[akanevsky]

Hello,

I have the following question:

Determine the standard enthalpy of formation for NO given the following information about the formation of NO₂ under standard conditions, and heat of formation for NO₂ = + 33.2 kJ/mol.

2 NO (g) + O₂(g) --> 2NO₂(g) --- Heat of Reaction = -114.2 kJ

My attempt at solution:

Half of the reaction:

NO (g) + 1/2 O₂(g) --> NO₂(g) --- Heat of Reaction = -57.1 kJ

To get at the standard enthalpy of formation for NO, we must destroy the bons in NO₂, which will take -33.2 kJ since there is just one mole, and then reverse the reaction, which will take 57.1 kJ (the inverse of the original reaction).

Summing up the two, I get 57.1 - 33.2 = 23.9 kJ.

The correct answer is 90.3 kJ/mol.

Where did I make a mistake?