# Chemistry Equilibrium Constant Question

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#1
Could someone help me on this question please...

If phosphorus pentachloride is heated in a sealed tube the following equilibrium is set up:

PCl5(g) ----> PCl3(g) + Cl2(g)

At a given temperature (T) a sample of phosphorus pentachloride is 40.0% dissociated, the total equilibrium pressure being 2.00 atm. Calculate the partial pressures of each of the components of the equilibrium mixture, to three significant figures. Hence calculate the value of Kp under these conditions.

So far I have worked out the mole ratio to be 0.6x (PCl5) and 0.2x each for both PCl3 and Cl2 (but I'm not sure if this is right). I don't know what to do from there.

Any help would be appreciated. Thanks in advance.
1
4 years ago
#2
(Original post by alde123)
Could someone help me on this question please...

If phosphorus pentachloride is heated in a sealed tube the following equilibrium is set up:

PCl5(g) ----> PCl3(g) + Cl2(g)

At a given temperature (T) a sample of phosphorus pentachloride is 40.0% dissociated, the total equilibrium pressure being 2.00 atm. Calculate the partial pressures of each of the components of the equilibrium mixture, to three significant figures. Hence calculate the value of Kp under these conditions.

So far I have worked out the mole ratio to be 0.6x (PCl5) and 0.2x each for both PCl3 and Cl2 (but I'm not sure if this is right). I don't know what to do from there.

Any help would be appreciated. Thanks in advance.
I think it should be 0.4 for cl2 and pcl3 and 0.6 for pcl5.
then work out mole fraction = moles of x/total moles so for example for pcl5= 0.6/1.4
then use the mole fractions to work out partial pressures= mole fraction x total pressure ( 2 in this case )
then use these partial pressure values in the kp equation = (partial pressure of cl2 x partiall pressure of pcl3)/ partial pressure of pcl5
0
#3
(Original post by Nightinwind)
I think it should be 0.4 for cl2 and pcl3 and 0.6 for pcl5.
then work out mole fraction = moles of x/total moles so for example for pcl5= 0.6/1.4
then use the mole fractions to work out partial pressures= mole fraction x total pressure ( 2 in this case )
then use these partial pressure values in the kp equation = (partial pressure of cl2 x partiall pressure of pcl3)/ partial pressure of pcl5

Thanks for replying. There is one thing that is confusing me - why is it 0.4 for Cl2 and PCl3 each if the total dissociation of the products is 40% - surely it should be evenly split between the two?
0
4 years ago
#4
(Original post by alde123)
Thanks for replying. There is one thing that is confusing me - why is it 0.4 for Cl2 and PCl3 each if the total dissociation of the products is 40% - surely it should be evenly split between the two?
as the ratio is 1 1 so if 0.4 of reactant dissociated it would produce the same amount of each product, if it was 0.2 the ratio would be 1:0.5:0.5
0
4 years ago
#5
in all honesty its not something i completely understand except from what i said above, its more just something i learnt ahaha
0
#6
(Original post by Nightinwind)
in all honesty its not something i completely understand except from what i said above, its more just something i learnt ahaha
Oh well, thanks anyway!
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