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Question on Kc/Kp!

Could somebody please explain this to me:

1) 2NO(g) (irreversible sign) N_2 (g) + O_2 (g)

At 1600c and 1.5atm pressure NO is 99% dissociated at equilibrium.

The moles at the beginning are :
NO:1
N_2:0
O_2:0

The moles at equilibrium are:
NO: 0.01
N_2: 0.005
O_2: 0.005

I’m wondering where the 0.01 and 0.005 comes from.

2) PCl_5 (g) (irreversible sign) PCl_3 (g) +Cl_2 (g)

At a given temperature T a sample of phosphorous pentachloride is 40% dissociated, the total equilibrium pressure being 2atm.

At start the moles are:
PCl_5: 1
PCl_3: 0
Cl_2: 0

At equilibrium the moles are:
PCl_5: 0.80
PCl_3: 0.40
Cl_2: 0.40

Why 0.80 and 0.40 and then why are they each divided by 1.40 for the mole fraction, and not 1???
Reply 1
Could somebody please explain this to me:

1) 2NO(g) (irreversible sign) N_2 (g) + O_2 (g)

At 1600c and 1.5atm pressure NO is 99% dissociated at equilibrium.

The moles at the beginning are :
NO:1
N_2:0
O_2:0

The moles at equilibrium are:
NO: 0.01
N_2: 0.005
O_2: 0.005

I’m wondering where the 0.01 and 0.005 comes from.

2) PCl_5 (g) (irreversible sign) PCl_3 (g) +Cl_2 (g)

At a given temperature T a sample of phosphorous pentachloride is 40% dissociated, the total equilibrium pressure being 2atm.

At start the moles are:
PCl_5: 1
PCl_3: 0
Cl_2: 0

At equilibrium the moles are:
PCl_5: 0.80
PCl_3: 0.40
Cl_2: 0.40

Why 0.80 and 0.40 and then why are they each divided by 1.40 for the mole fraction, and not 1???
Reply 2
First question:

The beginning moles of N2 and O2 have to be zero as they are the products.

Second question:

When 40% dissociates there should only be 60% left, hence 0.60, and not 0.80.

Because the steiochiometric equation is equal, the amount reacted for the products of this reaction should be +0.60 moles. And not 0.40 moles.
Reply 3
First question:

The beginning moles of N2 and O2 have to be zero as they are the products.

Second question:

When 40% dissociates there should only be 60% left, hence 0.60, and not 0.80.

Because the steiochiometric equation is equal, the amount reacted for the products of this reaction should be +0.60 moles. And not 0.40 moles.
ahh i just love kc and kp so beautiful ok... let me see here. Well basically you are told 99 % of NO is dissociated correct ? now what is 99 % of 1 mol 1.0 * 099 = 0.99 <---- This is the amount reacted hence the amount reacted at eqm is 1-0.99 which is 0.01 simple. Now you need to look at the ratio of NO to n2 and o2 there are two lots of no and 1 lot of n2 and o2 so as 99 % of no was dissociated it means half of 99 i.e. 49.5 % of n2 and o2 were dissociated because of the 2:1:1 ratio (divide by 2) when you take 49.5 % of n2 and o2 you get 0.99 of n2 and o2 so your moles at eqm are wrong (2.0 * 0.495) does this make sense ? i hope i have made it clear. im PREtTY sure i got this right. I take it you know how to do the remainder of the question. Ok now q2 for one thing 40 % of 1.0 is not 0.80 its 0.40 or 0.4 so basically 0.60 at eqm (who provided you with this info they were wrong!!) your values for pcl3 and cl2 look fine at eqm. Now you need to remember that the mole fraction if the no.of moles of the element divided by the total number of moles the total number of moles here is 0.60 + 0.60 +0.60 = 1.8 (remember in all of em 40% dissociated as there is a 1:1:1 ratio) so you should be dividing by 1.8 not 1.40. Again im pretty sure im right..so i hope this helps good luck
ahh i just love kc and kp so beautiful ok... let me see here. Well basically you are told 99 % of NO is dissociated correct ? now what is 99 % of 1 mol 1.0 * 099 = 0.99 <---- This is the amount reacted hence the amount reacted at eqm is 1-0.99 which is 0.01 simple. Now you need to look at the ratio of NO to n2 and o2 there are two lots of no and 1 lot of n2 and o2 so as 99 % of no was dissociated it means half of 99 i.e. 49.5 % of n2 and o2 were dissociated because of the 2:1:1 ratio (divide by 2) when you take 49.5 % of n2 and o2 you get 0.99 of n2 and o2 so your moles at eqm are wrong (2.0 * 0.495) does this make sense ? i hope i have made it clear. im PREtTY sure i got this right. I take it you know how to do the remainder of the question. Ok now q2 for one thing 40 % of 1.0 is not 0.80 its 0.40 or 0.4 so basically 0.60 at eqm (who provided you with this info they were wrong!!) your values for pcl3 and cl2 look fine at eqm. Now you need to remember that the mole fraction if the no.of moles of the element divided by the total number of moles the total number of moles here is 0.60 + 0.60 +0.60 = 1.8 (remember in all of em 40% dissociated as there is a 1:1:1 ratio) so you should be dividing by 1.8 not 1.40. Again im pretty sure im right..so i hope this helps good luck
turtle2
Could somebody please explain this to me:

1) 2NO(g) (irreversible sign) N_2 (g) + O_2 (g)

At 1600c and 1.5atm pressure NO is 99% dissociated at equilibrium.

The moles at the beginning are :
NO:1
N_2:0
O_2:0

The moles at equilibrium are:
NO: 0.01
N_2: 0.005
O_2: 0.005

I’m wondering where the 0.01 and 0.005 comes from.

99% dissociated => 1% of NO remaining
Mole ratios for NO:N2 and NO:O2

Therefore at equilibrium, mole values are;
NO: 1 - 0.99 = 0.01
N2: 0 + 0.5(0.99) = 0.495
O2: 0 + 0.5(0.99) = 0.495
turtle2
Could somebody please explain this to me:

1) 2NO(g) (irreversible sign) N_2 (g) + O_2 (g)

At 1600c and 1.5atm pressure NO is 99% dissociated at equilibrium.

The moles at the beginning are :
NO:1
N_2:0
O_2:0

The moles at equilibrium are:
NO: 0.01
N_2: 0.005
O_2: 0.005

I’m wondering where the 0.01 and 0.005 comes from.

99% dissociated => 1% of NO remaining
Mole ratios for NO:N2 and NO:O2

Therefore at equilibrium, mole values are;
NO: 1 - 0.99 = 0.01
N2: 0 + 0.5(0.99) = 0.495
O2: 0 + 0.5(0.99) = 0.495
Reply 8
Thanks!!
Reply 9
Thanks!!
your welcome glad to be of help :smile:
your welcome glad to be of help :smile: