Consider an aqueous reversible reaction of two chemicals $\ce{A <-> B}$, where $\ce{A}$ is red and $\ce{B}$ is yellow. The reaction is in equilibrium when there is $\pu{1 dm^3}$ of A with concentration $\pu{1 M}$ and $\pu{0.5 dm^3}$ of $\ce{B}$ with concentration $\pu{1 M}$, such that $K_C = 0.5$ and the solution is orange (due to the equal parts of red and yellow).
If I were to add $\pu{1 mol}$ of $\ce{A}$ to the system without increasing the total volume of the solution, Le Châtelier's principle predicts that the equilibrium position would move to the right, hence there would be more of $\ce{B}$ than $\ce{A}$ and the solution would become more yellow, however this does not seem to agree with my understanding:
$K_C = 0.5$, and is constant when the concentrations of chemicals are changed, so some of $\ce{A}$ must be converted to $\ce{B}$ by the forwards reaction to keep $K_C$ as $0.5$. This will make the concentration of $\ce{A}$ $\pu{\frac{5}{3} M}$ and the concentration of $\ce{B}$ $\pu{ \frac{5}{6} M}$ at equilibrium. The chemicals are still in the same ratio, so shouldn't the colour remain the same instead of becoming more yellow as Le Châtelier suggests?
This is a simplified problem, and I am wondering if the logic applies to more complicated cases (if it's correct in the first place of course!) Essentially, in my understanding, if you increase the concentration of a chemical in a dynamic equilibrium, when equilibrium is re-established, some but not all of the added chemical would be converted into the products on the opposite side of the reaction. This means the amounts of chemical on both sides of the reaction have increased, so it doesn't make any sense to say that the equilibrium position moves to one side of the other - both sides have increased in concentration, and in terms of coloured chemicals, the colour would remain the same.
Please use this more complicated reaction for demonstrative purposes:
$$\ce{Fe^{3+}_{(aq)} + 3\,SCN^-_{(aq)} <--> Fe(SCN)3_{(aq)} }$$
where $\ce{Fe^{3+}}$ is yellow, $\ce{SCN^-}$ is colourless and $\ce{Fe(SCN)3}$ is red. How would the reaction behave if the concentration of $\ce{Fe^{3+}}$ ions is increased (e.g. add iron III nitrate).
Apologies in advance if this question is a little convoluted, I'm not a professional chemist (I'm studying A-level chemistry), and I'm quite confused, so I'm struggling to form a coherent question.
Addition (edit): In the case of the more complicated reaction, my understanding is as follows:
- If I add $\ce{Fe^{3+}}$, the forwards reaction will convert some of the added $\ce{Fe^{3+}}$ and some $\ce{SCN^-}$ (already present) into $\ce{Fe(SCN)3}$.
- Not all of the added $\ce{Fe^{3+}}$ will be converted (the equilibrium cannot completely counteract the change, as this would involve converting all the $\ce{Fe^{3+}}$ into $\ce{Fe(SCN)3}$, and this would be an increase in the amount of $\ce{Fe(SCN)3}$ which the reaction would have to counteract).
- Therefore the concentration of reactants and products increases, because the added $\ce{Fe^{3+}}$ is split between some which is converted into $\ce{Fe(SCN)3}$, and some that stays as $\ce{Fe^{3+}}$.
- Since this gives an overall increase in the concentrations of both chemicals, neither has increased relative to the other, and the equilibrium position hasn't moved, so the colour doesn't change.