# 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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