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1) When can't the rate-determining step be applied, and why? How do you recognize these cases?

2) When can't the steady state approximation be applied, and why?

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When can't the rate-determining step be applied, and why? How to recognize those mechanisms?

The rate determining step (often abbreviated as "rds") of a reaction is the slowest step in a reaction. Every reaction has a "slowest step", therefore, every reaction has an rds. The slowest step will be the step where we pass over a transition state with the highest energy, e.g. the highest energy point along the reaction coordinate. Look at the figure below, the reaction coordinate shown in the top left diagram

enter image description here

illustrates a reaction with a transition state, but no intermediate. There is only one step in this reaction, passing over that transition state (marked by the double-headed arrow) is the rds for that reaction. The drawing in middle and right side of the top row show a reaction with an intermediate, and consequently, two transition states. In each case the double-headed arrow identifies the rds - the highest energy point along the reaction coordinate. The 3 drawings in the bottom row illustrate the case of a reaction with 2 intermediates (and therefore 3 transition states) along the reaction coordinate - in each case the arrow identifies the transition state with the highest energy, the rds.

When can't the steady state approximation be applied, and why?

Consider the following reaction $$\ce{A <=>C[{k_1}]\ [B] <=>C[{k_2}]\ C}$$ $\ce{A}$ is the starting material, $\ce{[B]}$ is an intermediate and $\ce{C}$ represents the product. Either the drawing in the middle or right in the top row represent the reaction coordinate for this reaction. The steady state assumption applies to reactions that have one or more intermediates along the reaction coordinate. The steady state assumption states that the rate of change of the concentration of the intermediate is approximately zero. In other words, the intermediate is being created at about the same rate at which it is $$\ce{\frac{d[B]}{dt} = 0}$$ being destroyed. The steady state assumption is best applied to situations where the intermediate is present in low concentration (e.g. $\ce{k_2 >> k_1}$). A general rule of thumb for the assumption to be valid is for $$\ce{\frac{k_2}{k_1} > 10 }$$ Looking back at the figure, we now see that this condition is met in the top row, center drawing. Here is a Wikipedia article on the steady state assumption, the section on "Validity" is nice and clear if you'd like to read further on the subject.

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  • $\begingroup$ All in all, the SSA can be applied when the rate of consumption of the intermediate is more than ten times faster than the rate of its formation. Thanks. $\endgroup$ – RBW Aug 21 '14 at 9:19
  • $\begingroup$ I know this topic is old but why do you want the K2 to be higher then k1 for the steady state? This will actually produce more P + E so it will deviate from a constant ES concentration?@ron $\endgroup$ – KingBoomie Nov 16 '16 at 20:30
  • $\begingroup$ @RickBeeloo k2>>k1 guarantees a low concentration of the intermediate - a requirement for the Steady State Assumption to be valid. $\endgroup$ – ron Dec 10 '16 at 15:37

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