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A mass action expression

  1. Oct 20, 2004 #1
    Hi all,

    I have the following equilibrium reaction [itex]\mathrm{2NO_{2} \leftrightharpoons N_{2} O_{4}}[/itex] which has a tempeture of
    [itex] \mathrm{t = 100 \ ^\circ C.}[/itex]
    The reactions takes place in a canister with a volume of 0,50 Liters, where the pressure at equilibrium is 1,6 bar.

    Futhermore the equilibrium constant is found to be [itex]\mathrm{K_{c} = 5,0 M^{-1}}[/itex]

    a/ First I write the mass action expression [itex]K_{c} = \frac{\mathrm{[N_{2}O_{4}]}}{\mathrm{[NO_{2}]^2}}[/itex]

    b/ Second I calculate the amount of mol substance then the reaction is in equilibrium:

    [itex]\mathrm{n_{equ}} = \frac{\mathrm{1,6bar \cdot 0,5 L}}{\mathrm{0,0831 \frac{bar \cdot L}{mol \cdot K}\cdot \ {373,0 K}}}= 2,6 \times 10^{-2} \ \mathrm{mol}[/itex]

    c/ How do I calculate [itex][\mathrm{NO_{2}}] \ \mathrm{and} \ [\mathrm{N_{2} O_{10}}][/itex] ??

    Many Thanks in advance.

    Last edited: Oct 21, 2004
  2. jcsd
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