Lecture 3 - Chemical Kinetics (Part 3) and Chemical Equilibrium!

Lecture Recording:


Lecture Summary/Notes:

We finished off chemical kinetics today, with enzyme-substrate reactions and started chemical equilibrium. Again, just click the lines that look like this to expand or minimize the notes for the related topic.

Enzyme-Substrate Reactions

All enzymes have the ability to react/work on one substrate. We talked about the mechanism and the rate of this reaction:

Enzyme-substrate reactions is a 2-step process. The enzyme and the substrate come together to form an enzyme-substrate complex (the intermediate). The intermediate then decomposes to leave the enzyme and some product. *Assume that ES is in steady state (it is being formed at the same rate that it is being consumed - refer to lecture 2 for more on steady state conditions).

So, since we assumed that the second step is the rate determining (slow) step, we use it to form the rate law (this is the purpose of everything that follows - to find the rate law!). The second line is using the steady state principle: ES is being formed at the same rate in which ES is being consumed (in the reverse step of the equilibrium and in the second step). We factor out the [ES].
The third line is saying that the concentration of the enzyme + the concentration of ES is equal to the initial concentration of enzymes. We can rearrange this line to say [E] = [E]0-[ES]. We substitute this into the second line (steady state equation), and we get:

The second line here is what we get when we foil out the first line, multiply it by [S], and rearrange it a bit. (So there should be one line in between the first and the second that goes  K1[E]0[S] - K1[ES][S] = [ES](K-1 + K2). Then in the third line, we factor out the [ES], then we isolate [ES]. We THEN sub [ES] back into the FIRST equation (the rate law) to get the rate. We did all this because we cannot have the rate law in terms of the intermediate.

Alright, now that we have the rate, we used Km to lump the K constants together, so that we can simplify the math a little bit. Then, we set two conditions (what could happen). If Km is a lot larger than [S], then the reaction will be first order for the substrate. Vice-versa, if [S] is a lot larger than Km, then the reaction will be zero-order for the substrate. This explains why enzyme-substrate reactions are sometimes first order, and at other times zero order.

Chemical Equilibrium - Introduction

We started  chemical equilibrium with an example which went something like this: imagine that all the girls in the class started at the right side of the classroom and all the boys on the left. Then, the whole class was blindfolded and told to run around in any direction (everyone had the ability to walk/run straight and bounce off walls). After a certain time, there would be roughly an equal number of girls and boys on each side of the room (there will be some running to the other side and others bouncing off of walls heading to the other side). After that time, there will be a dynamic equilibrium, which is where the number of students of each gender on either side does not change, but everyone is still moving. The following is a graph representing this example:


So if we put this example into an equation, there is an equilibrium of the concentration of either boys or girls the right side [A] and the left side [B] (there are as many people going to the left side as there are going to the 
 
We find the rate law for both the people going from A to B (forward reaction), and for those going from B to A (reverse reaction). We will say that the rate constant for the forward reaction is K1 and the rate constant for the reverse reaction is K-1. The rates are equal, so we say that K1[A] = K-1[B]. We use this to find the equilibrium constant which is [B]/[A]

This example shows a general way to find the equilibrium constant of any aqueous equilibrium reaction (concentrations of the products to the power of their coefficients over the reactants to the power of their coefficients). Also, it shows that you can get an equilibrium constant from overall reactions, though you cannot get rates. However, the concentrations are just a part of what makes up ACTIVITY, which is what really makes up the equilibrium constant.

Activity

The following shows an example of finding the equilibrium constant using the method learned above, and what it really represents:

Just in case you can't read my writing... I wrote that the concentrations are REALLY activities... not just concentrations! So the activity of the products, raised to the power of their coefficients over the reactants to their powers are what make up K. Activity is unitless and ranges from 0-1. For aqueous solutions, activity can be found by using just the concentrations, but as we will learn later on, it is not the case for gaseous or pure liquid/solid states.

As a solution becomes more dilute, the activity coefficient approaches 1. This is why we can use just the concentrations when finding the activity of aqueous solutions.  

Pretty straight forward. Except for the last rule, when finding the activity for a gas, you can just say that activity of a gas = the partial pressure of that gas (in atm).

An example involving gases.

*See example 1 in the textbook examples section

Manipulation of Equations

• If you have multiple equations that add up resulting in an overall equation, the equilibrium constant of an overall equation = the product of individual equilibrium constants.

• If you multiply an equation through by a constant, you raise the equilibrium constant to the power of that constant (e.g.  2 times A+B <=> C+D, K^2)

• If you reverse an equation, you take the reciprocal of the equilibrium constant (e.g. C+D<=> A+B, 1/K)

*See example 2 in the textbook examples section

Gases (Equilibrium Constant)

The following is a generalization of how to find the equilibrium constant for equations involving gases (using partial pressures).

We then sought a way to find the equilibrium constant for equations involving gases with concentrations, as we did before.

Hopefully this is all pretty straight forward, and is easy to follow along. Remember that delta n involves only the coefficients of gas molecules!

*See example 3 in the textbook examples section

Textbook Examples Used in Class

Example 1 - Find the concentration of [Hg2]2+ using the given concentrations and the equilibrium constant.

Example 2 - find the equilibrium constant of the second equation given the first equation and its equilibrium constant.

Example 3 - Find the pressure-based equilibrium constant of the given equation given the concentration-based equilibrium constant and the temperature.

Example 4 - Find the concentration based equilibrium constant of the given equation (remember that pure liquids and solids have an activity of 1).

Example 5 - Find the pressure-based and concentration-based equilibrium constants of the given equation.