔH = -1273kJ
The Ostwald process for the commercial production of nitric acid from ammonia and oxygen involves the following steps.

4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g) -908kJ
2 NO(g) + O2(g) 2 NO2(g) -112 kJ
3 NO2(g) + H2O(l) 2 HNO3(aq) + NO(g) -140kJ

***Write the overall equation for the production of nitric acid by the Ostwald process by combining the preceding equations. 

_ __(g) + _ O2(g) --> _ ___(aq) + _ __(g) + _H2O(g)
***Is the overall reaction exothermic or endothermic? Why?		  

 

 

 

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