The Student Room Group

I need help on an AS Level Paper 1 Chemistry question...

It's from winter 2013, P11, CIE As Level question 9:

9) Nitrogen reacts with hydrogen to produce ammonia.

N2(g) + 3H2(g) <=> 2NH3(g)

A mixture of 2.00 mol of nitrogen, 6.00 mol of hydrogen and 2.40 mol of ammonia is allowed to come to reach equilibrium in a sealed vessel of volume 1dm3 under certain conditions. It was found that 2.32 mol of nitrogen were present at equilibrium.

What is the value of Kc under these conditions?

A) (1.76)^2/(2.32)(6.96)^3

B) (1.76)^2/(2.32)(6.32)^3

C) (2.08)^2/(2.32)(6.96)^3

D) (2.40)^2/(2.32)(6.00)^3

The answer is A but I don't see how an extra mol can be made if the container is sealed (the initial starting mols were 10.4 but A's is 11.04...)? If someone could explain, I'd be grateful! :smile:

Scroll to see replies

Original post by Giulia Z
It's from winter 2013, P11, CIE As Level question 9:

9) Nitrogen reacts with hydrogen to produce ammonia.

N2(g) + 3H2(g) <=> 2NH3(g)

A mixture of 2.00 mol of nitrogen, 6.00 mol of hydrogen and 2.40 mol of ammonia is allowed to come to reach equilibrium in a sealed vessel of volume 1dm3 under certain conditions. It was found that 2.32 mol of nitrogen were present at equilibrium.

What is the value of Kc under these conditions?

A) (1.76)^2/(2.32)(6.96)^3

B) (1.76)^2/(2.32)(6.32)^3

C) (2.08)^2/(2.32)(6.96)^3

D) (2.40)^2/(2.32)(6.00)^3

The answer is A but I don't see how an extra mol can be made if the container is sealed (the initial starting mols were 10.4 but A's is 11.04...)? If someone could explain, I'd be grateful! :smile:


Heya, I'm just going to pop this into the Chemistry forum for you as you're likely to get a response there. :smile:
Reply 2
The total number of moles can change, since there are four on the left and two on the right of the equation.

Do an ICE diagram. You know three of the starting amounts and one of the equilibrium amounts. How much has that one changed by? How much have the other two changed by? Voila.
Reply 3
The total number of moles can change, since there are four on the left and two on the right of the equation.

Do an ICE diagram. You know three of the starting amounts and one of the equilibrium amounts. How much has that one changed by? How much have the other two changed by? Voila.
Reply 4
Original post by Puddles the Monkey
Heya, I'm just going to pop this into the Chemistry forum for you as you're likely to get a response there. :smile:


Thanks! :biggrin:
Reply 5
Original post by Pigster
The total number of moles can change, since there are four on the left and two on the right of the equation.

Do an ICE diagram. You know three of the starting amounts and one of the equilibrium amounts. How much has that one changed by? How much have the other two changed by? Voila.


I tried that way but I must have done something wrong because I didn't get A:

N2 + 3H2 <=> 2NH3

Initial Mols: 2.00 6.00 2.40

Equilibrium: 2.32 6.00 + x 2.40 - 2x

So x= - 0.32, then that means the equilibrium mols for H2 is 6.32 and NH3 is 1.76, which is B?
Original post by Giulia Z
I tried that way but I must have done something wrong because I didn't get A:

N2 + 3H2 <=> 2NH3

Initial Mols: 2.00 6.00 2.40

Equilibrium: 2.32 6.00 + x 2.40 - 2x

So x= - 0.32, then that means the equilibrium mols for H2 is 6.32 and NH3 is 1.76, which is B?


The equilibrium moves to the LHS.

You are correct that x = 0.32, which means that hydrogen in creases by 3 x 0.32 = 0.96, so the equilibrium moles of hydrogen = 6.96

So the answer is A
Reply 7
Original post by charco
The equilibrium moves to the LHS.

You are correct that x = 0.32, which means that hydrogen in creases by 3 x 0.32 = 0.96, so the equilibrium moles of hydrogen = 6.96

So the answer is A


Oh I see. What a rookie mistake! And sorry, the formatting of my answer (ie the lining up of the equation and mols), it was changed when it was posted.
Reply 8
Hi again! I was wondering if I could have some more help with the Summer 11, QP11 question 9?

