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.
  • $\begingroup$ I think that this question is interesting. May also have to do something with optics. Here's a video to make you think: youtube.com/watch?v=LgJGTH5Xp0o Adding Fe 3+ made the solution darker. $\endgroup$ Commented May 15, 2017 at 19:37
  • $\begingroup$ Thanks for pointing me towards that video :) @QuantumAMERICCINO. The video clearly contradicts my reasoning (which I expected to happen; Le Chatelier's is obviously going to be correct), so I'll have to do some more digging to figure out what I'm misunderstanding. $\endgroup$ Commented May 15, 2017 at 19:56

2 Answers 2


In the general case of $\ce{A + B <=>C}$ if there are initially $a_0, b_0,c_0$ amounts present , then if an amount x of A and B react to produce C then the equilibrium constant $K_e$ is given by $$ K_e=\frac{k_1}{k_{2}}=\frac{c_0+x_e}{(a_0-x_e)(b_0-x_e)}$$ where $k_1, k_2$ are the rate constants for the forward and reverse reaction and $x_e$ is the amount reacted at equilibrium. Solving the equation and for (algebraic) simplicity let $c_0=0$, gives the amount at equilibrium $$x_e=\frac{1}{2}\left(a_0+b_0+K_e^{-1}-\sqrt{(a_0-b_0)^2+2(a_0+b_0)/K_e +K_e^{-2}}\right)$$ Using this the amount of each species can be obtained as for example the equilibrium amount of A is $a_e=a_0-x_e$. However, in your question you ask about the colour not the concentrations and this alters things a little. The quantity that is measured optically as in a spectrophotometer is the optical density of for example species A is $D=\epsilon_\lambda[A]l$ where $\epsilon_\lambda$ is the extinction coefficient of the molecule at wavelength $\lambda $ and is effectively how strongly the molecule absorbs at that wavelength, $\ce{[A]}$ is the concentration and l the path-length of the sample through which the light passes.

The total absorption is the sum of that for all species so depending on the values of $\epsilon$ and how the concentrations change the absorption may go up, down or remain constant at any given wavelength.

Furthermore, if you do not use a spectrophotometer and measure optical density but you are looking through the sample you will observe the transmitted light which is by the Beer-Lambert law given by $I_{trans}=I_0e^{-D}$ for each species present at each wavelength, so what you may see becomes pretty complicated.

  • $\begingroup$ Thank you very much for this great answer, I've got my head round this problem now :). If I could tick two answers, I would, but since I had to choose, I ticked "a-cyclohexane-molecule"'s, since I found their use of actual numerical values helpful, and it was a little more accessible to me. $\endgroup$ Commented May 16, 2017 at 18:23
  • $\begingroup$ I would up vote your answer but I haven't earned that privilege yet. $\endgroup$ Commented May 16, 2017 at 18:29

Below is the final bullet point in your question (emphasis mine):

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.

I will show that this is incorrect for the reaction $$\ce{A + B <=> C},$$ even though it is correct for the reaction $$\ce{A <=> B}.$$

Let $K = 1$, $\ce{[A]}=0.2$, $\ce{[B]}=5$, and $\ce{[C]}=1$ (dimensionless concentration units). Then $\ce{[C]/[A]}=5$. Now increase the concentration of A to $\ce{[A]}=1$, leaving the concentrations of B and C unchanged. We can solve the resulting quadratic equation to find the equilibrium concentration of all three species: $$1 = K = \frac{1+x}{(5-x)(1-x)} \quad \Longleftrightarrow \quad x = 0.63,$$ so that $\ce{[C]}=1.63$, $\ce{[A]}=0.37$, and $\ce{[C]/[A]}=4.40\neq5$.

I'm not sure how relevant the above point is, but it's something worth considering.

It's also important to note that the color is dependent not solely on the relative concentration of one species to another, but also on the absolute concentration of a species: consider a solution of water to which red food coloring is added. The more drops of red food coloring we add, the stronger the color. We start out with a solution that looks vaguely pinkish and end up with a solution that looks very red. I think this might be the more important point.

  • $\begingroup$ Thanks for your answer, I was stuck thinking about this for hours, so you've been really helpful! I was basically using the colours as a tangible way to understand the movement of the equilibrium position, so the above point was very relevant to me, despite colour not being an exact indicator of equilibrium position. (I would up vote your answer but I haven't earned that privilege yet). $\endgroup$ Commented May 16, 2017 at 18:28

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