At a certain temperature, the equilibrium constant, ????c, for this reaction is 53.3. H2(g)+I2(g)↽−−⇀2HI(g)????c=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?

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:

[tex][HI]={[HI]}_{0}+2x\\{[H_{2}]}={[H_{2}]}_{0}-x\\{[I_{2}]}={[I_{2}]}_{0}-x\\[/tex]

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:

[tex]K_{c}=\frac{(2x)^{2}}{(0.5-x)(0.5-x)}=\frac{(2x)^{2}}{(0.5-x)^{2}}=53.3[/tex]

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

[tex]53.3*(0.5-x)*(0.5-x)=(2x)^{2} \\53.3*(0.25-x+x^{2})=4x^{2}\\49.3x^{2} -53.3x+13.325=0\\x=0.3925, x=0.6886[/tex]

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.