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In this unit, we will be looking at the relationship between Kc

Â the equilibrium constant with respect to concentration

Â and Kp, the equilibrium constant with respect to pressure.

Â Our goal of this unit is to understand the formula

Â and to be to calculate one value form the other.

Â When we have a reaction in which one or more

Â substances is in the gas phase

Â we can write either a Kc expression

Â or a Kp expression.

Â It just depends on what information we know

Â or what information we are trying to find.

Â We can look at the Kc expression

Â for our reaction, C cubed over AB squared

Â or we look over the Kp expression for our reaction, C cubed over AB squared

Â or we look over the Kp expression

Â and in this case we still see the same thing we saw or we look over the Kp expression

Â and in this case we still see the same thing we saw

Â before, that the coefficients are used

Â as powers

Â 0:49

we still see we have products over reactants

Â the only thing that differs, is that instead of having we still see we have products over reactants

Â the only thing that differs, is that instead of having

Â the concentration of C, we have the pressure of C. the only thing that differs, is that instead of having

Â the concentration of C, we have the pressure of C.

Â Likewise we have the concentrations of A and B

Â and the pressures of A and B. Likewise we have the concentrations of A and B

Â and the pressures of A and B.

Â Note also, that we have a

Â p as a subscript for Kp

Â and c as a subscript

Â this distinguishes on K value from the other

Â because they will have different numerical values.

Â There is a way to derive the mathematic relationship between these two values

Â using the ideal gas law, but we are not going to show that here There is a way to derive the mathematic relationship between these two values

Â using the ideal gas law, but we are not going to show that here

Â and instead, we are just going to show the final result of that

Â which shows us Kp and instead, we are just going to show the final result of that

Â which shows us Kp

Â the equilibrium constant with respect to pressure

Â equals Kc

Â times RT to the delta n.

Â Note: that RT is in parentheses times RT to the delta n.

Â Note: that RT is in parentheses

Â and so that delta n goes with

Â everything inside those parentheses and so that delta n goes with

Â everything inside those parentheses

Â R is our ideal gas constant

Â 0.08206 liters-atmospheres per mole kelvin. R is our ideal gas constant

Â 0.08206 liters-atmospheres per mole kelvin.

Â And our temperature, as usual, must be in unit of kelvin. 0.08206 liters-atmospheres per mole kelvin.

Â And our temperature, as usual, must be in unit of kelvin.

Â 1:58

when we look at the delta n value what we see is

Â delta n is a change in the moles of gas

Â and only the gas phase substances.

Â We will ignore solids, liquids, and aqueous solutions

Â because change in amounts of those do not significantly change the pressure.

Â However, changing the moles of gas results in a very large

Â change of pressure in some cases.

Â So we have to take that into consideration.

Â So when I look at my reaction I have

Â 2 SO_2 + 1 O_2 yields 2 SO_3

Â so my moles of product is 2

Â my moles of reactant is 2 + 1 so 3 so my moles of product is 2

Â my moles of reactant is 2 + 1 so 3

Â and I am left delta n of -1

Â on my next example I see and I am left delta n of -1

Â on my next example I see

Â N_2 O_4 going to 2 NO_2 on my next example I see

Â N_2 O_4 going to 2 NO_2

Â and I see that my products side is 2

Â my reactant side is 1 so I have a delta n equal to 1 and I see that my products side is 2

Â my reactant side is 1 so I have a delta n equal to 1

Â we can have positive of negative values

Â for that delta n, it just depends on the specifics of our reaction,

Â 2:54

However, sometimes we will find that K_c actually equals K_p

Â and this is from a mathematical calculation.

Â One thing we know, is that when delta n is 0

Â RT to the delta n will be equal to 1 but anything

Â to the zero power is equal to 1.

Â So anytime we have n equal to zero to the zero power is equal to 1.

Â So anytime we have n equal to zero

Â the K_p value and the K_c value will be the same.

Â For example, hydrogen gas plus iodine gas

Â yielding hydrogen iodide gas

Â we have delta n of 2 - 2 = 0.

Â So in this case, the K_p of the reaction

Â will be equal to the K_c of the reaction.

Â 3:34

Lets look at an example.

Â A container at 800 degrees Celsius

Â was filled with NOCl gas

Â which decomposes to fform NO gas and chlorine gas.

Â What reaction represents this decomposition? which decomposes to fform NO gas and chlorine gas.

Â What reaction represents this decomposition?

Â 3:48

Hopefully you answered C.

Â we know we have NOCl gas as our reactant.

Â NO gas is one product, and Cl gas is the other product.

Â Remember that Chlorine always exists

Â as a diatomic in nature so we have to write that as Cl_2.

Â Then we need to go back and balance our equation.

Â We have 2 chlorines on the right, we need 2

Â chlorine on the left, which gives us a coefficient of 2.

Â That also increases the number of N's and O's to 2.

Â So we need to put a 2 in front of the Nitrogen Monoxide on the right side.

Â Now, we can check that we have balanced chemical equation

Â that represents the decomposition of NOCl gas.

Â 4:38

When I look at my reaction I had

Â 2 NOCl in the gas phase

Â going to 2 NO in the gas phase

Â plus Cl_2 also in the gas phase. going to 2 NO in the gas phase

Â plus Cl_2 also in the gas phase.

Â Now because all my substances were in the gas phase plus Cl_2 also in the gas phase.

Â Now because all my substances were in the gas phase

Â they will all be included in th calculation.

Â I have 2 moles of NO and 1 mole of Cl_2

Â on my products side for a total of 3 moles of gas.

Â Minus the 2 moles of gas on my reactant side

Â and so delta n is equal to 1. Minus the 2 moles of gas on my reactant side

Â and so delta n is equal to 1.

Â 5:15

Now, we know the reaction, we know the value of delta n.

Â Now we can actually find the value of K_c Now, we know the reaction, we know the value of delta n.

Â Now we can actually find the value of K_c

Â since we are give the value of K_p.

Â 5:41

We look at our equation

Â K_p = K_c (RT)^âˆ†n that we were

Â calculated earlier that was derived.

Â We look at our K_p value that was given.

Â We are trying to find K_c.

Â We know our R value is 0.08206

Â and our temperature is in units of Kelvin, we has 800 degree Celsius

Â and we had to add 273

Â to get our 1073

Â in units of kelvin, so we can use that in our formula.

Â I do the calculation

Â and then I have to rearrange I

Â am going to take my 1.8 x 10^-2

Â divided by

Â my term here divided by

Â my term here

Â and what I end up with

Â K_c = 2.0 x 10 ^ -4.

Â 6:29

Now that we have done some practice calculating equilibrium constants

Â now we are going to look at what happens when we are given

Â concentrations of reactants, or possibly products

Â and have to figure out either the equilibrium constant

Â or later we will look at example where we are give initial concentrations

Â and we have find out what those final concentrations are

Â based on that equilibrium constant.

Â