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Schrodinger
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I'm partly there, but I get stuck at a particular point in. It's kind of Thermochemistry/Stoichiometry. (Question will be posted below.)
 2 years ago
 2 years ago
Schrodinger Group Title
I'm partly there, but I get stuck at a particular point in. It's kind of Thermochemistry/Stoichiometry. (Question will be posted below.)
 2 years ago
 2 years ago

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Schrodinger Group TitleBest ResponseYou've already chosen the best response.1
Consider the reaction: \[4CO _{(g)}+2NO _{2(g)} \rightarrow 4CO _{2(g)}+N _{2(g)}\] Using the following information, find ΔH° at 25° C. \[NO _{(g)}, \Delta H ^{°} _{f} = 91.3 kJ.mol\]\[CO _{2(g)}, \Delta H _{f} ^{°} = 393.5 kJ/mol\]\[2NO _{(g)}+O _{2(g)} \rightarrow 2NO _{2(g)}, \Delta H _{f} ^{°} = 116.2 kJ/mol\]\[2CO _{(g)}+O _{2(g)} \rightarrow 2CO _{2(g)}, \Delta H _{f} ^{°} = 566.0 kJ/mol\]
 2 years ago

Schrodinger Group TitleBest ResponseYou've already chosen the best response.1
I'm applying Hess's Law and doing this: 4CO(g)+2NO2(g)→4CO2(g)+N2(g) is what I want as a final product. Therefore, 2x [ 2CO(g)+O2(g)→2CO2(g) ] So I can get 4CO on the reactant side. 2*(566.0) Flip [ 2NO(g)+O2(g)→2NO2(g) ] So I can get 2NO2 on the reactant side. (116.2) subtract 2x the standard formation enthalpy of [ NO(g) ] to get rid of the unnecessary 2NO in the reaction. Ignore any values regarding N2, as it is in its natural state. Then I have an extra O2 molecule, after cancelling out one of them. How do I deal with this using the equations given without adding unnecessary elements to the reaction that I will also have to cancel out, but won't be able to without not getting the result?!?
 2 years ago

Kryten Group TitleBest ResponseYou've already chosen the best response.1
did you get to any answer? i got dH = 1198,4 kJ/mol
 2 years ago

Schrodinger Group TitleBest ResponseYou've already chosen the best response.1
Well, I couldn't, because I couldn't get rid of Oxygen. I can't arrive at an answer until I eliminate all the unnecessary species in the reaction, and I get stuck with a single O2 molecule.
 2 years ago

Kryten Group TitleBest ResponseYou've already chosen the best response.1
ok so this is what i did: dH = sum(products)  sum(reactants) first reaction: 2NO(g)+O2(g)→2NO2(g), ΔH∘f=−116.2kJ/mol sum(products) = dH + sum(reactants) 2*dHf(NO2) = 116,2 + 2*91,3 dHf(NO2) = 33,2 kJ mol1 second reaction: 2CO(g)+O2(g)→2CO2(g), ΔH∘f=−566.0kJ/mol 2*dHf(CO) = 2*(393,5)  ( 556,0) dHf(CO) = 110,5 kJ mol1 now its simple just calculate for requested equation: 4CO(g)+2NO2(g)→4CO2(g)+N2(g) dH = [4*(393,5)]  [4*(110,5) + 2*33,2 ] dH = 1198,4 kJ mol1
 2 years ago

Schrodinger Group TitleBest ResponseYou've already chosen the best response.1
I'm reading your response, but just a heads up, apparently your answer is wrong. (Mine obviously is, too.) One second.
 2 years ago

Schrodinger Group TitleBest ResponseYou've already chosen the best response.1
PS, is "D" delta? It seems to be so, but I don't like assumptions.
 2 years ago

Kryten Group TitleBest ResponseYou've already chosen the best response.1
yes d is delta
 2 years ago

Schrodinger Group TitleBest ResponseYou've already chosen the best response.1
Oh, pardon me. You were right. Sorry about that, didn't mean to make you doubt yourself. Yup, you were right. Now I just need to figure out what you did, XD.
 2 years ago
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