dHrxn = (n*dHo products) - (n*dHo reactants)
Substitute and solve for dH rxn. Post your work if you get stuck.
What is the standard heat of reaction (H0) for the combustion of ethane, C2H6(g), to form carbon dioxide gas and water?
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l)
Standard heats of formation (∆Hf⁰): C2H6 = –84.68 kJ
O2(g) = 0.00 kJ CO2(g) = –393.5 kJ H2O(l) = –285.5 kJ
3 answers
@DrBob222 what does dHrxn and n*dHo mean?
dHrxn is delta H for the reaction as written and is the standard heat of reaction that the problem is asking for. n is the mols (the coefficients) in the balanced equation. dHo is delta Ho given to you in the problem.
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l)
So delta for the reaction is
(4*delta Hof + 6*dHo H2O) - (2*dHo C2H6 + 7*dHo O2)
I write dHo because it takes to long to type in delta Hof.
Hope this helps. BTW, if you are using multiple screen names, please don't. It helps us help you if you stick to the same name. Thanks. Another BTW. dHo for O2 is 0.
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l)
So delta for the reaction is
(4*delta Hof + 6*dHo H2O) - (2*dHo C2H6 + 7*dHo O2)
I write dHo because it takes to long to type in delta Hof.
Hope this helps. BTW, if you are using multiple screen names, please don't. It helps us help you if you stick to the same name. Thanks. Another BTW. dHo for O2 is 0.
0 answers