Equilibrium Constant Expressions

[Chris127]

Hey all, I have a quick question. This is driving me insane, and my book doesnt help.

I have a test Friday and cant wait to ask my prof tomorrow because this is driving me crazy. Ok,

Chem 1202 btw, _ = subscript

When writing Equilibrium Constant Expressions, say for 20_3 ---->30_2, the expression would be

Kc = [0_2]^3 / [0_3]^2

Simple enough, right?

The thing that is driving me crazy is when asked to calculate this value. You see, my book will simply take an expression like that and say it equals something like 4.4537383 X 10^-5 or something. But, where do you get these values? Is there a method to compute them, or do they have to be given, like a constant?

Again, I will ask my prof tomorrow, but I want to get a lot done tonight, but I cant figure this out. Thanks,

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[princessd3]

Hey all, I have a quick question. This is driving me insane, and my book doesnt help.

I have a test Friday and cant wait to ask my prof tomorrow because this is driving me crazy. Ok,

Chem 1202 btw, _ = subscript

When writing Equilibrium Constant Expressions, say for 20_3 ---->30_2, the expression would be

Kc = [0_2]^3 / [0_3]^2

Simple enough, right?

The thing that is driving me crazy is when asked to calculate this value. You see, my book will simply take an expression like that and say it equals something like 4.4537383 X 10^-5 or something. But, where do you get these values? Is there a method to compute them, or do they have to be given, like a constant?

Again, I will ask my prof tomorrow, but I want to get a lot done tonight, but I cant figure this out. Thanks,

Those are constants. You don't necessarily compute them they should be given to you. You're usually asked to solve the other part of the equation, such as the concentration of the reactants or products.

 

[SeventhSon]

Hey all, I have a quick question. This is driving me insane, and my book doesnt help.

I have a test Friday and cant wait to ask my prof tomorrow because this is driving me crazy. Ok,

Chem 1202 btw, _ = subscript

When writing Equilibrium Constant Expressions, say for 20_3 ---->30_2, the expression would be

Kc = [0_2]^3 / [0_3]^2

Simple enough, right?

The thing that is driving me crazy is when asked to calculate this value. You see, my book will simply take an expression like that and say it equals something like 4.4537383 X 10^-5 or something. But, where do you get these values? Is there a method to compute them, or do they have to be given, like a constant?

Again, I will ask my prof tomorrow, but I want to get a lot done tonight, but I cant figure this out. Thanks,

This value will be given to you. It can also be calculated if you Know the gibbs free energy (deltaG) of a reaction at a given temperature and pressure, or equivalently, gibbs energy of formation of all reactants and products. Again, these values are in a table. dG(rxn) = sum[coeff *(dG(products)] - sum[coeff*dG(reactants)] where dG is the Gf.

in case you don't remember, dG(rxn) = - RT * ln K . Also there are "different" equilibrium constants (concentration basis, pressure basis, etc.) and if you are using the notation Kc i assume you are doing concentration.

 

[DesiMcatAcer]

Hey all, I have a quick question. This is driving me insane, and my book doesnt help.

I have a test Friday and cant wait to ask my prof tomorrow because this is driving me crazy. Ok,

Chem 1202 btw, _ = subscript

When writing Equilibrium Constant Expressions, say for 20_3 ---->30_2, the expression would be

Kc = [0_2]^3 / [0_3]^2

Simple enough, right?

The thing that is driving me crazy is when asked to calculate this value. You see, my book will simply take an expression like that and say it equals something like 4.4537383 X 10^-5 or something. But, where do you get these values? Is there a method to compute them, or do they have to be given, like a constant?

Again, I will ask my prof tomorrow, but I want to get a lot done tonight, but I cant figure this out. Thanks,

Hi,
Kc's are constant for a particular reaction(but remember they are a very strong function of temperature, and a weaker function of pressure as well). They could either give you the values of the concentrations, in which case you can calculate the Kc OR they can provide you with a Kc value at a particular temperature and pressure for a reaction and you can get the rest OR you can get it from delta G values like the posters above mentioned. Hope it helps! Good luck! 🙂

 

[MedicineNutt]

Hey all, I have a quick question. This is driving me insane, and my book doesnt help.

I have a test Friday and cant wait to ask my prof tomorrow because this is driving me crazy. Ok,

Chem 1202 btw, _ = subscript

When writing Equilibrium Constant Expressions, say for 20_3 ---->30_2, the expression would be

Kc = [0_2]^3 / [0_3]^2

Simple enough, right?

