Testing a Coloured Equilibrium Against Three Disturbances
A student studies the equilibrium Fe3+(aq) + SCN−(aq) ⇌ FeSCN2+(aq) by mixing equal volumes of 0.0200 M Fe(NO3)3 and 0.0200 M KSCN, so the initial concentrations after mixing are 0.0100 M in each reactant. At equilibrium, a spectrophotometer calibrated for the deep-red FeSCN2+ ion reads [FeSCN2+] = 2.0 × 10−4 M. The student then splits the equilibrium mixture into three portions. Portion (i) receives solid NaF, which binds Fe3+ strongly. Portion (ii) is diluted tenfold with distilled water. Portion (iii) is cooled in an ice bath. The student observes that (i) the red color fades, (ii) the red color fades, and (iii) the red color deepens. A classmate argues that dilution should deepen the color because there are more particles on the reactant side, and that the cooling result proves the forward reaction is endothermic.
1. Calculate Kc from the equilibrium data, showing the concentration changes clearly.
Read the worked answer
Initial concentrations after mixing: [Fe3+]0 = [SCN−]0 = 0.0100 M. At equilibrium, [FeSCN2+] = 2.0 × 10−4 M. From the 1:1:1 stoichiometry, both reactants have decreased by the same amount:
[Fe3+]eq = [SCN−]eq = 0.0100 − 0.00020 = 0.00980 M.
Kc = [FeSCN2+]/([Fe3+][SCN−]) = (2.0 × 10−4)/(0.00980)2.
(0.00980)2 = 9.604 × 10−5, so Kc = 2.0 × 10−4 / 9.604 × 10−5 ≈ 2.08.
If units are retained, Kc has units M−1; in the usual thermodynamic convention it is written as a dimensionless value relative to standard states. The important point is the size: Kc is of order 1, so neither side is overwhelmingly favoured under these conditions.
2. Evaluate the dilution observation. A classmate says dilution should deepen the colour because the reactant side has more particles. Identify the flaw in that reasoning and predict the direction of the shift quantitatively.
Read the worked answer
The classmate is counting particles on each side of the equation instead of comparing the reaction quotient Q with K. For this equilibrium, the reactant side has two solute particles and the product side has one, but dilution changes Q in a specific way.
After tenfold dilution, every concentration is divided by 10. Then
Q = ([FeSCN2+]/10) / (([Fe3+]/10)([SCN−]/10)) = 10 × [FeSCN2+]/([Fe3+][SCN−]) = 10Kc.
Since Q > Kc, the reaction shifts left, consuming FeSCN2+ and producing Fe3+ and SCN−. The red colour therefore fades, as observed. The “more particles” argument fails because it ignores the concentration dependence of Q; dilution does not simply favour the side with more particles, it changes Q by a factor that depends on the stoichiometric coefficients.
3. Decide whether the NaF and cooling observations are consistent with the same equilibrium model. State the sign of ΔH for the forward reaction and correct the classmate’s endothermic claim.
Read the worked answer
NaF removes free Fe3+ by forming a stable complex. Lowering [Fe3+] makes Q larger than Kc, so the equilibrium shifts left and FeSCN2+ is consumed; the colour fades. This is consistent with the model and does not require a new explanation.
Cooling deepens the colour, so [FeSCN2+] increases and the equilibrium shifts right as temperature falls. For the forward reaction, a decrease in temperature increasing K means the forward reaction is exothermic, ΔH < 0. The classmate’s claim that cooling proves endothermic behaviour has the sign reversed.
All three observations fit one model: the forward reaction is exothermic, and the equilibrium responds to Fe3+ removal and to dilution by shifting left. The cooling result is the only one that identifies ΔH, and it identifies ΔH as negative.