Calculating K
Give initial concentrations and one equilibrium concentration

Consider a system where potassium dichromate is dissolved in water in known concentration.
The Cr2O72– ion reacts with water occurs according to the equation given. 
 
Cr2O72–(aq) + H2O  2H+(aq) + 2CrO42–(aq)

The equilibrium concentraton of Cr2O72–  is measured.  These data, this reaction equation and the equilibrium expression can be used to calculate Kc for the reaction.
 
K =
[CrO42–]2[H+]2  [Cr2O72–]

 
Why is there no [H2O] in the K expression?  The H2O reactant is also the solvent in this reaction.  Thus H2O is present in large excess and reaction to restore/reach equilibrium does not change the amount of water present to a significant extent.  Therefore [H2O] is constant.

First construct a table having
four rows (reaction, initial, change, equilibrium). 
a column for each component in the equilibrium constant expression. 
 Second enter given data
Initial amounts are assumed to be zero if not present before reaction.
 
 Cr2O72–(aq) 2H+(aq) CrO42–(aq)
 initial/
given given given
 change/
 -x +2x  +x
 equilibrium/
 given    

Third calculate the change in one component from the given data and deduces changes in other components from the calculated one.
The ratio of the changes is the ratio of coefficients in the balanced equation.
The sign of a change is negative for substances consumed in reaction to reach equilibrium.
Fourth calculate the equilibrium concentrations of other substances
equilibrium = change + initial

Fifth calculate Kc by substituting equilibrium concentrations into the expression.
Dr Sheila Woodgate
© The University of Auckland 2002