Question

Please help me! I don't know how to do this at all.

Calculate the enthalpy for the reaction:

NO (g) + O (g) → NO2

Given the following known reactions:
NO(g)+O3 (g)→NO2 (g)+O2 (g) 03 (g)→1.5O2 (g) O2 (g)→2O(g)

ΔH°=-198.9kJ
ΔH°=-142.3kJ
ΔH°=495.0kJ

Answer 1

It would help if you wrote questions that could be interpreted. You might try looking at the post to see how it is shown. What you have makes no sense; I have hit the enter button to try and separate your one reaction into three separate reactions. I hope I hit the enter key at the right place.


NO(g)+O3(g)→NO2(g)+O2(g) dH = -198.9 kJ
03(g)→1.5O2(g) dH = -142.3 kJ
O2(g)→2O(g) dH = 495.0 kJ

Add equation 1 to the reverse of equation 2 to the reverse of 1/2 of equation 3.

If you multiply the equation remember to multiply the dH value by the same number. If you reverse an equation change the sign of dH.
Add all of the new dH values to find the overall dH for the desired reaction.