At a certain temperature, the equilibrium constant, , for this reaction is 53.3. h2(g)+i2(g)↽−−⇀2hi(=53.3 at this temperature, 0.500 mol h2 and 0.500 mol i2 were placed in a 1.00 l container to react. what concentration of hi is present at equilibrium?

[HI]=0.785 M

Explanation:

Kc=53.3

First, write the equation of the equilibrium constant:

This equation is the only one that we will need to solve the problem, but thinking on a new variable; let's call the variable 'x' and it will mean: moles of H2 that reacts (or moles of I2 that react, the result will be the same). If x moles of H2 reacts, the molar change will be x M (because the volume is 1 liter), so the molar concentration of each compound will be:

The last was written having in mind the stoichiometry of the reaction; The subscript  zero means the initial concentration, which is zero for the case of HI, and 0.5 for I2 and H2. Let's replace these equations in the equilibrium constant equation:

The only unknown from this is 'x', so let's solve the equation:

Two roots are obtained, but x=0.6886 does not have sense, because we just put 0.5 M on the container, it is impossible that more than that reacts.

So, x=0.3925 M.

And the equilibriums concentrations can be calculated:

[HI]=2*x=0.785 M

[H2]=[I2]=0.5-X=0.1075 M

You can replace that concentrations on the equilibrium constant equation, and you will see that it gives the constant 53.3 as a result.

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