Chem II

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2 NOCl(g) --> 2 NO(g) + Cl2(g)

Let’s assume that at a given temperature the equilibrium constant is 2.25: Also, the equilibrium concentration of the NOCl is 0.04M. Determine the concentration of the NO and the Cl2

CLUE: the concentration of both of the products must be equal if we started only from NOCl, since they are in a 1:1 ration. Since you don’t know the concentration of either of the products, use the variable ‘x’ to represent their concentration.

  • Chem II -

    First, I don't buy that NO and Cl2 are equal and for exactly the reason the hint points out. The products are NOT a 1:1 ratio. (NO) must be twice the (Cl2).
    .........2NOCl(g) --> 2NO(g) + Cl2(g)
    equilib....0.04M.........2x.......x

    K = 2.25 = (NO)^2(Cl2)/(NOCl)^2
    2.25 = (2x)^2(x)/(0.04)^2
    Solve for x = (Cl2), then 2x = (NO).

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