I'm attempting to answer a question from my textbook, however the answer that the textbook provides doesn't make sense to me.

The question:

Laura, a VCE chemistry student, loves graphs and wanted to represent an equilibrium reaction by using graphs. The reaction she chose was:

$\ce{Fe^{3+}(aq) + SCN-(aq) <=> FeSCN^{2+}(aq) \ \ \ \ \ \ \Delta H=-\nu e}$

Draw the rate-time graph for the reaction when the system is diluted at constant temperature.

In my mind, this dilution would instantly decrease the concentration of the reactants and products. But as it decreases the concentration of each species present by the same factor this will not disturb equilibrium, and simply result in a slower rate of reaction.

Given my assumptions I drew a graph where the forwards and reverse reaction both dropped vertically then continued on at the same, lower rate of reaction in equilibrium.

But the answer provided by the textbook is as follows

The answer

I just came across another question similar to this, and supposedly the system partially opposes the increase in concentration by favouring the reaction that produces the least amount of particles (in the same way as a change in pressure would cause the system to react). But if this is the case, and the concentrations are still balanced after the dilution, wouldn't it change the equilibrium constant? Which I have been told only temperature can change.

I'm assuming then that my understanding is flawed somewhere in regards to the system remaining balanced after the dilution, but I can't seem to see how diluting the system would unbalance the concentrations as they're all being decreased by the same factor.

Is the answer provided by the textbook wrong? Or is there something that I'm missing?

  • 2
    $\begingroup$ the value of $K_\mathrm{eq}$ doesn't change but $\ce{[Fe^3+]}$, $\ce{[SCN-]}$ and $\ce{[FeSCN^2+]}$ all change. Unless the change is very precise, there would be a change in point of equilibrium.. $\endgroup$ Commented Apr 30, 2021 at 13:46

1 Answer 1


The rate forwards and back are the same at equilibrium and differ before this is attained. Let the reaction be $A+B=C$. If each concentration is suddenly halved then the reaction is not necessarily any longer at equilibrium and in this case the rates forwards and back differ until a new equilibrium is reached. The figure shows this. The red curve is the forwards rate $k_1[A][B]$ and the green one $k_{-1}[C]$ the reverse. As you see the new equilibrium takes a short while to re-establish. (You can ignore the scales they depend on the particular rate constants and concentrations used but the behaviour is general.)

The figure you show is not fully supported by the calculation. The red curve (forwards rate) agrees but the blue one does not. If you do not halve the product concentration you get the blue curve in the figure.

AB=C then by 2

  • $\begingroup$ @ porphyrin :Do the rate of the back reaction more than the rate of forward reaction before new equilibrium ?Why? $\endgroup$ Commented Apr 30, 2021 at 19:23
  • 2
    $\begingroup$ @AdnanAL-Amleh ...because the number of particles is higher for the back reaction (2) compared to the forward reaction (1). $\endgroup$
    – Karsten
    Commented Apr 30, 2021 at 20:13
  • 1
    $\begingroup$ Never mind I've figured it out. I was previously thinking that the ratios of the concentrations were all that mattered. But looking at the equilibrium constant of any equation where there is an imbalance of moles on either side of the equation, that isn't the case. The concentration ratios will remain constant after the dilution, but that doesn't necessarily mean the system will remain in equilibrium. $\endgroup$
    – Jack
    Commented May 1, 2021 at 2:07
  • $\begingroup$ "If you do not halve the product concentration you get the blue curve in the figure." Are you referring to the OPs figure? $\endgroup$
    – Buck Thorn
    Commented May 18, 2021 at 14:23
  • $\begingroup$ yes, the OP's figure $\endgroup$
    – porphyrin
    Commented May 18, 2021 at 14:56

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