14.4.4 Steady State Kinetics Flashcards
Steady State Kinetics
- In some reactions, the rate-determining step depends on the concentrations of reactants.
- By the steady state approximation, the concentrations of reaction intermediates are assumed to be constant during the course of the reaction.
- Steady state approximations can be used to predict how reaction rates change under different conditions.
- Steady state approximations can be applied to reactions involving enzymes.
note
- In some reactions, the rate-determining step depends on the concentrations of reactants.
- In the decomposition reaction of dinitrogen pentoxide (N 2 O 5 ), either the first or second elementary step can be the rate-determining step, depending on the concentrations of reactants.
- In this reaction mechanism, M represents any gas particle, such as an N 2 O 5 molecule or an atom of an inert gas added to the reaction. When an N 2 O 5 molecule collides with a gas particle with sufficient kinetic energy, the N 2 O 5 becomes excited (N 2 O 5 *).
- Determining which step is limiting under a given set of
conditions requires the steady state approximation. - By the steady state approximation, the concentrations of reaction intermediates are assumed to be constant during the course of the reaction.
- For example, in the decomposition reaction of dinitrogen pentoxide, the dinitrogen pentoxide excited state (N 2 O 5 *) is assumed to be at steady state. This means that the rate of change of the concentration of N 2 O 5 * (d[N 2 O 5 *]/dt) is zero.
- N 2 O 5 * is formed by the first elementary step, and is consumed by the reverse of the first elementary step and by the second elementary step. The rates of these reactions can be used to express the concentration of N 2 O 5 * in terms of the concentrations of N 2 O 5 and M.
note 2
- The reaction rate for the overall reaction is proportional to the change in the concentration of nitrogen dioxide (NO 2 ).
- By the second elementary step, the rate of change in the concentration of nitrogen dioxide equals k 2 [N 2 O 5 *].
- Substituting the formula for the concentration of N 2 O 5 * yields a rate law in terms of the concentrations of reactants (N 2 O 5 and M). There are two extremes for this rate law.
- If the concentration of M is small such that k –1 [M] is much smaller than k 2 , the denominator of the rate law reduces to simply k 2 . This yields a simpler rate law, in which the reaction rate depends on the concentrations of both N 2 O 5 and M. In this case, the first elementary step is rate determining.
- This is analogous to a sandwich shop in the morning, when the rate of production of happy customers depends both on the number of sandwich makers and on the number of customers.
- If the concentration of M is large such that k –1 [M] is much larger than k 2 , the denominator of the rate law reduces to k –1 [M]. This yields a simpler rate law, in which the reaction rate depends only on the concentration of N 2 O 5 . In this case, the second elementary step is rate determining.
- This is analogous to a sandwich shop at noon, when the rate of production of happy customers depends only on the number of sandwich makers.
- Steady state approximations can be applied to reactions involving enzymes (E) with their substrates (S) to form products (P).
- The reaction of an enzyme and a substrate produces an enzyme-substrate reaction intermediate (ES). The
concentration of the reaction intermediate is at steady state. - As in the decomposition of dinitrogen pentoxide, the rate of reaction depends on the concentration of substrate, with different rate laws at the two extremes.
What does the variable k−1 refer to?
The rate constant for the reverse reaction
What are enzymes?
Catalysts for biological reactions
In enzyme kinetics, the ratio, Δ[ ES] / Δt, is equivalent to zero in the reaction mechanism of an enzyme and a substrate to form an enzyme-substrate complex (ES). Which term best describes the state of the enzyme-substrate complex in the reaction?
steady
Which of the following statements is true?
The change in the concentration of an intermediate over the change in time is equal to zero.
Which of the following is true regarding an intermediate in a reaction mechanism?
I = intermediate
I is a reactant and product.
If two or more steps in a reaction mechanism are slow, what can one use to predict the rate law?
The steady state approximation
Which of the equations below forms a rate law?
rate = k[A] ^m [B]^ n…
If [S] is large in the equation, rate = k2 [ E] + [S] / (k−1 + k2 / k1 ) + [S], then the rate is proportional to what?
k2 [ E]
In the reaction, N2O5(g) → 2NO2(g) + O2(g), the first step, N2O5 + Ar ↔ N2O5* + Ar, is fast. Which best describes the concentration of N2O5*?
Constant
