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2.04 Relationship Between Kc and Kp

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Química Avançada

Universidade de Kentucky

4.7 (424 classificações) | 51K alunos inscritos

A chemistry course to cover selected topics covered in advanced high school chemistry courses, correlating to the standard topics as established by the American Chemical Society. Prerequisites: Students should have a background in basic chemistry including nomenclature, reactions, stoichiometry, molarity and thermochemistry.

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4.7 (424 classificações)

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KO

Jul 17, 2017

Excellent content and delivery!\n\nThis course has many well-explained examples and practice questions to reinforce the understanding of the taught concepts. Thank you!

CW

Sep 17, 2015

I used this course as a significant portion of my preparation for the CLEP Chemistry exam, and passed today with an A! Thank you!

Na lição

Chemical Equilibrium

This unit introduces the concept of chemical equilibrium and how it applies to many chemical reactions. The quantitative aspects of equilibrium are explored thoroughly through discussions of the law of mass action as well as the relationship between equilibrium constants with respect to concentrations and pressures of substances. Much of the discussion explores how to solve problems to find either the value of the equilibrium constant or the concentrations of substances at equilibrium. ICE (initial-change-equilibrium) tables are introduced as a problem-solving tool and multiple examples of their use are included. From a qualitative standpoint, Le Châtelier’s principle is used to explain how various factors affect the equilibrium constant of a reaction along with the concentrations of all species.

2.01 Dynamic Equilibrium6:28

2.02 Law of Mass Action7:34

2.03 Law of Mass Action for Combined Reactions4:36

2.04 Relationship Between Kc and Kp6:51

2.04a Relationship Between Kc and Kp2:34

2.05 Calculating the Equilibrium Constant13:09

2.05a Finding Kc5:04

2.06 Reaction Quotient5:55

2.06a Predicting Reaction Progress with the Reaction Quotient1:55

2.07 Calculating Equilibrium Concentrations12:54

2.07a Finding Equilibrium Concentrations, part 14:59

2.07b Finding Equilibrium Concentrations, part 23:20

2.07c Finding Equilibrium Concentrations, part 34:22

2.08 Le Chatelier's Principle Part A9:55

2.08a Le Chatelier's Principle Part B19:50

2.08b How Changes Shift the Equilibrium5:18

Ministrado por

Dr. Allison Soult
Lecturer
Dr. Kim Woodrum
Sr. Lecturer

Transcrição

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

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.

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,

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.

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?

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.

Now what we need to look at is

what is the change in n?

What is the delta n for this reaction?

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.

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.

and C is our answer so the value of K_c

2.0 x 10 ^ -4.

Now the next slide I am going to show you how we work

that problem out and get to that answer.

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.

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.

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