The equilibrium constant `K_c` for the reaction `H_2(g)+I_2(g)hArr 2HI(g)` is `54.3` at `400^@C`. If the initial concentrations of `H_2`, `I_2` and `HI` are `0.00623M`, `0.00414M`, and `0.0224M`, respectively, calculate the concentrations of these species at equilibrium.

Explanation

The only difference is that now we have non-stoichiometric amounts of reactants. Also, find the value of `Q_c` to decide the direction of reaction.
Solution: Let x be the decrease in concentration `(mol L^(-1))` for `H_2` and `I_2` at equilibrium. Then, according to the stoichiometry of the reaction, the increase in concentration for `HI` must be `2x`. Summarize the changes n concentrations as follows:
`{:(," "H_(2) (g)" " +" "I_(2)(g)" " hArr 2HI(g)),("Initial (M)", " 0.00623 0.00414 0.0224"),("Change (M)", " -x -x +2x"),("Equilibrium (M)", bar((0.00623-x) (0.00414-x) (0.0224+2x))):}`
Note that we have considered the net reaction in the forward direction because `Q_c(19.45)` is less than `K_c(54.3)`.
Step 2: Writing the equilibrium constant expression
`K_c=(C_(HI)^2)/(C_(H_2)C_(I_2))`
Substituting, we get
`K_c=((0.0224+2x)^2)/((0.0063-x)(0.00414-x))`
It is not possible to solve this equation by the square root shortcut as the left side of this equation is not a perfect square, because the starting concentrations of `H_2` and `I_2` are unequal. Instead, we can arrange this quardratic equation into the standard form by carrying out the multiplications:
`54.3(0.00623-x)(0.00414-x)=(0.0224+2x^2)`
`54.3(2.58xx10^5-0.0104x+x^2)=5.02xx10^-4+0.0896x+4x^2`
Collecting terms, we get
`50.3x^2-0.65x+8.98xx10^-4=0`
This is a quardratic equation of the form `ax^2+bx+c=0`, which can be solved by the quardratic formula
`x=(-b+-sqrt(b^2-4ac))/(2a)`
In this case, we have
`a=50.3, b=-0.654`, and `c=8.98xx10^-4`
Substituting these values gives
`x=(0.654+-sqrt((-0.654)^2-4(50.3)(8.98xx10^-4)))/(2xx50.3)`
`implies x=0.0114M` or `x=0.00156M`
The first solution is physically impossible since the amounts of `H_2` and `I_2` reacted would be more than those originally present. The second solution gives the correct answer. Note that in solving a quardratic equation of this type, one answer is always physically impossible. Thus, the choice of which value to use for x is easy to make.
Step 3: The equilibrium concentrations are
`C_(H_2)=(0.00623-0.00156)M=0.00467M`
`C_(I_2)=(0.00414-0.00156)M=0.00258M`
`C_(HI)=(0.0224+2xx0.00156)M=0.0255M`