9) 50 cm3 of 2.50 mol dm–3 hydrochloric acid was placed in a polystyrene beaker of negligible heat
capacity. Its temperature was recorded and then 50 cm3 of 2.50 mol dm–3 NaOH at the same
temperature was quickly added, with stirring. The temperature rose by 17 °C.

The resulting solution may be considered to have a specific heat capacity of 4.2 J g–1 K–1.

What is an approximate value for the molar enthalpy change of neutralisation of hydrochloric acid
and sodium hydroxide from this experiment?

A (5.2x050.0)/(17x2.4x50) J mol-1

B (5.2x10.0)/(17x2.4x50) J mol -1

C (5.2x050.0)/(17x2.4x100) J mol -1

D (5.2x50)/(17x2.4x100) J mol -1

The answer is C though I don't understand seeing as the mols of the two reactants need to be added together?
Reply 9
Why do you want to add numbers of moles? Do you mean that because you mixed 125 mmoles of NaOH and 125 mmoles of HCl there were 250 mmoles neutralized? That's wrong, enthalpy refers to the reaction, written as

H+ + OH- -> H2O

when 1 mole reacts with 1 mole it means there was 1 mole of "reaction as written".
Reply 10
Original post by Borek
Why do you want to add numbers of moles? Do you mean that because you mixed 125 mmoles of NaOH and 125 mmoles of HCl there were 250 mmoles neutralized? That's wrong, enthalpy refers to the reaction, written as

H+ + OH- -> H2O

when 1 mole reacts with 1 mole it means there was 1 mole of "reaction as written".


Oh thanks that totally slipped my mind (as usual). I'm generally very frazzled when it comes to Chemistry. Sorry for the late reply, just threw myself at my statistics exam and now, P1 on Tuesday then I am finished! I was wondering if I could have a little bit more help?

10) Tanzanite is used as a gemstone for jewellery. It is a hydrated calcium aluminium silicate mineral
with a chemical formula Ca2Al xSiyO12(OH).6½H2O. Tanzanite has Mr of 571.5.

Its chemical composition is 14.04 % calcium, 14.17 % aluminium, 14.75 % silicon, 54.59 % oxygen
and 2.45 % hydrogen.

(Ar values: H = 1.0, O = 16.0, Al = 27.0, Si = 28.1, Ca = 40.1)

What are the values of x and y?

A) x = 1, y = 1
B) x = 3, y = 3
C) x = 3, y = 3
D) x = 6, y = 1

How would you go about solving this?

Oh and this one if you could be so kind as well:

11) 0.144 g of an aluminium compound X react with an excess of water, to produce a gas. This gas burns completely in O2 to form H2O and 72 cm3 of CO2 only. The volume of CO2 was measured at room temperature and pressure.

What could be the formula of X?
[C = 12.0, Al = 27.0; 1 mole of any gas occupies 24 dm3 at room temperature and pressure]


A) Al2C3
B) Al3C4
C) Al4C3
D) Al5C3
Reply 11
Many ways of solving the first one. For example - if 14.17% of the mass of a compound with molar mass of 571.5 g/mol is Al, how many moles of Al per mole of that compound?

For the second one - how many moles of carbon dioxide? How many moles of carbon in 0.144 g of the compound? Assuming it is C3 - what mass left for Al? Does it yield reasonable number of moles of Al (reasonable meaning in a correct ratio to carbon)?
Original post by Giulia Z
Oh thanks that totally slipped my mind (as usual). I'm generally very frazzled when it comes to Chemistry. Sorry for the late reply, just threw myself at my statistics exam and now, P1 on Tuesday then I am finished! I was wondering if I could have a little bit more help?

10) Tanzanite is used as a gemstone for jewellery. It is a hydrated calcium aluminium silicate mineral
with a chemical formula Ca2Al xSiyO12(OH).6½H2O. Tanzanite has Mr of 571.5.

Its chemical composition is 14.04 % calcium, 14.17 % aluminium, 14.75 % silicon, 54.59 % oxygen
and 2.45 % hydrogen.

(Ar values: H = 1.0, O = 16.0, Al = 27.0, Si = 28.1, Ca = 40.1)

What are the values of x and y?

