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Steady State Approximation

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Jetzt kostenlos anmeldenSo, you're standing in the pasta aisle of the grocery store, and you're feeling spicy. You decide to try a new type of pasta with twists and turns you never imagined in a food. Once home, you boil some water and drop it in. You don't want to just sit and watch it, so you consult the ancient texts (the back of the pasta box).

The box states it will take 7-9 minutes to cook. Cooking times are a very helpful, pre-determined quantity, that have saved many lives. This is an example of kinetics, or essentially, how quickly reactions take place. This concept can also be applied to single-step or multistep reactions. In chemistry, we can derive an equation called the rate law, which describes the speed of a reaction.

Multistep reactions can be very complex, so deriving the rate law isn't always easy. When a reaction mechanism has several steps with comparable rates, it can be hard to figure out the rate-determining step. When this happens, we utilize the **steady-state approximation**.

In this article, we will be learning all about this approximation, such as what it is and how to use it.

- This article covers the
**Steady-State Approximation** - First, we will learn what this approximation is and do a brief overview of kinetics
- Then, we will learn how to use the approximation and apply it to some examples

Let's start by looking at the definition of **steady state approximation**.

The **steady-state approximation **(also called the quasi-steady-state approximation or pseudo-steady-state approximation)** **is a way to simplify the derivation of the rate equation. It is based on the assumption that one intermediate in the mechanism will be produced as fast as it is consumed (i.e., it is at a **steady state**).

In a multistep reaction, there will be species that are **intermediates.**

An **intermediate** is a species that isn't one of the initial reactants or final product(s). It is produced during the mechanism and will be completely consumed by the end.

The steady-state approximation assumes that at some point this intermediate will not have a change in concentration.

We can illustrate this steady state using the equation:

$$\frac{d[I]}{dt}=0$$

Where the concentration of the intermediate is [*I *].

In this article, we will be diving into the concept of the steady-state approximation and see how and where it can be used.

We note that a reaction’s rate is dependent on three things:

the rate constant (k) for the reaction,

the concentration of the reactants,

and the reaction order of the reactants.

The rate constant tells us how fast or slow a reaction is, the smaller the constant, the slower the reaction. A reactant's "order" shows the direct relationship between the rate and that reactant's concentration. At this point let us ask: if reactant A is doubled, will the rate also double, will it quadruple, or will it not change at all? Here is a general reaction:

$$A+B\xrightarrow{k_1}C$$

For this reaction, our rate equation looks like this:

$$\text{rate}=k_1[A]^x[B]^y$$

Where, k, is the rate constant, and the superscripts (x & y) represent the order of the reaction. Now, this is for an elementary, or "one-step" reaction, but what about a multistep reaction? Here is a general reaction mechanism:

$$A+B\underset{k_{-1}} {\stackrel{k_1}{\rightleftharpoons}}I\,\text{(fast)}$$

$$B+I\xrightarrow{k_2}C\,\text{(slow)}$$

All reactions have an energy barrier called **activation energy **that they need to overcome for a reaction to occur.

The activation energy barrier is based on the energy of the reactants and products relative to the energy of the transition state.

A system wants to be as stable as possible, so if a reaction produces a higher energy (i.e less stable) product, then it is going to have a larger activation energy since the system would rather stay as the stable reactants.

In our equation, we see that they are labeled "slow" and "fast". A "slow" reaction has a very high activation energy, so it takes a while for the system to get enough energy to proceed. This slow step is called the **rate-determining step**.

The

**rate-determining step**is what determines the rate for the whole reaction, and is also the step we write the rate equation for.

Think of it like being stuck in traffic: All the cars are stuck going the same speed as the slow truck in front, even if they can go faster. Here is how we would write the rate equation for this set of reactions:

We follow our basic formula from before, and set up the equation using:

$$A+B\underset{k_{-1}} {\stackrel{k_1}{\rightleftharpoons}}B+I\xrightarrow{k_2} C$$

Then,

$$rate=k_2[B][I]$$

Now here is where we have a problem, we cannot leave an intermediate in a rate reaction. We need to express the intermediate in terms of the reactants. We can do this by setting up the expression for the **equilibrium**** constant** for the equilibrium step:

The **equilibrium constant (K _{eq}) **shows whether a reaction favors the "forward" reaction (forms products) or the "backward" reaction (forms reactants). For a general equation:

$$A+B\rightleftharpoons C+D$$

The expression is:

$$K_{eq}=\frac{[C][D]}{[A][B]}$$

The constant can also be expressed in terms of the rate constants. For a general equation:

$$A+B\underset{k_{-1}} {\stackrel{k_1}{\rightleftharpoons}}C+D$$

The expression is:

$$K_{eq}=\frac{k_1}{k_{-1}}$$

Back to our reaction. We can express the equilibrium step in terms of the equilibrium constant:

$$A+B\underset{k_{-1}} {\stackrel{k_1}{\rightleftharpoons}}I\,\text{(fast)}$$

Then,

\(K_{eq}=\frac{[I]}{[B][A]}\)\(K_{eq}=\frac{k_1}{k_{-1}}=K_1\)or,\(K_1=\frac{[I]}{[B][A]}\)rearranging we then get, \(K_1[B][A]=[I]\)or,\([I]=K_1[A][B]\)

Now that we have our intermediate in terms of the reactants, we can substitute, \([I]=K_1[A][B]\), into our rate equation.

$$B+I\xrightarrow{k_2}C\,\text{(slow)}$$

\(\text{rate}=k_2[B][I]\)

\(\text{rate}=k_2[B](K_1[B][A])=k_2K_1[B]^2[A]\)

\(\text{rate}=k[B]^2[A]\)

Now we have our final rate equation. We combined our two constants, k_{2} and K_{1} , into "k" for simplicity.

