2 deleted 1 character in body edited Apr 8 '17 at 0:59 Melanie Shebel♦ 3,50577 gold badges3535 silver badges7474 bronze badges 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 it'sits 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-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. 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 it's 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. 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. 1 answered Apr 7 '17 at 18:17 airhuff 15.2k77 gold badges3232 silver badges159159 bronze badges 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 it's 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.