# Equilibria QuestionsWatch

#1
A 36.8g sample of N2O4 was heated in a closed flask of volume 16.0 dm3. An equilibrium was established at a constant temperature according to the following equation.

N2O4 (g) â‡‹ 2NO2 (g)

The equilibrium mixture was found to contain 0.180 mol of N2O4.

i) Calculate the number of moles of N2O4 in the 36.8g sample.

ii) Calculate the number of moles of NO2 in the equilibrium mixture.
0
4 years ago
#2
(Original post by kayleigh.williams)
A 36.8g sample of N2O4 was heated in a closed flask of volume 16.0 dm3. An equilibrium was established at a constant temperature according to the following equation.

N2O4 (g) â‡‹ 2NO2 (g)

The equilibrium mixture was found to contain 0.180 mol of N2O4.

i) Calculate the number of moles of N2O4 in the 36.8g sample.

ii) Calculate the number of moles of NO2 in the equilibrium mixture.
What have you done so far? If you don't know where to start, you might want to create a table of the number of moles of N2O4 and NO2 before and at equilibrium.
0
#3
(Original post by Chlorophile)
What have you done so far? If you don't know where to start, you might want to create a table of the number of moles of N2O4 and NO2 before and at equilibrium.
Here is the whole question and what i have done, does it sound right?

b) A 36.8g sample of N2O4 was heated in a closed flask of volume 16.0 dm3. An equilibrium was established at a constant temperature according to the following equation.

N2O4 (g) â‡‹ 2NO2 (g)

The equilibrium mixture was found to contain 0.180 mol of N2O4.

i) Calculate the number of moles of N2O4 in the 36.8g sample.

Moles = Mass / Mr
= 36.8 / (14 + 14 + 16 + 16 + 16 + 16)
= 0.4

ii) Calculate the number of moles of NO2 in the equilibrium mixture.
Mole ratio N2O4 : NO2 = 1 : 2
Moles of NO2 = 2 x 0.4
= 0.8

iii) Write an expression for Kc and calculate its value under these conditions.

Expression for Kc:
Kc = [NO2]2 / [N2O4]

Calculation:
Concentration of NO2 = Mass / Volume
= 36.8 / 16
= 2.3 mol dm-3

Kc = 2.32 / 0.18
= 29.39 mol dm-3
0
4 years ago
#4
(Original post by kayleigh.williams)
ii) Calculate the number of moles of NO2 in the equilibrium mixture.
Mole ratio N2O4 : NO2 = 1 : 2
Moles of NO2 = 2 x 0.4
= 0.8
This is not right. The reaction started out with 0.4mol N2O4 and 0mol NO2, and we know that it finished (ie found an equilibrium) with 0.180mol N2O4. This means that 0.22mol N2O4 reacted, and therefore formed 0.22*2 = 0.44mol NO2.

You may also want to check your formula for calculating concentration.

(From a quick calculation I found Kc to be 0.0672moldm-3, if you have another attempt and want to check then it should be around this figure)
0
4 years ago
#5
(Original post by kayleigh.williams)
Here is the whole question and what i have done, does it sound right?

b) A 36.8g sample of N2O4 was heated in a closed flask of volume 16.0 dm3. An equilibrium was established at a constant temperature according to the following equation.

N2O4 (g) â‡‹ 2NO2 (g)

The equilibrium mixture was found to contain 0.180 mol of N2O4.

i) Calculate the number of moles of N2O4 in the 36.8g sample.

Moles = Mass / Mr
= 36.8 / (14 + 14 + 16 + 16 + 16 + 16)
= 0.4

ii) Calculate the number of moles of NO2 in the equilibrium mixture.
Mole ratio N2O4 : NO2 = 1 : 2
Moles of NO2 = 2 x 0.4
= 0.8

iii) Write an expression for Kc and calculate its value under these conditions.

Expression for Kc:
Kc = [NO2]2 / [N2O4]

Calculation:
Concentration of NO2 = Mass / Volume
= 36.8 / 16
= 2.3 mol dm-3

Kc = 2.32 / 0.18
= 29.39 mol dm-3
Part i is correct, part ii isn't. The formula does not say that the N2O2:NO2 ratio is 1:2, it simply means that 1 mole of N2O4 forms 2 moles of NO2. To work out the number of moles of NO2 formed, you firstly need to work out how many moles of N2O4 actually decomposed and use THAT figure to work out the number of moles of NO2 formed.
1
#6
(Original post by Gaiaphage)
This is not right. The reaction started out with 0.4mol N2O4 and 0mol NO2, and we know that it finished (ie found an equilibrium) with 0.180mol N2O4. This means that 0.22mol N2O4 reacted, and therefore formed 0.22*2 = 0.44mol NO2.

You may also want to check your formula for calculating concentration.

(From a quick calculation I found Kc to be 0.0672moldm-3, if you have another attempt and want to check then it should be around this figure)
Does this sound more reasonable? =D

A 36.8g sample of N2O4 was heated in a closed flask of volume 16.0 dm3. An equilibrium was established at a constant temperature according to the following equation.

N2O4 (g) â‡‹ 2NO2 (g)

The equilibrium mixture was found to contain 0.180 mol of N2O4.

i) Calculate the number of moles of N2O4 in the 36.8g sample.

