Chem

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The initial concentration for the compounds involved in the reaction displayed were determined to be [NO2(g)] = 1.243 mol/L, [N2O4(g)] = 0.2578 mol/L. Calculate the value of the equilibrium constant (Kc) at 500.0 K if the equilibrium concentration of NO2(g) was 0.7629 mol/L.


2NO2(g) = N2O4(g)

  • Chem -

    ..............2NO2 ==> N2O4
    initial......1.243...0.2578
    change.......-2x........+x
    equil.......0.7629......0.2578+x

    So 2x = 1.243-0.7829 = ?
    and x = ?
    (NO2) = 0.7629
    (N2O4) = 0.2578+x
    Substitute in Kc expression and solve for Kc. Post your work if you get stuck.

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