Rocket Engine Fundamentals and Design Part 1: Thrust and Combustion 

Use this to play Kerbal Space Program ^^

As it accelerates out the product gas loses temperature and therefore cool down (the air outside is cooler 2 so there os also heat transfer after) . Be advised i am an ammetur/noob and have no prof whatsoever of what i am talking about.

Remember that the simplest state of the chemical is considered to have 0 entalpy . So that is 0 for N2 and O2.

@@MakakunaruLoco The original comment, I believe, is talking about why the sensible enthalpies are cancelled out in all of the reactants. This is simply because the 'h bar with the little circle' is the sensible enthalpy at 25 degrees celsius, and 'h bar' is the sensible enthalpy at the given temperature in the problem. However, since the problem's given temp is also 25 degrees, they are the same and cancel out.

Remember that the simplest state of the chemical is considered to have 0 entalpy . So that is 0 for N2 and O2.