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
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?