chemistry help please! I really need it

Question

Given the following enthalpies of reaction

N2(g)+2O2(g)-->2NO2(g)        ?H°1=+66.4kJ

2NO(g)+O2(g)-->2NO2(g).     ?H°2=-114.2kJ

    

 Calculate the enthalpy for the reaction for

       N2(g)+O2(g)-->2NO(g)

Answer 1

Hello Justis,

 

This is a classic Hess' Law problem, you want all unneeded reactants or products to cancel out when you add runs 1 and 2.

 

By flipping the second rxn we get 

 

N2(g)+2O2(g)-->2NO2(g)  ?H°1=+66.4kJ
2NO2(g)--> 2NO(g)+O2(g)  ?H°2=+114.2kJ

 

Notice since we flip the rxn the enthalpy also flips signs.

 

Now we simply add the rxns to get...

 

N2(g)+2O2(g)-->2NO2(g) ?H°1=+66.4kJ
2NO2(g)--> 2NO(g)+O2(g) ?H°2=+114.2kJ

 

All that remains is N2(g) + O2(g) --> 2NO(g), which is what we want.

 

Now we add the new enthalpies which is 66.4kJ + 114.6 kJ = 181kJ. 

 

Hope this all helped!