Equilibrium

[Studyinghard]

PCl3 + Cl2 -----> PCl5

2 mol of PCl5 is added to an empty 10 L vessel at 200 deg.
When equilibrium is reached there is 1.7mol PCl5 left. calculate the equilibrium constant at 200 deg

[kakar0t]

k=[PCl3][CL2]/[PCl5]

[PCL5]=1.7/10= 0.17 M
[CL2]= 0.3mol/10L = 0.03M
[PCl3]= 0.03M

I wrote the k like that because since the reaction starts at the right side, the left side will give products

etc.

[fady_22]

k=[PCl3][CL2]/[PCl5]

[PCL5]=1.7/10= 0.17 M
[CL2]= 0.3mol/10L = 0.03M
[PCl3]= 0.03M

I wrote the k like that because since the reaction starts at the right side, the left side will give products

etc.

The working out for the concentrations is correct, however K should not be written that way. The product is supposed to be PCl5, as the equation is set out. I can see your reasoning, but you want to find the equilibrium constant for the equation that is specified.

[Studyinghard]

So i do 0.17/(0.03)2 = 188.89

?

[98.40_for_sure]

k=[PCl3][CL2]/[PCl5]

[PCL5]=1.7/10= 0.17 M
[CL2]= 0.3mol/10L = 0.03M
[PCl3]= 0.03M

I wrote the k like that because since the reaction starts at the right side, the left side will give products

etc.

The working out for the concentrations is correct, however K should not be written that way. The product is supposed to be PCl5, as the equation is set out. I can see your reasoning, but you want to find the equilibrium constant for the equation that is specified.

can't you k^-1 to find the reverse equation?

[fady_22]

k=[PCl3][CL2]/[PCl5]

[PCL5]=1.7/10= 0.17 M
[CL2]= 0.3mol/10L = 0.03M
[PCl3]= 0.03M

I wrote the k like that because since the reaction starts at the right side, the left side will give products

etc.

The working out for the concentrations is correct, however K should not be written that way. The product is supposed to be PCl5, as the equation is set out. I can see your reasoning, but you want to find the equilibrium constant for the equation that is specified.

can't you k^-1 to find the reverse equation?

Yep. But its not necessary if you can find K directly.

[fady_22]

So i do 0.17/(0.03)2 = 188.89

?

Yep, that looks right.

[Studyinghard]

In the reaction 2SO3(g) -----> 2SO2(g) + O2

0.10 mol SO3 is added to a 1L vessel. At equilibrium, there is 0.08 mol of the SO3 remainin. Calculate value of K

So will it be
K = (0.02)^2  x (0.01)
     ------------------
            0.08^2

[Russ]

^^
yes, looks right

[98.40_for_sure]

Yeah, thats correct