Standard Enthalpy of formation. Getting 2 different answers

[TinySeahorse]

I am getting two different answers for the standard enthalpy of formation of CO. Here are my two answers and how I got them:

C(graphite) + O2(g) --->CO2(g) Std enthalpy of rxn: -393.5 kJ/mol
CO(g) + (1/2)O2(g) ---> CO2 (g) Std enthalpy of rxn: -283.0 kJ/mol

First method:
CO2(g) ---> CO(g) + (1/2)O2(g) 283.0 kJ/mol
(multiplying everything by 0.5 in the following rxn)
(0.5)C(graphite) + (0.5)O2(g) ---> (0.5)CO2(g) (-393.5)/2 = -196.75 kJ/mol
Final Equation:
(1/2)CO2(g) + (1/2)C(s) ---> CO (g) Std enthalpy of formation = 86.25 kJ/mol

Second method:
CO2(g) ---> CO(g) + (1/2)O2(g) 283.0 kJ/mol
C(graphite) + O2(g) ---> CO2(g) -393.5 kJ/mol
Final Equation:
C(graphite) + (1/2)O2(g) ---> CO(g) Std Enthalpy of rxn: -110.5 kJ/mol

The second answer is the correct one according to my textbook. However, I don't understand what is wrong about the first answer. Can you help me? Thanks

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[NextStepTutor_1]

Hi @TinySeahorse ,

It looks like there's a stoichiometry issue here. The first clue to this is the fact that the standard enthalpy of formation value for the first equation is the same in both methods, but the standard enthalpy of formation for the section equation differs by a factor of two between the two methods (-393.5 = 2(-196.75)). Put simply, it looks like you multiplied the second equation by 0.5 in method 1 when you didn't need to. In method 1, note that the first equation has CO2(g) on the reactant side but 0.5 CO2(g) on the product side, which indicates a mismatch. These kinds of mistakes can be very easy to make (I know I have definitely made stoichiometry mistakes in the past!), so it's worth thinking about how you can build in 'stoichiometry checks' to your problem-solving process by setting aside 10-15 seconds or so to check that everything matches. (Depending on how neat you tend to be, another thing to consider is working on how you use your scratch paper -- can you write things out on the paper in a way that will help you see potential stoichiometry mismatches?).

Hope this helps & best of luck in your studying!

 

[TinySeahorse]

Thank you for your reply. I am still not quite seeing the mismatch though. In the end, I am still ending up with CO(g). By having (0.5)CO2 on the products side, I was able to cancel it out and end up with (0.5)CO2 on the reactant side in the final equation. Hmm...

 

[AdaptPrep]

I am getting two different answers for the standard enthalpy of formation of CO. Here are my two answers and how I got them:

C(graphite) + O2(g) --->CO2(g) Std enthalpy of rxn: -393.5 kJ/mol
CO(g) + (1/2)O2(g) ---> CO2 (g) Std enthalpy of rxn: -283.0 kJ/mol

First method:
CO2(g) ---> CO(g) + (1/2)O2(g) 283.0 kJ/mol
(multiplying everything by 0.5 in the following rxn)
(0.5)C(graphite) + (0.5)O2(g) ---> (0.5)CO2(g) (-393.5)/2 = -196.75 kJ/mol
Final Equation:
(1/2)CO2(g) + (1/2)C(s) ---> CO (g) Std enthalpy of formation = 86.25 kJ/mol

Second method:
CO2(g) ---> CO(g) + (1/2)O2(g) 283.0 kJ/mol
C(graphite) + O2(g) ---> CO2(g) -393.5 kJ/mol
Final Equation:
C(graphite) + (1/2)O2(g) ---> CO(g) Std Enthalpy of rxn: -110.5 kJ/mol

The second answer is the correct one according to my textbook. However, I don't understand what is wrong about the first answer. Can you help me? Thanks[/QUOT

Thank you for your reply. I am still not quite seeing the mismatch though. In the end, I am still ending up with CO(g). By having (0.5)CO2 on the products side, I was able to cancel it out and end up with (0.5)CO2 on the reactant side in the final equation. Hmm...

Hi, TinySeaHorse-
Standard enthalpy of formation must be calculated from pure elements in their standard state. CO2 is not in the pure standard state, and therefore, your first method is not a correct procedure for calculating standard enthalpy of formation. Graphite and Oxygen gas are the pure elements, so your final equation must have those on the reactant side. I think that is why you are getting a different answer.
I hope that helps.

 

[TinySeahorse]

Makes sense now. Thanks for your help!

 

[AdaptPrep]

Makes sense now. Thanks for your help!

You are welcome. Happy studying!