A) x = 1, y = 1
B) x = 3, y = 3
C) x = 3, y = 3
D) x = 6, y = 1

How would you go about solving this?

Oh and this one if you could be so kind as well:

11) 0.144 g of an aluminium compound X react with an excess of water, to produce a gas. This gas burns completely in O2 to form H2O and 72 cm3 of CO2 only. The volume of CO2 was measured at room temperature and pressure.

What could be the formula of X?
[C = 12.0, Al = 27.0; 1 mole of any gas occupies 24 dm3 at room temperature and pressure]


A) Al2C3
B) Al3C4
C) Al4C3
D) Al5C3


1. Divide the percentage values by the relative mass of each element to get a mole ratio for the compound.

2. Use the values for carbon dioxide to find the mass of carbon and from this the moles of carbon.
Subtract the mass of carbon from the mass of the compound to get the mass of aluminium and then the moles of aluminium.
You now have a ratio of moles carbon to moles aluminium.
Reply 13
Thanks for help on the first one! Could you solve the second one for me so I can see? I can't figure it out myself.
Original post by Giulia Z
Thanks for help on the first one! Could you solve the second one for me so I can see? I can't figure it out myself.


Follow the steps I have outlined.

If you can't understand a specific step tell me ...
Reply 15
Alright, finding the mols of CO2 and then finding the mass of carbon from it. I'm doing it this way:

73cm3 = 0.073 dm3 so 0.073/24 = 0.0030417 mols of CO2.

mass/12+16*2 (the Ar of CO2) = 0.0030417 mols, which means the mass is 32.0365g.

Percentage composition of carbon in CO2 is 12/12+16*2 = 0.272727... and 0.2727 * 32.0365 = 8.737g of carbon. But I can't subtract the mass of carbon from the mass of the aluminium compound, seeing as it's only 0.144g. I tried the way where I didn't convert 73cm3 into dm3, just 73cm3 = 73g but that didn't work either...
Original post by Giulia Z
Alright, finding the mols of CO2 and then finding the mass of carbon from it. I'm doing it this way:

73cm3 = 0.073 dm3 so 0.073/24 = 0.0030417 mols of CO2.

mass/12+16*2 (the Ar of CO2) = 0.0030417 mols, which means the mass is 32.0365g. :frown:

Percentage composition of carbon in CO2 is 12/12+16*2 = 0.272727... and 0.2727 * 32.0365 = 8.737g of carbon. But I can't subtract the mass of carbon from the mass of the aluminium compound, seeing as it's only 0.144g. I tried the way where I didn't convert 73cm3 into dm3, just 73cm3 = 73g but that didn't work either...


73 or 72?

72cm3 = 0.072 dm3 so 0.072/24 = 0.0030 mols of CO2. :smile:
this equals 0.003 x 44g = 0.132g

of this 12/44 are carbon = 0.036g
Reply 17
Oh wow it's 72...silly me.
Would you say I'm overcomplicating things then? Let's get this straight:

1) Change the volume from cm3 to dm3 so you can divide by 24dm3 to get the mols of CO2.
2) Find the mass of the gas (mass/ar = no. of mols)
3) Multiply the proportion of carbon in CO2 with the mass thereby finding the mass of carbon
4) Subtracting the mass of the carbon from the mass of the aluminium compound then finds the mass of Al which you can work out the moles from
5) Find the mole ratio

Right?
Original post by Giulia Z
Oh wow it's 72...silly me.
Would you say I'm overcomplicating things then? Let's get this straight:

1) Change the volume from cm3 to dm3 so you can divide by 24dm3 to get the mols of CO2.
2) Find the mass of the gas (mass/ar = no. of mols)
3) Multiply the proportion of carbon in CO2 with the mass thereby finding the mass of carbon
4) Subtracting the mass of the carbon from the mass of the aluminium compound then finds the mass of Al which you can work out the moles from
5) Find the mole ratio

Right?


correct :smile:

... you can even take a shortcut by combining steps 2 and 3 as you know that one carbon dioxide molecule contains one carbon atom. You just multiply moles of carbon dioxide by 12 to get mass of carbon.
(edited 9 years ago)
Reply 19
Okay, I sat my paper today! I thought it went okay, I tried my best! Now I'm off exams *phew*

Thank you all so much for your help! :biggrin:

Quick Reply

Latest