achiever test chem 43

[pistolpete007]

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)

i got to 2CO2 -><- 2CO + O2
1.5m - -
-2X +2X +X
1.5-2X +2X +X

IM LOST FROM HERE

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[sl2obel2ts]

2CO2 <-> 2CO + O2
1.5M 0 0 <-initial
-2x +2x x

so K = (2CO)^2(O2)/(1.5-2x)^2
since it dissociated 20%, that leaves ur initial CO2 1.5M to 1.2M (20% reduction). that makes X = 0.15.

K = .3^2*.15/1.2^2

 

[osimsDDS]

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)

i got to 2CO2 -><- 2CO + O2
1.5m - -
-2X +2X +X
1.5-2X +2X +X

IM LOST FROM HERE

You should just try to use straight up logic for this it works better than using formulas and crap...so you have CO2 that dissociated only 20% so that means 80% of it didnt dissociate...so try and find the molarity for each first...

CO2 = (3moles)x(80%)= 2.4 moles and then divide by 2 liters to find molarity which will = 1.2M

CO = (3moles)x(20%) = .6 moles and then divide by 2 liters to find molarity which will = .3M

O2 = (3moles) x(20%) = .6 moles but divide by 2moles because in the equation it only gives 1 mole compared to the other 2...now you will get .3 moles and to find molarity again divide by 2 liters and you will get .15M

So now use the equation Keq=products/reactants

Keq= [O2][CO]^2/[CO2]^2....plug in the correct molarity and your done