Kc Calculations

If anyone could help me with how to go about working these out please (apologies for lazy formulae):

2.0 moles of PCl5 vapour are heated to 500 K in a vessel of volume 20 dm3. The equilibrium mixture contains 1.2 moles of chlorine. Calculate Kc for the equilibrium at 500K.

PCl5(g) \leftrightarrow PCl3(g) + Cl2(g)

So I'd work out the initial moles and moles made/used and this is where I get stuck (sadly).

- - - - - - - - PCl5 - - - PCl3 - - - - Cl2
Initial moles - 2.00 - - - ? - - - - - -1.2
Made/ used - - ? - - - - ? - - - - - - ?
Eq moles - - - -? - - - - ? - -- - - - ?
Eq conc - - - - 20 - - - -20 - - - - - 20

I know how to write out the equation after that, just not how to get all the constituents.

Help/guiding most appreciated. :smile:

Reply 1

charco

Study Forum Helper

18

Holsy

If anyone could help me with how to go about working these out please (apologies for lazy formulae):

2.0 moles of PCl5 vapour are heated to 500 K in a vessel of volume 20 dm3. The equilibrium mixture contains 1.2 moles of chlorine. Calculate Kc for the equilibrium at 500K.

PCl5(g) \leftrightarrow PCl3(g) + Cl2(g)

So I'd work out the initial moles and moles made/used and this is where I get stuck (sadly).

- - - - - - - - PCl5 - - - PCl3 - - - - Cl2
Initial moles - 2.00 - - - ? - - - - - -1.2
Made/ used - - ? - - - - ? - - - - - - ?
Eq moles - - - -? - - - - ? - -- - - - ?
Eq conc - - - - 20 - - - -20 - - - - - 20

I know how to write out the equation after that, just not how to get all the constituents.

Help/guiding most appreciated. :smile:

work out the concentrations AT equilibrium

if there are 1.2 moles of cholorine there are also 1.2 moles of PCl3

how many moles of PCl5 must have reacted?
How many are left?

suss this out and you are on the way

Reply 2

Holsy

OP

10

charco

work out the concentrations AT equilibrium

if there are 1.2 moles of cholorine there are also 1.2 moles of PCl3

how many moles of PCl5 must have reacted?
How many are left?

suss this out and you are on the way

Thanks for the reply, Charco.

Okay - so if there are 1.2 moles of chlorine there 1.2 moles of PCl3 at equilibirum.

So would there also be 1.2 moles of PCl5 at equilbibrum and 0.8 has reacted?

Reply 3

modini

15

Holsy

Thanks for the reply, Charco.

Okay - so if there are 1.2 moles of chlorine there 1.2 moles of PCl3 at equilibirum.

So would there also be 1.2 moles of PCl5 at equilbibrum and 0.8 has reacted?

No, there would be 0.8 moles of PCl5 at equilibrium as 1.2 moles have become CL2 and PCl3 :smile:

EDIT: This should look more like this:-

- - - - - - - - PCl5 - - - PCl3 - - - - Cl2
Initial moles - 2.00 - - - 0 - - - - - 0
used - - - - - 1.2 - - - - 0 - - - - - - 0
made - - - - - 0 - - - - - 1.2 - - - - - 1.2
Eq moles - - - 0.8 - - - - 1.2 - -- - - - 1.2
Eq conc - - - - 0.8/20 - - - -1.2/20 - - - - - 1.2/20

Hope this helps :biggrin:

Reply 4

GoCh95

1

What I'm a bit confused about is how do you start with 2 moles but then end up with 3.2 moles in total at equilibrium?

Reply 5

Pigster

20

Original post by GoCh95

What I'm a bit confused about is how do you start with 2 moles but then end up with 3.2 moles in total at equilibrium?

Wowsers, you've resurrected an 11 year old thread to ask that? Surely no one who was here 11 years ago would still be here, never mind still interested in helping people with their chemistry. Well, if there was anyone like that, they must be well old by now.

Anyhoo. How about this for a comparison. CaCO3 + heat -> CaO + CO2. Surely 1 mol of CaCO3 can make 2 mol of products (1 of CaO and 1 of CO2).

So 2 mol making 3.2 isn't too much of a stretch.

Kc Calculations

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