Question
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?
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?
Answers
DrBob222
delta Hrxn = [delta H products] - [delta H reactants]
Texts have a set of tables. Look up delta H<sup>o</sup><sub>f</sub> for each and subtract the products - reactants to give delta H for the reaction. Elements in their standard state (O2 for example) are zero.
Texts have a set of tables. Look up delta H<sup>o</sup><sub>f</sub> for each and subtract the products - reactants to give delta H for the reaction. Elements in their standard state (O2 for example) are zero.
Anonymous
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