The thing that is driving me crazy is when asked to calculate this value. You see, my book will simply take an expression like that and say it equals something like 4.4537383 X 10^-5 or something. But, where do you get these values? Is there a method to compute them, or do they have to be given, like a constant?

Again, I will ask my prof tomorrow, but I want to get a lot done tonight, but I cant figure this out. Thanks,

what exactly is the question?? cuz i'm kinda gettin confused...if you can scan or write out the question that would be a lot better...btw i just learned this not too long ago!! lol

 

[princessd3]

In other words, that value is the constant for that particular reaction.

 

[Chris127]

Thanks everyone, that helped alot. I was pulling my hair out trying to figure out where to obtain these values.

The thing is, in my study packet, there is a problem sismilar to this.

He gives me two chemical equations, and tells me the Kp of one is 6.77 X 10^-5.

So I go, ok, no big deal, I just set up my two equillibrium constant expressions. The thing is, he gave me no constants or values to plug in. I was afarid I was missing something.

 

[MedicineNutt]

The step i would take to solve this is...

1) Define initial concentration of each compound (if they dont state the initial molar and liter or just the concentration (M), then just use 0 M)
2) the x-value or the disassociated amount of the compound
3) initial + x-value (or disassociated amount)

After you have all those figured out...

--I would use that constant (Kc) into the equilibrium equation (with the plugged initial + x-value (disassociated amt)...

Kc = equilibrium equation

AFTER THIS STEP YOU SHOULD GET THAT ANSWER!!! lol

 

[Rafa]

The fact that none of this makes any sense to me suggests I'd do well to look over Gen.Chem. before next April.

 

[Chris127]

Exact question:

The equilibrium constant for the reaction

N_2 (g) + 3H_2 (g) <-----> 2NH_3 (g)

is Kp=6.77 X 10^-5 at some temperature. What is the value of Kp for the reaction 1/3N_2 (g) + H_2 (g) <-----> 2/3NH_3 (g)


After setting up my equlibrium constant equations like in the OP, I have no values or constants to plug in.


IE,

k1 = [NH_3]^2 / [N_2][H_2]^3 for the first equation, but where do I get the values for these? Hopefully they are given

 

[DesiMcatAcer]

The step i would take to solve this is...

1) Define initial concentration of each compound (if they dont state the initial molar and liter or just the concentration (M), then just use 0 M)
2) the x-value or the disassociated amount of the compound
3) initial + x-value (or disassociated amount)

After you have all those figured out...

--I would use that constant (Kc) into the equilibrium equation (with the plugged initial + x-value (disassociated amt)...

Kc = equilibrium equation

AFTER THIS STEP YOU SHOULD GET THAT ANSWER!!! lol

And just to add to that a little if your value of Keq is very very small, then you can simplify the quadratic equation into a simple algebraic equation...

 

[MedicineNutt]

Exact question:

The equilibrium constant for the reaction

N_2 (g) + 3H_2 (g) <-----> 2NH_3 (g)

is Kp=6.77 X 10^-5 at some temperature. What is the value of Kp for the reaction 1/3N_2 (g) + H_2 (g) <-----> 2/3NH_3 (g)


After setting up my equlibrium constant equations like in the OP, I have no values or constants to plug in.


IE,

k1 = [NH_3]^2 / [N_2][H_2]^3 for the first equation, but where do I get the values for these? Hopefully they are given

Value of Kp is the square root of 6.77 x 10^-5...i honestly couldnt tell you why, but this is just from memory!!! and if you have a third equation its 1/Kp...

 

[MedicineNutt]

And just to add to that a little if your value of Keq is very very small, then you can simplify the quadratic equation into a simple algebraic equation...

yea...exactly!!! that's a lot easier to grasp...im a horrible explainer

 

[MedicineNutt]

.

 

[kirexhana]

i miss g chem 😍 . pchem is a pain in the @$$... 😡

 

[jebus]

I use these expressions all the time in my personal life.
"No baby, you've got to do it more. Think about the equilibrium of your mouth and my---" Well, I've said enough. I don't want to give away my whole game.
Chemistry rox.

 

[MedicineNutt]

i miss g chem 😍 . pchem is a pain in the @$$... 😡

seriously, how bad is pchem??? what makes it painful?

