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Schrodinger

  • 2 years ago

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

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  1. Schrodinger
    • 2 years ago
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    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. Schrodinger
    • 2 years ago
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    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?!?

  3. Kryten
    • 2 years ago
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    did you get to any answer? i got dH = -1198,4 kJ/mol

  4. Schrodinger
    • 2 years ago
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    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.

  5. Kryten
    • 2 years ago
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    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 mol-1 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 mol-1 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 mol-1

  6. Schrodinger
    • 2 years ago
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    I'm reading your response, but just a heads up, apparently your answer is wrong. (Mine obviously is, too.) One second.

  7. Schrodinger
    • 2 years ago
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    PS, is "D" delta? It seems to be so, but I don't like assumptions.

  8. Kryten
    • 2 years ago
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    yes d is delta

  9. Schrodinger
    • 2 years ago
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    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.

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