Asked by Dan
Change in enthalpy problem. Consider the reaction represented below. Referring to the data in the table below, calculate the standard enthalpy change for the reaction at 25 degrees C.
O3(g)+NO(g)-->O2(g)+NO2(g)
Standard enthalpy of formation in kJ/mol:
O3(g) 143
NO(g) 90
NO2(g) 33
I am sure this isn't a hard problem, but I can't really remember how to do it.
O3(g)+NO(g)-->O2(g)+NO2(g)
Standard enthalpy of formation in kJ/mol:
O3(g) 143
NO(g) 90
NO2(g) 33
I am sure this isn't a hard problem, but I can't really remember how to do it.
Answers
Answered by
DrBob222
delta H<sub>reaction</sub> = delta H<sub>products</sub> - delta H<sub>reactants</sub>
Answered by
Dan
Ok, so I have DeltaHrxn=(X+90)-(143+90). How do you calculate the enthalpy of O2 (X above) when it is not given to you. Am I even doing that right?
Answered by
DrBob222
No. Re-read my answer. I said
PRODUCTS - REACTANTS.
You have one of the reactants on the product side.
DHrxn = DH products - DH reactants.
DHrxn = (O2 + NO2) - (O3 + NO)
DHrxn = (0 + 33) - (143 + 90) = ??
PRODUCTS - REACTANTS.
You have one of the reactants on the product side.
DHrxn = DH products - DH reactants.
DHrxn = (O2 + NO2) - (O3 + NO)
DHrxn = (0 + 33) - (143 + 90) = ??
Answered by
yea
drbob222 is correct . the answer is 200. DUh!!
Answered by
Nope
It's -200. not 200