Changes in concentration: Some hydrogen and iodine are mixed at `229^@C` in a 1.00-liter container. When equilibrium is estabilished, the following concentrations are present: `C_(H_2)=0.080M`, `C_(I_2)=0.060M`, and `C_(HI)=0.490M`. If an additional `0.300` mol of `HI` is then added, what concentrations will be present when the new equilibrium is established?

Explanation

Strategy: Use the initial equilibrium concentrations to calculate the value of `K_c`. Then determine the new concentrations after some `HI` has been added and calculate `Q_c`. The value of `Q_c` tells us which reaction is favored. Represent the new equilibrium concentrations, substitute these representations into the `K_c` expression, and solve for the new equilibrium concentrations.
Note that it is obvious that adding some `HI` favors the reaction to the left. If more than one substance is added to the reaction mixture, it might not be obvious which reaction will be favored. Calculating Q always lets us make the decision.
Solution:
Calculate the value of `K_c` from the first set of equilibrium concentrations:
`K_c=(C_(HI)^2)/(C_(H_2)C_(I_2))=((0.490)^2)/((0.080)(0.060))=50`
When we add `0.300` mol of `HI` to the `1.00-L` container, the concentration of `HI` instantaneously increases by `0.300M`:
`{:(,H_(2) (g)+I_(2)(g)hArr 2HI(g)),("Equilibrium conc.(M)", " 0.080 0.060 0.490"),("Added conc.(M)", " 0 0 +0.300"),("New initial conc.(M)", bar("0.080 0.060 0.790")):}`
Calculate the value of reaction quotient by substituting these new initial concentrations:
`Q_c=(C_(HI)^2)/(C_(H_2)C_(I_2))=((0.790)^2)/((0.080)(0.060))=130`
Since `Q_c gt K_c`, the net reaction proceeds to the left to establish a new equilibrium. To determine the new equilibrium concentrations, let us assume that `2x mol L^(-1)` of `HI` is consumed. Then `x mol L^(-1)` of `H_2` is formed and `x mol L^(-1)` of `I_2` is formed. (Note that equal concentrations of `H_2` and `I_2` must be formed by the new progress of the reaction.)
`{:(," "H_(2) (g)+" "I_(2)(g)hArr" "2HI(g)),("Initial (M)", " 0.080 0.060 0.790"),("Change (M)", " +x +x -2x"),("New equilibrium (M)", bar((0.080+x)(0.060+x)(0.790-2x))):}`
Calculated x by substituting these new equilibrium concentrations into the equilibrium constant expression:
`K_c=50=(C_(HI)^2)/(C_(H_2)C_(I_2))`
`=((0.790-2x)^2)/((0.080+x)(0.060+x))`
`=(0.624-3.16x+4x^2)/(0.0048+0.14x+x^2)`
`0.24+7.0x+50x^2=0.624-3.61x+4x^2`
`46x^2+10.2x-0.38=0`
Using the quadratic formula provides
`x=0.032` and `-0.25`
Because x cannot be less than zero in this case, `x=-0.25` is the irrelevant root. Because the reaction is proceeding towards the left, it does not consume a negative quantity of `HI`. Thus, `x=0.032` is the root with physical meaning. Hence, the new equilibrium concentrations are
`C_(H_2)=(0.080+0.032)M=0.112M`
`C_(I_2)=(0.060+0.032)M=0.092M`
`C_(HI)=(0.790-0.064)M=0.726M`