Moles N2O4 = Mass / Mr
= 36.8 / (14 + 14 + 16 + 16 + 16 + 16)
= 0.4
At start of reaction:
Moles of N2O4 = 0.4
Moles of NO2 = 0

At equilibrium:
Moles of N2O4 = 0.4 - 0.180
= 0.22

ii) Calculate the number of moles of NO2 in the equilibrium mixture.
Mole ratio N2O4 : NO2 = 1 : 2
Moles of NO2 = 2 x 0.22
= 0.44

iii) Write an expression for Kc and calculate its value under these conditions.

Expression for Kc:
Kc = [NO2]2 / [N2O4]

Calculation:
Concentration of NO2 = Moles / Volume
= 0.44 / 16
= 0.0275 mol dm-3

Concentration of N2O4 = Moles / Volume
= 0.22 / 16
= 0.0275 mol dm-3
Kc = 0.02752 / 0.01375
= 0.055 mol dm-3
0
4 years ago
#7
(Original post by kayleigh.williams)
Does this sound more reasonable? =D

A 36.8g sample of N2O4 was heated in a closed flask of volume 16.0 dm3. An equilibrium was established at a constant temperature according to the following equation.

N2O4 (g) â‡‹ 2NO2 (g)

The equilibrium mixture was found to contain 0.180 mol of N2O4.

i) Calculate the number of moles of N2O4 in the 36.8g sample.

Moles N2O4 = Mass / Mr
= 36.8 / (14 + 14 + 16 + 16 + 16 + 16)
= 0.4
At start of reaction:
Moles of N2O4 = 0.4
Moles of NO2 = 0

At equilibrium:
Moles of N2O4 = 0.4 - 0.180
= 0.22

ii) Calculate the number of moles of NO2 in the equilibrium mixture.
Mole ratio N2O4 : NO2 = 1 : 2
Moles of NO2 = 2 x 0.22
= 0.44

iii) Write an expression for Kc and calculate its value under these conditions.

Expression for Kc:
Kc = [NO2]2 / [N2O4]

Calculation:
Concentration of NO2 = Moles / Volume
= 0.44 / 16
= 0.0275 mol dm-3

Concentration of N2O4 = Moles / Volume
= 0.22 / 16
= 0.0275 mol dm-3
Kc = 0.02752 / 0.01375
= 0.055 mol dm-3
Very close!

It says in the question that at equilibrium there are 0.18 moles of N2O4 - you've said there are 0.22 moles! What I meant was that 0.22 moles of N2O4 had been lost from the original 0.4 moles (0.4 - 0.18 = 0.22), and therefore this has reacted to form 0.44 moles of NO2.

Well done on the concentration formula - if you use 0.18 instead you should get it right.

You gave the Kc formula correctly, but didn't use it properly! Remember to square the concentration of NO2.
0
#8
(Original post by Gaiaphage)
Very close!

It says in the question that at equilibrium there are 0.18 moles of N2O4 - you've said there are 0.22 moles! What I meant was that 0.22 moles of N2O4 had been lost from the original 0.4 moles (0.4 - 0.18 = 0.22), and therefore this has reacted to form 0.44 moles of NO2.

Well done on the concentration formula - if you use 0.18 instead you should get it right.

You gave the Kc formula correctly, but didn't use it properly! Remember to square the concentration of NO2.
So for concentration of N2O4 i use 0.18 instead of 0.22 moles? Therefore concentration = 0.18 / 16 = 0.01125 mol dm-3?

And kc = (0.01125)2 / 0.01375
= 9.2 x 10-3

Ugh, it's been a long time since i did any of these calculations
0
4 years ago
#9
(Original post by kayleigh.williams)
So for concentration of N2O4 i use 0.18 instead of 0.22 moles? Therefore concentration = 0.18 / 16 = 0.01125 mol dm-3?

And kc = (0.01125)2 / 0.01375
= 9.2 x 10-3

Ugh, it's been a long time since i did any of these calculations
You're still not following your own formula!

(Original post by kayleigh.williams)
iii) Write an expression for Kc and calculate its value under these conditions.

Expression for Kc:
Kc = [NO2]2 / [N2O4]
Your concentration of NO2 is 0.44/16. You square this, then divide by the concentration of N2O4, which you just worked out as 0.18/16

Basically, you keep getting NO2 and N2O4 mixed up, so make sure you keep track of which is which and you'll get it right easily
1
#10
(Original post by Gaiaphage)
You're still not following your own formula!

Your concentration of NO2 is 0.44/16. You square this, then divide by the concentration of N2O4, which you just worked out as 0.18/16

Basically, you keep getting NO2 and N2O4 mixed up, so make sure you keep track of which is which and you'll get it right easily
Aha! I see where i went wrong now...

Calculation:
Concentration of NO2 = Moles / Volume
= 0.44 / 16
= 0.0275 mol dm-3

Concentration of N2O4 = Moles / Volume
= 0.18 / 16
= 0.01125 mol dm-3
Kc = 0.0275^2 / 0.01125
= 0.0672 mol dm-3

Thank you so much!! =D Now i can go to bed haha =D
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