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I can appreciate that the rate determining step affects the rate of reaction. What I take issue with however is the exact wording given in the definition of the RDS. From here:
The rate determining step is the slowest step of a chemical reaction that determines the speed (rate) at which the overall reaction proceeds.
The last part about the RDS determining the rate baffles me. The RDS would affect the rate by slowing down a succeeding step, but in what way does it determine it?
While it looks like a merger of the comment by @Sawarnik (+1) and the answer by @airhuff (+1), maybe an illustration of this quasi-steady state approximation or Bodenstein principle is of benefit. A brief search on the internet yielded an excerpt of a textbook of biochemistry, made freely available Wiley-Blackwell, with these two figures and a reference to the very original works about this phenomenon:
Please note that the width of the arrow $\ce{A -> B}$ is slightly different to the one relating $\ce{B -> C}$ in left-hand figure 1.12; but not in right-hand figure 1.13. You may see this as a hint of slightly different reaction rates leading to an increase of B.
According to the author of the book, Chapman and Underhill pioneered the work (DOI: 10.1039/CT9130300496), which was expanded by the one by Bodenstein (literally "A theory of photochemical reaction rates", DOI: 10.1515/zpch-1913-0112) and still is of importance in view of biochemical reactions.
For the purpose of this post, the figures found were rearranged. The are located on page 13 (first chapter) of said Fundamentals of Enzyme Kinetics, 4th edition (2012), authored by Athel Cornish-Bowden, ISBN 978-3-527-33074-4.
Think of a chemical reaction as a series of processes that must happen in order. If reaction $3$ is the slowest (rate determining) step of some overall reaction, then the intermediates formed in reaction $2$ just accumulate and wait for their turn to participate in the reaction. Depending on the order and kinetics of the reaction, the rate of reaction $3$ can be increased by this build-up of reactants, but there will always be an excess waiting to participate in this slow step. Once the reactants get past step $3$, they breeze on through the remaining faster steps of the reaction. Since reactions $4$ and above are fast compared to $3$, it makes little difference how fast they are.
Note that it is somewhat of an approximation to say that the overall rate is equal to that of the rate determining step. As mentioned in the example above, depending on the order and kinetics of the reaction, a buildup of intermediates from reaction $2$ could increase the rate of $3$, thus the rate of $2$ becomes somewhat relevant. Likewise, it's possible that a buildup of products from reaction $3$ could inhibit its rate depending on the rate of reaction step $4$. Wikipedia's definition of the rate-determining step of a reaction includes this mention that it is an approximation:
In chemical kinetics, the overall rate of a reaction is often approximately determined by the slowest step, known as the rate-determining step (RDS) or rate-limiting step. For a given reaction mechanism, the prediction of the corresponding rate equation (for comparison with the experimental rate law) is often simplified by using this approximation of the rate determining step.
