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Lena772

  • one year ago

The reaction below was observed in a 4.0 L flask at 100oC. N2O4 (g) ↔ 2 NO2(g) Initially, 0.2 mol of N2O4 and 0.2 mol of NO2 were placed in the flask. If the kp = 11 atm , will the reaction proceed toward products or reactants? Toward reactants Toward products It's at equilibrium

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  1. Lena772
    • one year ago
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    I tried towards reactants and that was wrong... Maybe it's equilibrium?

  2. Lena772
    • one year ago
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    @Photon336

  3. Abhisar
    • one year ago
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    You can calculate molar concentrations of each of them and then find \(\sf Q_c\). Compare it with \(\sf K_c\). If \(\sf Q_c>K_c\) then reaction will be backwards or forward otherwise. You can find \(\sf K_c\) by using the equation \(\sf K_p = K_c \times (RT)^{\Delta n}\)

  4. Photon336
    • one year ago
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    agree with abhisar, you can figure out Qc which is relatively easy \[Qc = \frac{ [NO_{2}]^{2} }{ [N_{2}O_{4}] } = [0.05M]^{2}/[0.05] = 1 \] However as abhisar said you would need to convert Kp and compare it to Kc then this would tell you what direction the reaction is going.

  5. Lena772
    • one year ago
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    Qc=1 Kc=0.359 1>0.359 so backwards. Wouldn't that mean more reactants are formed? :/ @Abhisar @Photon336

  6. Photon336
    • one year ago
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    So technically we know that Qc = 1 so 0.359 is the value of Kc right? Right away because Qc and Kc are not equal you know that the reaction has not reached equilibrium or should I say at equilibrium. so if Q>K that means that there are too many products, so the equilibrium will shift to form more reactants until the Q = K and equilibrium is established.

  7. Lena772
    • one year ago
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    Thank you! @Photon336

  8. Photon336
    • one year ago
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    No problem

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