When calculating eq. constant...

cwag · 10 Oct 2008, 11:24 AM

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Quote:

Originally Posted by Undermyskin

It's correct, I think. I've never dealt with a K questions that have solid in it, to my ignorance, I suppose.

But revising what we need to do, I can prove it.

What we need to calculate K are the []s of chemicals involved. These can only be applied to aqueous solutions or gases only, right? Have you ever had a concentration of any solids? This is because the 'concentration' of a solid is preserved all the time. Taking a crystal of salt as an example. The number of molecules of NaCl/ the volume that crystal occupies = constant, yeah? Because even if adding more salt into an already concentrated solution, the extra settles to the bottom, joining the solid particles. And since the density of solid salt is the same at certain temperature and pressure, 'the concentration' is also the same. The more salt you add, the larger volume it occupies with the more molecules in there.

Yea that makes sense.... thanks everyone

Almatari · 9 Nov 2008, 2:50 PM

Units depends on the K expression

units for [] is mol/L
but if it is []^2, then the units are squared as well (mol/L)^2
and so on.......

chuboy · 1 Nov 2009, 4:38 PM

Quote:

Originally Posted by Almatari

Units depends on the K expression

units for [] is mol/L
but if it is []^2, then the units are squared as well (mol/L)^2
and so on.......

No, fail.

The K value is the ratio of products to reactants in an equilibrium system. OBVIOUSLY NO UNITS.

boxhunter91 · 1 Nov 2009, 4:51 PM

Quote:

Originally Posted by chuboy

No, fail.

The K value is the ratio of products to reactants in an equilibrium system. OBVIOUSLY NO UNITS.

LOL mate this was ages ago.

chuboy · 1 Nov 2009, 6:06 PM

True >.<

Well even so, this should be of some relevance since the exam is tomorrow.

Aerath · 1 Nov 2009, 6:32 PM

Quote:

Originally Posted by chuboy

No, fail.

The K value is the ratio of products to reactants in an equilibrium system. OBVIOUSLY NO UNITS.

Actually, Almatari is correct.

For instance, for Haber process K = [NH3]^2 / [h2]^3 [N2]

K has units M^-2.

That's outside the scope of the syllabus.

9 Nov 2008, 2:50 PM	#17 (permalink)
Almatari New Member HSC: N/A Gender: Undisclosed Join Date: Oct 2007 Posts: 14 Last Activity: 12 Nov 2008, 1:25 PM	Re: When calculating eq. constant... Units depends on the K expression units for [] is mol/L but if it is []^2, then the units are squared as well (mol/L)^2 and so on.......

1 Nov 2009, 6:06 PM	#20 (permalink)
chuboy Assistant Member HSC: 2009 Gender: Male Location: Northern NSW Join Date: May 2009 Posts: 97 Last Activity: Today, 9:17 PM	Re: When calculating eq. constant... True >.< Well even so, this should be of some relevance since the exam is tomorrow. __________________ 2009 HSC: AdvEng, Ext2 Math, Chem, Physics, Music1

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