Gen chem equilibrium question

[Erhatstil]

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What is the equilibrium constant, K, at 400K if 3.00 moles of CO2(g) introduced into a 2 liter container are 20.0% dissociated at equilibrium?
2 CO2(g) « 2CO(g) + O2(g)


answer key says:

1. CO formed: (.2)(3.00moles)=.60
[CO]=.60/2= .3M
2. O2 formed: (.2)(3.00moles)/2=.30
[O2]=.30/2= .15M
3. CO2 (non-dissociated): (.8)(3.00)=2.4 m
[CO2]= 2.4/2= 1.2M
then this is plugged in to equilibrium constant equation but i cant figure out why the oxygen formation is divided by 2 after we take in to account 20% dissociation....

maybe i'm missing something really simple
HELP!

 

[MNova]

What is the equilibrium constant, K, at 400K if 3.00 moles of CO2(g) introduced into a 2 liter container are 20.0% dissociated at equilibrium?
2 CO2(g) « 2CO(g) + O2(g)


answer key says:

1. CO formed: (.2)(3.00moles)=.60
[CO]=.60/2= .3M
2. O2 formed: (.2)(3.00moles)/2=.30
[O2]=.30/2= .15M
3. CO2 (non-dissociated): (.8)(3.00)=2.4 m
[CO2]= 2.4/2= 1.2M
then this is plugged in to equilibrium constant equation but i cant figure out why the oxygen formation is divided by 2 after we take in to account 20% dissociation....

maybe i'm missing something really simple
HELP!

I could be wrong, but I believe the divided by 2 that you have highlighted in red takes into account the mole ratio (2CO2:2CO:O2). The second divided by two (on the subsequent line) takes into account the fact that the gas is introduced into a 2 liter container, and you want the quantities on a per liter basis.

But, if someone else can verify, that would be great 🙂

 

[nze82]

I think here's what you're missing:

2CO2<-->2CO + 1O2

Now let's look at the ICE table (Initial, Change, Equilibrium):

NOTE: In the following calculations, x = 3(0.2) = 0.6, because the question says that CO2 is 20% dissociated.

[CO2] = 3moles...[CO2]= 3-2x moles = 3 - 2(0.6) = 2.4mole...2.4/2L=1.2M

[CO] = 0moles...[CO] = 2x moles = 2(0.6) = 1.2mole...1.2/2L=0.6M

[O2] = 0moles...[O2] = x moles = 0.6mole...0.6/2L=0.3M

Now, you can plug in the "Green" numbers and calculate Keq.

Keq = [CO]^2 x [O2] / [CO2]^2 = (0.6)^2 X (0.3) / (1.2)^2 = 0.075

Hope this helps!

 

[Erhatstil]

according to achiever the answer is K= (0.30)^2(0.15)/(1.20)^2

i definitely think ice chart is the best way to solve this but your final [CO] and [O2] are different from the key which are [CO]=.3 and [O2]=.15

the [O2] division by 2 highlighted in red i still can't figure out either

 

[nze82]

according to achiever the answer is K= (0.30)^2(0.15)/(1.20)^2

i definitely think ice chart is the best way to solve this but your final [CO] and [O2] are different from the key which are [CO]=.3 and [O2]=.15

the [O2] division by 2 highlighted in red i still can't figure out either

Hmmm. Then I have no idea how to solve it
Sorry!
I hate achiever anyways!

 

[tdkyun]

There is no need to setup an ICE table

__2CO2 <-> 2CO + O2

__2.4mol__0.6mol 0.3mol

1. 3mol - (3mol*0.20) = 2.4 mols of CO2 left
2. 3mol * 0.20 = 0.6 mol of 2CO (b/c it's 1:1 ratio with CO2)
3. 3mol * 0.20 / 2 = 0.3 mol of O2 (b/c it's 2:1 ratio)

1) 2.4/2L = 1.2M = [CO2]
2) 0.6/2L = 0.3M = [CO]
3) 0.3/2L = 0.15M = [O2]

K=[CO]^2 * [O2] / [CO2]^2
so K = (0.30)^2(0.15)/(1.20)^2