chemisty

when 2.75 mol HI(g) placed in 1L container and allowed to dissociate into
2HI(g) <==> H2(g)+I2(g)
Final H2 concin .275M. What is Ka for rxn

The answer is .0156. But I don't know why
Thanks

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  1. ..........2HI ==> H2 + I2
    initial...2.75.....0....0
    change...-2x........x....x
    equil....2.75-2x....x.....x

    The problem tells you H2 (which I've let = x above) = 0.275; therefore, I2 must be 0.275 and HI must be 2.75-(2*0.275) = ?
    Now just substitute into Ka expression and solve. I get 0.0156 also.

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  2. could you do it in moles instead of molars

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  3. and in the K eqn is the M for HI squared

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  4. ?

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  5. 2.75 moles in 1L = 2.75M so the solution I posted is both moles and molar.
    Yes, (HI) is squared.
    Ka = (H2)(I2)/(HI)^2
    Ka = (0.275)(0.275)/(2.20)^2
    Ka = 0.0156

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