Asked by Austin
                Hydrazine, N2H4, is a colorless liquid used in rocket fuel. What is the enthalpy change for the process in which hydranzine is formed from its elements? 
N2(g) + H2(g) => N2H4(l)
Use the following reactions and enthalpy changes:
N2H4 + O2 => 2H2O DeltaH = -622.2KJ
H2 + 1/2O2 => H2O DeltaH = -285.5KJ
So how would you do this?
            
        N2(g) + H2(g) => N2H4(l)
Use the following reactions and enthalpy changes:
N2H4 + O2 => 2H2O DeltaH = -622.2KJ
H2 + 1/2O2 => H2O DeltaH = -285.5KJ
So how would you do this?
Answers
                    Answered by
            bobpursley
            
    Your first reaction is not quite complete:
N2H2 + O2 >> N2 + 2H2O deltaH=-622.2kJ/mol
H2+ 1/2 O2>>H2O deltaH=-285.5KJ
Well, if we double the second reaction, and subtract that from the first reaction
N2H2 -2H2 >> N2 deltaH=-622.2+2(285.5)
or
N2H2 >> N2+2H2 deltaH=above about-50kj/mol
reversing the reaction...
N2+H2>>N2N2 deltaH= - above. So, the heat of formation is a +value, an endothermic process.
    
N2H2 + O2 >> N2 + 2H2O deltaH=-622.2kJ/mol
H2+ 1/2 O2>>H2O deltaH=-285.5KJ
Well, if we double the second reaction, and subtract that from the first reaction
N2H2 -2H2 >> N2 deltaH=-622.2+2(285.5)
or
N2H2 >> N2+2H2 deltaH=above about-50kj/mol
reversing the reaction...
N2+H2>>N2N2 deltaH= - above. So, the heat of formation is a +value, an endothermic process.
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