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What does it mean when the equilibrium shifts to the right Watch

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Size:  26.8 KB the pressure increases so the equilibrium shifts to the right

    The mark scheme says that NO2 concentration increases more than N2O4. I'm confused


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Size:  174.4 KB there's the actual question and me writing in the correct answers


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    Are you familiar with Le Chatelier's principle?

    If you increase the overall pressure of the equilibrium, the system moves to decrease the overall pressure - in this instance by producing less moles of gas (since an ideal gas has a fixed volume at a given temperature).

    Equilibrium constants are only affected by changes in temperature.
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    Have a read of this:

    http://www.chemguide.co.uk/physical/...chatelier.html
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    (Original post by EierVonSatan)
    Are you familiar with Le Chatelier's principle?

    If you increase the overall pressure of the equilibrium, the system moves to decrease the overall pressure - in this instance by producing less moles of gas (since an ideal gas has a fixed volume at a given temperature).

    Equilibrium constants are only affected by changes in temperature.
    But why does NO2 increase? there's more moles of NO2 then there is of N2O4


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    (Original post by rambo1168)
    But why does NO2 increase? there's more moles of NO2 then there is of N2O4
    It helps to stick some imaginary numbers in. Say that Kc = 4 and that [NO2] = 1 and [N2O4] = 4 to start with:

    K_c = \frac{[N_2O_4]}{[NO_2]^2} = \frac{4}{1^2}

    The concentration's double briefly, meaning that the [NO2]2 increases more than [N2O4, and the fraction becomes more bottom heavy (smaller):

     \frac{[N_2O_4]}{[NO_2]^2} = \frac{8}{2^2} = 2

    Now the equilibrium shifts to restore the original value of 4 (or close enough :p:) by converting 2NO2 into N2O4:

    K_c = \frac{[N_2O_4]}{[NO_2]^2} = \frac{8.28}{1.44^2} = 3.99
 
 
 
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