 

[91946]

seriously, how bad is pchem??? what makes it painful?

pchem is math based so it really depends on where your comfort zone with math. When i took it, it was alot of thermodynamics proofs, entropy, enthalpy and the whole ten yard. A little bit of quantum mechanics also, but not too much.

 

[DesiMcatAcer]

edit

 

[MedicineNutt]

Well, let's think about the second equation you stated:
1/3N_2 (g) + H_2 (g) <-----> 2/3NH_3 (g), if I multiply both sides by 3, then I get the same reaction as the first one. So, it's the same reaction as the first one, hence the equilibrium CONSTANT will be the same(at same temperature and pressure of course unless your question stated you wanted at different temperature which you didnt mention?). Since the reaction is the same, so will be the Keq at same temperature. However, be careful because sometimes the reaction might not be the same or might have different ratio of coefficients, in which case you might need to add to reactions to get to that reaction, and you multiply the K's of the individual reactions to get the K of the reaction you are looking for.

hmmm...how sure are you about this? because im pretty confident about square-rooting the kp...

 

[DesiMcatAcer]

hmmm...how sure are you about this? because im pretty confident about square-rooting the kp...

Why the square root? Maybe someone else can confirm 🙂

 

[Chris127]

Why the square root? Maybe someone else can confirm 🙂

Yes, my prof said something about taking the square root

BUT that not the problem I am having, for k1 = [NH_3]^2/[N_2][H_2]^3, where do I get these values? Are they supposed to be given to me?

 

[MedicineNutt]

Yes, my prof said something about taking the square root

BUT that not the problem I am having, for k1 = [NH_3]^2/[N_2][H_2]^3, where do I get these values? Are they supposed to be given to me?

only the kp should be given for that equation...from what ive seen so far

 

[DesiMcatAcer]

Yes, my prof said something about taking the square root

BUT that not the problem I am having, for k1 = [NH_3]^2/[N_2][H_2]^3, where do I get these values? Are they supposed to be given to me?

well according to the question, it already gave you Kp=6.77 X 10^-5

 

[MedicineNutt]

so is this gen chem 1 or 2??

 

[SeventhSon]

this isn't hard guys...correct answer.... raise your original Kp to the one-third power... It's the same stiochiometric equation

There is no "plugging in". He is testing your knowledge of this rule regarding equilibrium constants: if you multiply a chemical equation by 'n' coefficient (in this case 1/3), you raise Kc to the nth power

silly freshmen

 

[Chris127]

this isn't hard guys...correct answer.... raise your original Kp to the one-third power... It's the same stiochiometric equation

There is no "plugging in". He is testing your knowledge of this rule regarding equilibrium constants: if you multiply a chemical equation by 'n' coefficient (in this case 1/3), you raise Kc to the nth power

silly freshmen

Yes, he said K2 = cubed root of K1, which = cubed root of 6.77 X 10^-5

So I just take the cubed root of this number? (answer is 4.07 X 10^2 btw)


EDIT: AH, I got it! Thats so easy 😳 Thanks Seventh (and everyone else 😀 )

 

[Mr. Tee]

so is this gen chem 1 or 2??

2

 

[MedicineNutt]

this isn't hard guys...correct answer.... raise your original Kp to the one-third power... It's the same stiochiometric equation

There is no "plugging in". He is testing your knowledge of this rule regarding equilibrium constants: if you multiply a chemical equation by 'n' coefficient (in this case 1/3), you raise Kc to the nth power

silly freshmen

does this stand true with every problem setup this way?

 

[thesauce]

Exact question:

The equilibrium constant for the reaction

N_2 (g) + 3H_2 (g) <-----> 2NH_3 (g)

is Kp=6.77 X 10^-5 at some temperature. What is the value of Kp for the reaction 1/3N_2 (g) + H_2 (g) <-----> 2/3NH_3 (g)

Hey Chris, the Kp for the new reaction is the cube root of the old Kp. Other variations on this idea are as follows:

1/2N_2 (g) + 3/2H_2 (g) <-----> 1NH_3 (g), newKp = (oldKp)^(1/2)

2N_2 (g) + 6H_2 (g) <-----> 4NH_3, newKp = (oldKp)^2

2NH_3 <-----> N_2 (g) + 3H_2 (g), newKp = (oldKp)^-1 = 1/oldKp

You can prove this to yourself by writing the old expression and the new expression and seeing that they are related by the given powers.