Now that we've brushed up on the multistep rate equation, let's look at why we need this approximation in the first place. Here's a basic equation:

$$A\xrightarrow{k_1}B$$

$$B\xrightarrow{k_2}C$$

The concentration of B is dependent on A, therefore the concentration of C is dependent on both A and B. The derivation for the concentration looks like this:

$$[C]=[A_0](1+\frac{k_2e^{-k_1t}-k_1e^{-k_2t}}{k_1-k_2})$$

The equation itself isn't important, this is just to show how complex these derivations can be.

Now let's apply the steady-state approximation. In this mechanism, B is our intermediate, so we set its change in concentration to zero. When a species is in a steady state, its rate of consumption is equal to its rate of creation.

Using this, we can calculate:

\(\frac{d[B]}{dt}=0\,\text{(this expression means change in concentration of, B, over time)}\)

Recalling that under the steady state approximation:

\(\text{rate of consumption}=\text{rate of creation}\)

Then, we get:

\(k_1[A]=k_2[B]\)

\([B]=\frac{k_1[A]}{k_2}\)

Now that we have the expression for the concentration of B, we can calculate the change in concentration of C.

\(\frac{d[C]}{dt}=k_2[B]\)

\(\frac{d[C]}{dt}=k_2(\frac{k_1[A]}{k_2})=k_1[A]\)

Solving for the concentration of C we get:

\([C]=[A_0](1-e^{-k_1t})\)

You will not need to derive this, this is just to show how much simpler the derivation is.

So as you can see, the steady-state approximation makes the final expression much simpler. Now that we know the basics, let's work on a problem:

For the following multistep reaction, set up the rate law:

$$A+B\underset{k_1} {\stackrel{k_1}{\rightleftharpoons}}I$$

$$I+B\xrightarrow{k_2}C$$

Our first step is to set up the steady-state approximation. As we saw before, this means that we set the rate of consumption equal to the rate of creation. There is one step where I is being made, which is step 1 forward. The two steps where I is being consumed are step 1 (reverse) and step 2. We would express this as:

\(\text{rate of consumption}=\text{rate of creation}\)

\(\text{rate of consumption}-\text{rate of creation}=0\)

\(\frac{d[I]}{dt}=0=(k_2[I][B]+k_{-1}[I])-k_1[A][B]\)

Next, we will solve for the concentration of I:

\(\frac{d[I]}{dt}=0=(k_2[I][B]+k_{-1}[I])-k_1[A][B]\)

\(k_2[I][B]+k_{-1}[I]=k_1[A][B]\)

\([I](k_2[B]+k_{-1})=k_1[A][B]\)

\([I]=\frac{k_1[A][B]}{k_2[B]+k_{-1}}\)

We can simplify this expression. We can assume that the rate constant for step 1 backward (k_{-1}) is much slower than the rate constant for step 2 forward (k_{2}). This means that k_{-1} << k_{2}, so we assume that the step 1 backward rate constant is approximately zero (k_{-1} ≈ 0). When a system at equilibrium is affected by a change, it will do what it can to negate that change. As, the intermediate, I, is being formed during step 1 (forwards), it is subsequently being consumed during step 2. This means that we are "losing" our step 1 product, so it will continue in the forward direction to make more. This also means that the reaction is unlikely to "flip" to step 1 backward, so that reaction will be so slow that it is negligible.

\([I]=\frac{k_1[A][B]}{k_2[B]+0}=\frac{k_1[A]}{k_2}\)

Lastly, we can substitute this for our rate equation for C.

\(\frac{d[C]}{dt}=k_2[I][B]=k_2(\frac{k_1[A]}{k_2})[B]\)

\(\frac{d[C]}{dt}=k_1[A][B]\)

- The
**steady-state approximation**is a way to simplify the derivation of the rate equation. It is based on the assumption that one intermediate in the mechanism will be produced as fast as it is consumed (i.e. it is at a**steady state**) - To use the steady-state approximation, we set the change in concentration of the intermediate equal to zero and use that to find the expression for the concentration of the product.
- We can also use the approximation to calculate the concentration of an intermediate.

**steady-state approximation **(also called the **quasi-steady-state approximation **or **pseudo-steady-state approximation**)** **is a way to simplify the derivation of the rate equation. It is based on the assumption that one intermediate in the mechanism will be produced as fast as it is consumed (i.e. it is at a **steady state**)

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