Finding equilibrium concentrations: A mixture of `0.50` mol `H_2` and `0.50` mol `I_2` is placed in a `1.00L` stainless steel container at `400^@C`. The equilibrium constant `K_c` for the reaction
`H_2(g)+I_2(g)hArr 2HI(g)`
is `54.3` at this temperature. Calculate the equilibrium concentrations of `H_2`, `I_2`, and `HI`.

Explanation

Strategy: First we find the initial concentrations. Then we write the reaction summary and represent the equilibrium concentrations algebraically. Finally, we substitute the algebric representations of equilibrium concentrations into the `K_c` expression and find the equilibrium concentrations.
Solution:
Step: The initial concentrations are
`CH_2=(n_(H_2))/(V_L)=(0.50mol)/(1.00L)=0.50 mol^-1`
`C_(I_2)=(n_(I_2))/(V_L)=(0.50 mol)/(1.00L)=0.50 mol L^-1`
The reaction can only proceed to the light because only reactants are present. The reaction summary includes the values, or symbols for the values, of (i) initial concentrations, (ii) changes in concentrations, and (iii) concentrations at equilibrium.
Step 2: According to the stoichiometry of the reaction, 1 mol `H_2` reacts with 1 mol `I_2` to produce 2 mol `HI`. Let x be the decrease in concentration `(mol L^(-1))` of either `H_2` or `I_2` at equilibrium. Then the equilibrium concentrations of `HI` must be `2x`. Summarizing the changes in concentrations, we get
`{:(,H_(2)(g)+ I_(2)(g)hArr 2HI(g)),("Initial (M)":,"0.50 0.50 0.00"),("Change (M)":,"-x -x +2x"),("Equilibrium",bar((0.50-x) (0.50) " " 2x)):}`
Step 2: According to the law of chemical equilibrium, the equilibrium constant is given by
`K_c=(C_(HI)^2)/(C_(H_2)C_(I_2))`
Now, `K_c` is known but concentrations are not. But the equilibrium concentrations have all been expressed in terms of a single variable x. We substitute the equilibrium concentrations (not the initial ones) into the `K_c` expressions and solve for x.
Substituting, we get
`54.3=((2x)^2)/((0.50-x)(0.50-x))`
`=((2x)^2)/((0.50-x)^2)`
Taking the square root of both sides, we get
`7.37=(2x)/((0.50-x))`
`(7.37)(0.50-x)=2x`
`3.685-7.37x=2x`
`3.685=2x+7.37x=9.37x`
`x=(3.685)/(9.37)=0.393M`
Step: Now, we know the value of x. So the equilibrium concentrations are
`C_(H_2)=(0.50-0.393)M=0.107M`
`C_(I_2)=(0.50-0.393)M=0.107M`
`C_(HI)=2xx0.393M=0.786M`
Step 4: To check our answers, we use the equilibrium concentrations to calculate `Q_c` and verify that its value is equal to `K_c`.
`Q_c=(C_(HI)^2)/(C_(H_2)C_(I_2))=((0.786)^2)/((0.107)(0.107))=54.0`