Question

what would happen to the values you calculated for kc in the experiment if the beer's law constant (k) value you used was high (e.g.you used 6000 M^-1 when the actual constant had a value of 5000 M^-1) ? Explain your answer.

Answer 1

Take the following reaction as an example:

Fe3+ + SCN− <---> FeSCN2+

K_C = [FeSCN2+ ]/[ Fe3+][ SCN−]

The reactants are colorless, but the FeSCN2+ ion is orange-red colored.

A = −log(T) = εbc              ; ε = beer's law constant, c = concentration, b = cell length

c = A/εb

From the knowledge of ε,b and A you can calculate the equilibrium concentration of FeSCN2+ in the solution.

Equilibrium concentration of Fe3+ = (initial concentration of Fe3+) - (equilibrium concentration of FeSCN2+ in the solution)

Similarly,

Equilibrium concentration of SCN- = (initial concentration of SCN-) - (equilibrium concentration of FeSCN2+ in the solution)

Now you can calculate, K_C. But if you use the value of ε too high, the value of c (equilibrium concentration of FeSCN2+ in the solution) will be low. As a result Equilibrium concentration of Fe3+ and SCN- calculated value will be higher than the original.

So at last the calculated value of K_C = [FeSCN2+ ]/[ Fe3+][ SCN−] will be lower than the original.