0:00
In this unit, we will be looking at the relationship between Kc
the equilibrium constant with respect to concentration
and Kp, the equilibrium constant with respect to pressure.
Our goal of this unit is to understand the formula
and to be to calculate one value form the other.
When we have a reaction in which one or more
substances is in the gas phase
we can write either a Kc expression
or a Kp expression.
It just depends on what information we know
or what information we are trying to find.
We can look at the Kc expression
for our reaction, C cubed over AB squared
or we look over the Kp expression for our reaction, C cubed over AB squared
or we look over the Kp expression
and in this case we still see the same thing we saw or we look over the Kp expression
and in this case we still see the same thing we saw
before, that the coefficients are used
as powers
0:49
we still see we have products over reactants
the only thing that differs, is that instead of having we still see we have products over reactants
the only thing that differs, is that instead of having
the concentration of C, we have the pressure of C. the only thing that differs, is that instead of having
the concentration of C, we have the pressure of C.
Likewise we have the concentrations of A and B
and the pressures of A and B. Likewise we have the concentrations of A and B
and the pressures of A and B.
Note also, that we have a
p as a subscript for Kp
and c as a subscript
this distinguishes on K value from the other
because they will have different numerical values.
There is a way to derive the mathematic relationship between these two values
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
using the ideal gas law, but we are not going to show that here
and instead, we are just going to show the final result of that
which shows us Kp and instead, we are just going to show the final result of that
which shows us Kp
the equilibrium constant with respect to pressure
equals Kc
times RT to the delta n.
Note: that RT is in parentheses times RT to the delta n.
Note: that RT is in parentheses
and so that delta n goes with
everything inside those parentheses and so that delta n goes with
everything inside those parentheses
R is our ideal gas constant
0.08206 liters-atmospheres per mole kelvin. R is our ideal gas constant
0.08206 liters-atmospheres per mole kelvin.
And our temperature, as usual, must be in unit of kelvin. 0.08206 liters-atmospheres per mole kelvin.
And our temperature, as usual, must be in unit of kelvin.
1:58
when we look at the delta n value what we see is
delta n is a change in the moles of gas
and only the gas phase substances.
We will ignore solids, liquids, and aqueous solutions
because change in amounts of those do not significantly change the pressure.
However, changing the moles of gas results in a very large
change of pressure in some cases.
So we have to take that into consideration.
So when I look at my reaction I have
2 SO_2 + 1 O_2 yields 2 SO_3
so my moles of product is 2
my moles of reactant is 2 + 1 so 3 so my moles of product is 2
my moles of reactant is 2 + 1 so 3
and I am left delta n of -1
on my next example I see and I am left delta n of -1
on my next example I see
N_2 O_4 going to 2 NO_2 on my next example I see
N_2 O_4 going to 2 NO_2
and I see that my products side is 2
my reactant side is 1 so I have a delta n equal to 1 and I see that my products side is 2
my reactant side is 1 so I have a delta n equal to 1
we can have positive of negative values
for that delta n, it just depends on the specifics of our reaction,
2:54
However, sometimes we will find that K_c actually equals K_p
and this is from a mathematical calculation.
One thing we know, is that when delta n is 0
RT to the delta n will be equal to 1 but anything
to the zero power is equal to 1.
So anytime we have n equal to zero to the zero power is equal to 1.
So anytime we have n equal to zero
the K_p value and the K_c value will be the same.
For example, hydrogen gas plus iodine gas
yielding hydrogen iodide gas
we have delta n of 2 - 2 = 0.
So in this case, the K_p of the reaction
will be equal to the K_c of the reaction.
3:34
Lets look at an example.
A container at 800 degrees Celsius
was filled with NOCl gas
which decomposes to fform NO gas and chlorine gas.
What reaction represents this decomposition? which decomposes to fform NO gas and chlorine gas.
What reaction represents this decomposition?
3:48
Hopefully you answered C.
we know we have NOCl gas as our reactant.
NO gas is one product, and Cl gas is the other product.
Remember that Chlorine always exists
as a diatomic in nature so we have to write that as Cl_2.
Then we need to go back and balance our equation.
We have 2 chlorines on the right, we need 2
chlorine on the left, which gives us a coefficient of 2.
That also increases the number of N's and O's to 2.
So we need to put a 2 in front of the Nitrogen Monoxide on the right side.
Now, we can check that we have balanced chemical equation
that represents the decomposition of NOCl gas.
4:38
When I look at my reaction I had
2 NOCl in the gas phase
going to 2 NO in the gas phase
plus Cl_2 also in the gas phase. going to 2 NO in the gas phase
plus Cl_2 also in the gas phase.
Now because all my substances were in the gas phase plus Cl_2 also in the gas phase.
Now because all my substances were in the gas phase
they will all be included in th calculation.
I have 2 moles of NO and 1 mole of Cl_2
on my products side for a total of 3 moles of gas.
Minus the 2 moles of gas on my reactant side
and so delta n is equal to 1. Minus the 2 moles of gas on my reactant side
and so delta n is equal to 1.
5:15
Now, we know the reaction, we know the value of delta n.
Now we can actually find the value of K_c Now, we know the reaction, we know the value of delta n.
Now we can actually find the value of K_c
since we are give the value of K_p.
5:41
We look at our equation
K_p = K_c (RT)^∆n that we were
calculated earlier that was derived.
We look at our K_p value that was given.
We are trying to find K_c.
We know our R value is 0.08206
and our temperature is in units of Kelvin, we has 800 degree Celsius
and we had to add 273
to get our 1073
in units of kelvin, so we can use that in our formula.
I do the calculation
and then I have to rearrange I
am going to take my 1.8 x 10^-2
divided by
my term here divided by
my term here
and what I end up with
K_c = 2.0 x 10 ^ -4.
6:29
Now that we have done some practice calculating equilibrium constants
now we are going to look at what happens when we are given
concentrations of reactants, or possibly products
and have to figure out either the equilibrium constant
or later we will look at example where we are give initial concentrations
and we have find out what those final concentrations are
based on that equilibrium constant.
Dr. Allison SoultLecturer
Dr. Kim WoodrumSr. Lecturer
