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KC Questions
I cant seem to workout the following two questions, having tried many times, I keep getting different answers to what the answers are supposed to be:
1.) Consider the equilibrium: N2O4(g) == 2NO2(g).
1 mol of dinitrogen tetroxide, N2O4, was introduced into a vessel of volume 10 dm3. At equilibrium 50% had dissociated. Calculate Kc for the reaction.
ANSWER : 0.2 moldm-3
2.) In an experiment, 9.0 moles of nitrogen and 27 moles of hydrogen were placed into a vessel of volume 10 dm3 and allowed to reach equilibrium. It was found that two thirds of the nitrogen and hydrogen were converted into ammonia. Calculate Kc for the reaction.
N2(g) + 3H2(g) == 2NH3(g)
ANSWER : 6.6 mol-2dm6
Thanks
1.) Consider the equilibrium: N2O4(g) == 2NO2(g).
1 mol of dinitrogen tetroxide, N2O4, was introduced into a vessel of volume 10 dm3. At equilibrium 50% had dissociated. Calculate Kc for the reaction.
ANSWER : 0.2 moldm-3
2.) In an experiment, 9.0 moles of nitrogen and 27 moles of hydrogen were placed into a vessel of volume 10 dm3 and allowed to reach equilibrium. It was found that two thirds of the nitrogen and hydrogen were converted into ammonia. Calculate Kc for the reaction.
N2(g) + 3H2(g) == 2NH3(g)
ANSWER : 6.6 mol-2dm6
Thanks
Scroll to see replies
tired
I cant seem to workout the following two questions, having tried many times, I keep getting different answers to what the answers are supposed to be:
1.) Consider the equilibrium: N2O4(g) == 2NO2(g).
1 mol of dinitrogen tetroxide, N2O4, was introduced into a vessel of volume 10 dm3. At equilibrium 50% had dissociated. Calculate Kc for the reaction.
ANSWER : 0.2 moldm-3
2.) In an experiment, 9.0 moles of nitrogen and 27 moles of hydrogen were placed into a vessel of volume 10 dm3 and allowed to reach equilibrium. It was found that two thirds of the nitrogen and hydrogen were converted into ammonia. Calculate Kc for the reaction.
N2(g) + 3H2(g) == 2NH3(g)
ANSWER : 6.6 mol-2dm6
Thanks
1.) Consider the equilibrium: N2O4(g) == 2NO2(g).
1 mol of dinitrogen tetroxide, N2O4, was introduced into a vessel of volume 10 dm3. At equilibrium 50% had dissociated. Calculate Kc for the reaction.
ANSWER : 0.2 moldm-3
2.) In an experiment, 9.0 moles of nitrogen and 27 moles of hydrogen were placed into a vessel of volume 10 dm3 and allowed to reach equilibrium. It was found that two thirds of the nitrogen and hydrogen were converted into ammonia. Calculate Kc for the reaction.
N2(g) + 3H2(g) == 2NH3(g)
ANSWER : 6.6 mol-2dm6
Thanks
show us how far you've got then...
For example in question 1 do you understand what is meant by '50% has dissociated'?
tired
I cant seem to workout the following two questions, having tried many times, I keep getting different answers to what the answers are supposed to be:
1.) Consider the equilibrium: N2O4(g) == 2NO2(g).
1 mol of dinitrogen tetroxide, N2O4, was introduced into a vessel of volume 10 dm3. At equilibrium 50% had dissociated. Calculate Kc for the reaction.
1.) Consider the equilibrium: N2O4(g) == 2NO2(g).
1 mol of dinitrogen tetroxide, N2O4, was introduced into a vessel of volume 10 dm3. At equilibrium 50% had dissociated. Calculate Kc for the reaction.
Moles at start of N2O4: 1
Moles at equilibrium of N2O4: 0.5 (50% dissociated)
Moles at start of NO2: 0
Moles at equilibrium of NO2: 0.5
I'm sure you can take it from there.
tired
2.) In an experiment, 9.0 moles of nitrogen and 27 moles of hydrogen were placed into a vessel of volume 10 dm3 and allowed to reach equilibrium. It was found that two thirds of the nitrogen and hydrogen were converted into ammonia. Calculate Kc for the reaction.
N2(g) + 3H2(g) == 2NH3(g)
2.) In an experiment, 9.0 moles of nitrogen and 27 moles of hydrogen were placed into a vessel of volume 10 dm3 and allowed to reach equilibrium. It was found that two thirds of the nitrogen and hydrogen were converted into ammonia. Calculate Kc for the reaction.
N2(g) + 3H2(g) == 2NH3(g)
Moles of N2 at start: 9.0
Moles of H2 at start: 27
Moles of NH3 at start: 0
Moles of N2 at eq: 3.0
Moles of H2 at eq: 9.0
Moles of NH3 at eq: 12
This is because we two thirds of the N2 and H2 became NH3, so 6 moles of N2 and H2 became NH3. Then use their ratios.
Then find equation for Kc, and solve it (using concentration = moles/volume.)
Im sooo sorrrry, ive made the stupidest mistale, i cancelled out the volumes.....and didnt looks carefully that i cudnt do so in these 2 questions. I was wondering why I couldnt do these but could do most of the rest lol
Also by the way in the first question the equilibrium moles of N02 would be 1.0 as it is 2N02.
