How do you use Hess's Law to calculate enthalpy for this reaction?

Given Data:
"N"_2(g) + "O"_2(g) -> 2"NO"(g)

DeltaH = +"180.7 kJ"

2"NO"(g) + "O"_2(g) -> 2"NO"_2(g)

DeltaH = -"113.1 kJ"

2"N"_2"O"(g) -> 2"N"_2(g) + "O"_2(g)

DeltaH = -"163.2 kJ"

Use Hess' Law to calculate DeltaH for the reaction:

"N"_2"O"(g) + "NO"_2(g) -> 3"NO"(g)

1 Answer
Mar 31, 2018

ΔH = 311.3 "kJ"

Explanation:

PC of cake

cancel"N"_2 + cancel"O"_2 " ----->" 2NO

2NO_2 " ----->" 2NO + cancel"O"_2

2N_2 O " ----->" cancel"2"N_2 + O_2
"---------------------------------"
Adding the 3 equations:

2NO_2 + 2N_2 O " -----> "4 NO + cancel"N"_2 + cancel"O"_2

Adding one more reaction
cancel"N"_2 + cancel"O"_2 " -----> "2 NO
"---------------------------------"
Adding previous 2 equations and dividing by 2

Finally,

NO_2 + N_2 O " -----> "3 NO

The Enthalpy is the sum of the all reactions:

"=>180.7 113.1 + (- 163.2) + 180.7 = 311.3 kJ