Question

N2(g) + 2O2() + 2NO2(g) + XKJ
2NO(g) + O2(g) + 2NO2(g) + YKJ
The enthalpy of formation of NO is
(A) (x - y)
(B) }(y-2)
(C) (2x - 2y)
(D) 2
- y​

Answer 1

Answer:

Ans.

Method to Solve :

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)+

2

O2(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