# Equilibrium constant, kc units

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#1
Hi
If the expression for eqm constant Kc has a fraction which cancels down to 0/1, does that mean there are no units???
0
4 years ago
#2
As in if the equilibrium constant cancels out to make 1/moldm^-3?
1
4 years ago
#3
Is 0/1 the actual value of the constant or the exponents of the concentration terms? The units should be independent of the value calculated so in the case of the former it'd be the reverse of the concentration units..
1
#4
(Original post by Deliciate)
As in if the equilibrium constant cancels out to make 1/moldm^-3?
The equilibrium constant cancels down to make 0/moldm^-3
0
4 years ago
#5
(Original post by kimmylee)
Hi
If the expression for eqm constant Kc has a fraction which cancels down to 0/1, does that mean there are no units???
In reality, the equilibrium constant depends on the activities of the species present and has no units.
1
4 years ago
#6
(Original post by kimmylee)
The equilibrium constant cancels down to make 0/moldm^-3
Can you post the expression on here prior to you cancelling it out?
0
4 years ago
#7
(Original post by kimmylee)
The equilibrium constant cancels down to make 0/moldm^-3
That seems pretty impossible. I'm assuming you mean there is a '1' left on the numerator. For example, 1/(moldm^-3) will become (mol^-1dm^3) which is just (dm^3mol^-1) If my assumption is wrong, can you post the step prior to you getting 0 on the numerator?
1
#8
(Original post by Deliciate)
Can you post the expression on here prior to you cancelling it out?
Hi
The equilibrium reaction is 2SO2 O2 =2SO3. (The = sign is meant to be the reversible reaction sign 0
4 years ago
#9
(Original post by kimmylee)
Hi
The equilibrium reaction is 2SO2 O2 =2SO3. (The = sign is meant to be the reversible reaction sign My assumption was right, ahah!
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#10
(Original post by Rohit_Rocks10)
That seems pretty impossible. I'm assuming you mean there is a '1' left on the numerator. For example, 1/(moldm^-3) will become (mol^-1dm^3) which is just (dm^3mol^-1) If my assumption is wrong, can you post the step prior to you getting 0 on the numerator?
Hi, I have just posted the equilibrium reaction but I have question: Why do we sometimes put dm^3mol^-1? Like why is it possible to put the dm^3 first and when do we do this?
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#11
(Original post by Rohit_Rocks10)
My assumption was right, ahah!
So does that mean there r no units??
0
4 years ago
#12
(Original post by kimmylee)
Hi, I have just posted the equilibrium reaction but I have question: Why do we sometimes put dm^3mol^-1? Like why is it possible to put the dm^3 first and when do we do this?
It really DOES NOT matter what unit comes first, as long as they are right. This is because the units are multiplied, so it doesn't really matter. (2x3 is the same as 3x2). We do this for the units to make more sense ('decimetres cubed per mole' makes more sense than 'per mole decimetres cubed') But if the switching of units confuses you, leave your units in the form you get when you simplify it! Hope that cleared your confusion!
1
4 years ago
#13
(Original post by kimmylee)
So does that mean there r no units??
NO, I never said that! The only time there will be no units is when the sum of the powers to which the units are raised to in the numerator is equal to the sum of the powers in the denominator!
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#14
(Original post by Rohit_Rocks10)
It really DOES NOT matter what unit comes first, as long as they are right. This is because the units are multiplied, so it doesn't really matter. (2x3 is the same as 3x2). We do this for the units to make more sense ('decimetres cubed per mole' makes more sense than 'per mole decimetres cubed') But if the switching of units confuses you, leave your units in the form you get when you simplify it! Hope that cleared your confusion!
Ah ok thanks,. So are the units moldm^-3?? Why is it this if there is 0 on the numerator like 0/1??
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4 years ago
#15
(Original post by kimmylee)
Ah ok thanks,. So are the units moldm^-3?? Why is it this if there is 0 on the numerator like 0/1??
The units are what I posted in Post No. 7
I know where the problem is in your understanding, this is just mathematical. You are assuming that there is a 0 on the top because there are no units on the numerator. That is wrong, because 2 of the units on the top cancel 3 of the units in the bottom. So the units are (1/moldm^-3) and not (0/moldm^-3). And, because you are NOT allowed to leave the units as (1/moldm^-3), you would write your final answer as either (mol^-1dm^3) or (dm^3mol^-1).

Let me explain this with a simple example. If you were to simplify 2/4, you would write it out as (2)/(2x2) where you would cancel the 2 on the top with one of 2s in the bottom, which becomes (1/2) and not (0/2).
This should hopefully clear things up for you now!
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#16
(Original post by Rohit_Rocks10)
The units are what I posted in Post No. 7
I know where the problem is in your understanding, this is just mathematical. You are assuming that there is a 0 on the top because there are no units on the numerator. That is wrong, because 2 of the units on the top cancel 3 of the units in the bottom. So the units are (1/moldm^-3) and not (0/moldm^-3). And, because you are NOT allowed to leave the units as (1/moldm^-3), you would write your final answer as either (mol^-1dm^3) or (dm^3mol^-1).

Let me explain this with a simple example. If you were to simplify 2/4, you would write it out as (2)/(2x2) where you would cancel the 2 on the top with one of 2s in the bottom, which becomes (1/2) and not (0/2).
This should hopefully clear things up for you now!
Ah, alright thanks, I get it now. So for a different eqm reaction, if there is 1unit on the top and 2units on the bottom, (1)/(1x2), would it just be 1/2 so the units is moldm^-3 ? 0
4 years ago
#17
(Original post by kimmylee)
Ah, alright thanks, I get it now. So for a different eqm reaction, if there is 1unit on the top and 2units on the bottom, (1)/(1x2), would it just be 1/2 so the units is moldm^-3 ? You're welcome. Nope, it would still be (1/moldm^-3) because 1 unit on the top cancels one of the 2 units on the bottom, and you are left with (1) on the top and (moldm^-3) on the bottom.
Picture 2 units on the top and 3 on the bottom. Now imagine cancelling 2 of the units with two of the 3 units on the bottom. You are left with (1) on the top and (moldm^-3) on the bottom.
Now, picture 1 unit on the top and 2 units on the bottom. Now imagine cancelling the only 1 unit on the top with one of the 2 units on the bottom. You are still left with (1) on the top and (moldm^-3) on the bottom. I can't explain this any further 0
#18
(Original post by Rohit_Rocks10)
You're welcome. Nope, it would still be (1/moldm^-3) because 1 unit on the top cancels one of the 2 units on the bottom, and you are left with (1) on the top and (moldm^-3) on the bottom.
Picture 2 units on the top and 3 on the bottom. Now imagine cancelling 2 of the units with two of the 3 units on the bottom. You are left with (1) on the top and (moldm^-3) on the bottom.
Now, picture 1 unit on the top and 2 units on the bottom. Now imagine cancelling the only 1 unit on the top with one of the 2 units on the bottom. You are still left with (1) on the top and (moldm^-3) on the bottom. I can't explain this any further Aha, tysm, I get it now. Yh there rly isn't much to say but I get how it works now so thanks again Have a good day! 😊
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4 years ago
#19
(Original post by kimmylee)
Aha, tysm, I get it now. Yh there rly isn't much to say but I get how it works now so thanks again Have a good day! 😊
Haha! No problems. You too! 😊
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