In this unit we are going to learn how to assemble a law of mass action. Which allows us to quantify the concentration of our products and reactants in an equilibrium reaction. The objective of this unit, is to understand how to assemble that, and how we use those numbers but also to see what happens when we have multiple reactions. How do we assemble the law of mass action or quantify that equilibrium constant for this new combination of reactions. So when we are looking at a a law of mass action it is based on a balance chemical equation. Like many of things we have seen in chemistry we want to look at a balance reaction. Now notice the K values do not have any units equilibrium constant do not have a unit associated with them. So what we are looking at here is we start with our balanced reaction where we have A moles of substance A. B mole s of substance B going to C moles of substance C and D moles of substance D. When I am writing my law of mass action what I am going to look at is the amount of products the concentration of products raised to the power that it equals their coefficient Divided by the concentration of the reactants raised to their coefficient as their power. The concentration of substance C to the C power, the concentration of substance D to the D power the concentration of substance A to the A power to the C power, the concentration of substance D to the D power the concentration of substance A to the A power and the concentration of substance B to the B power. Now I have a general formula for assembling these law of mass actions. So what we want to do is look at a specific example Now I have a general formula for assembling these law of mass actions. So what we want to do is look at a specific example for a different reaction. What I can look at here is I have a balanced equation, 2 hydrogen plus oxygen yields two water. Note, that we are looking at all these in the gas phase we will talk in just a second, about how we deal when we do not have everything in the gas phase. Here we are looking at water as out product and I raise it to the second power, because our coefficient is two on the denominator, we have hydrogen as a reactant. Again, raised to the power of 2 because its coefficient is 2. Oxygen and it does not have a power because coefficient is one. So now what I have is an equilibrium constant calculation or my law of mass action for this particular reaction. Given some information about either the value K or the value of concentration of water, hydrogen, and or oxygen. I could figure an unknown value. When we get to solids we actually omit them from out equilibrium expression because the concentration of a solid is constant. as I increase the amount of solid I am also increasing the mole and the volume by a proportional amount. So since since the concentration is constant we exclude that from our equilibrium expression, or law of mass action. Now when I reaction below I see I have a solid as a reactant and I have two gases as a product. I can still write a law of mass action but it is only going to include those non-solid substances. So here is an example, I have two carbon monoxide going to carbon dioxide plus carbon solid. In this case, my solid is one of my products and I do not include that in the equilibrium expression I write the concentration of CO2 again no power, because my coefficient is one. Over the concentration of carbon monoxide to the second power because my coefficient is two. Note also, that I have added a subscript here to the K value and so C stands for concentration here to the K value and so C stands for concentration because one thing we are going to look at as we talk about equilibrium is the fact that we can talk about the equilibrium constant with respect to concentration or the equilibrium constant with respect to the pressure of any gases present. When we are dealing with gases we can use either Kc or Kp we do need to indicate which one we are calculating because the values can be different. The other thing I exclude from my law of mass action is for pure liquid again, just like with solids the concentration of pure liquids does not change. So we have a reaction here 2 H202 going to oxygen and water, so the decomposition of Hydrogen Peroxide So we have a reaction here 2 H202 going to oxygen and water, so the decomposition of Hydrogen Peroxide and when I look at my law of mass action I have my oxygen as a product I notice water, because it is a pure liquid is omitted on the bottom I have H2O2 to the second power for the coefficient there and note this is aqueous on the bottom I have H2O2 to the second power for the coefficient there and note this is aqueous remember that aqueous mean dissolved in water. And so I do include that in my law of mass action. So the only thing I omit, are solids And so I do include that in my law of mass action. So the only thing I omit, are solids and my pure liquids. Once I am assembled my law of mass action I can use that information to calculate that equilibrium constant. The value of that equilibrium constant can tell something about whether products are favored or whether reactants are favored. If I have a value of K much much greater then 1 one thing I can look at is to remember that I have products over reactants. that I have products over reactants. So if I have a value that is greater that one what that tell me is that the concentration of products is going to be much greater then the concentration of reactants. I don't know how far the reaction is going to proceed to form products but the fact that it is greater then one tells me I am going to have more products then reactants present. The reverse if also true. When I look at a K value much much less then 1 I can see that reactants are favored. When I look at a K value much much less then 1 I can see that reactants are favored. Here, again I have my products on the top my reactants on the bottom and if this number is much much less than 1 that tell me that the numerator is a much smaller number then the denominator and therefore that show that my reactants are going to be the predominant species and what we see is that in fact we have a little bit of product forming here but predominantly what present in this mixture is the reactant. Now every reaction has its own K value and that can be affect by a variety of factors. What we are worried about right now is how can I set up a law of mass action how can I calculate the value of K. So lets look at an example here what is the law of mass action for the reaction of A to B. So you have had a change to answer to answer that question hopefully you saw that it should be answer B where we have B over A. Because B is our product, and A is out reactant, and so we want to set our law of mass action up to indicate products over reactant. Because the reaction we are giving has coefficients of 1 for both substances we do not need to include a power for either of those numbers. What we will look at next is figuring out how we can take the law of mass action for multiple reactions and combine them all into one expression and get one value of K simply by knowing the law of mass actions for some individual reactions that make up that overall reaction.
