delta Hrxn = [delta H products] - [delta H reactants]
Texts have a set of tables. Look up delta Hof for each and subtract the products - reactants to give delta H for the reaction. Elements in their standard state (O2 for example) are zero.
Calculate the enthalpy change for the following equation:
NO(g)+ 1/2O2(g) -----> NO2(g)
I have no clue how to do this. I just know the answer is supposed to be approximately -57.2 kJ. Can some explain to me what to do?
2 answers
Find ΔH, ΔS, and ΔG. Is this reaction exothermic and is it spontaneous? Show your solution (10 pts.)
NO (g) + ½ O2 (g) —> NO2 (g)
Given:
NO (g) + ½ O2 (g) —> NO2 (g)
Δ S° J/K mol: 211 205.0 240.5
ΔHf° kJ/mol: + 90 0 + 34
NO (g) + ½ O2 (g) —> NO2 (g)
Given:
NO (g) + ½ O2 (g) —> NO2 (g)
Δ S° J/K mol: 211 205.0 240.5
ΔHf° kJ/mol: + 90 0 + 34
1 answer