Thanks for the help anyways!
Also by the way in the first question the equilibrium moles of N02 would be 1.0 as it is 2N02.
Thanks for the help anyways!
Original post by charco
show us how far you've got then...
For example in question 1 do you understand what is meant by '50% has dissociated'?
For example in question 1 do you understand what is meant by '50% has dissociated'?
Does "50% dissociated" 50% will be left in solution or 50% produced as NO2? Why not the second option?
Original post by ps1265A
Does "50% dissociated" 50% will be left in solution or 50% produced as NO2? Why not the second option?
You are measuring with respect to the reactant. It means that 50% of the reactants changed.
Original post by charco
You are measuring with respect to the reactant. It means that 50% of the reactants changed.
Thanks!
Original post by charco
show us how far you've got then...
For example in question 1 do you understand what is meant by '50% has dissociated'?
For example in question 1 do you understand what is meant by '50% has dissociated'?
Why isn't it 24 mol of NH3 left at equilibrium? I followed the same working out as I did for all other equilibrium questions
Original post by ps1265A
Why isn't it 24 mol of NH3 left at equilibrium? I followed the same working out as I did for all other equilibrium questions
Read through the thread from the beginning ...
Original post by charco
Read through the thread from the beginning ...
"This is because we two thirds of the N2 and H2 became NH3, so 6 moles of N2 and H2 became NH3"
I understand why 6 moles of N2 became NH3, but why 6 moles of NH3?
Another Q: Why does the yield of oxygen in any reaction increase if we increase the size of the container, taking into account temperature and amount of starting reactants are the same?
Original post by ps1265A
"This is because we two thirds of the N2 and H2 became NH3, so 6 moles of N2 and H2 became NH3"
I understand why 6 moles of N2 became NH3, but why 6 moles of NH3?
Another Q: Why does the yield of oxygen in any reaction increase if we increase the size of the container, taking into account temperature and amount of starting reactants are the same?
I understand why 6 moles of N2 became NH3, but why 6 moles of NH3?
Another Q: Why does the yield of oxygen in any reaction increase if we increase the size of the container, taking into account temperature and amount of starting reactants are the same?
It comes down to the stoichiometry of the equation ...
Original post by charco
It comes down to the stoichiometry of the equation ...
I sort of understand it...
I'm confused about why even when the temperature is the same, and the reactants are the same, why does KC change?
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(edited 9 years ago)
Original post by charco
It comes down to the stoichiometry of the equation ...
Q2 is at the same T, why does KC change?
Posted from TSR Mobile
Original post by ps1265A
Been away... just got back.
Kc is referring to a specific equilibrium which is expressed as an equation. If the equation is changed then so does Kc. It is actually a different "Kc" and will be related to other Kc values for other representative equations by a simple proportion.
Remember that an equilibrium can also be expressed in reverse in which case the Kc value is the reciprocal.
Original post by charco
It comes down to the stoichiometry of the equation ...
for ammonia question start of reaction:
N2= 9 moles
3H2= 27 moles
2NH3=0 moles
at equilibrium 2/3 dissociated:
2/3 of N2 (9)= 3
2/3 of 3H2 (27)= 9
to work out 2NH3 at equilibrium: 9+3= 12 moles
calculate concentration: divide each of the answers by 10 dm3 to get N2 (0.3), 3H2 (0.9) and 2NH3 (1.2)
calculate Kc. products/reactants
(1.2)2/(0.3) (0.9)3 = 6.58 or 6.6 mol-2 dm6
N2= 9 moles
3H2= 27 moles
2NH3=0 moles
at equilibrium 2/3 dissociated:
2/3 of N2 (9)= 3
2/3 of 3H2 (27)= 9
to work out 2NH3 at equilibrium: 9+3= 12 moles
calculate concentration: divide each of the answers by 10 dm3 to get N2 (0.3), 3H2 (0.9) and 2NH3 (1.2)
calculate Kc. products/reactants
(1.2)2/(0.3) (0.9)3 = 6.58 or 6.6 mol-2 dm6
Original post by BlackRose09
for ammonia question start of reaction:
N2= 9 moles
3H2= 27 moles
2NH3=0 moles
at equilibrium 2/3 dissociated:
2/3 of N2 (9)= 3
2/3 of 3H2 (27)= 9
to work out 2NH3 at equilibrium: 9+3= 12 moles
calculate concentration: divide each of the answers by 10 dm3 to get N2 (0.3), 3H2 (0.9) and 2NH3 (1.2)
calculate Kc. products/reactants
(1.2)2/(0.3) (0.9)3 = 6.58 or 6.6 mol-2 dm6
N2= 9 moles
3H2= 27 moles
2NH3=0 moles
at equilibrium 2/3 dissociated:
2/3 of N2 (9)= 3
2/3 of 3H2 (27)= 9
to work out 2NH3 at equilibrium: 9+3= 12 moles
calculate concentration: divide each of the answers by 10 dm3 to get N2 (0.3), 3H2 (0.9) and 2NH3 (1.2)
calculate Kc. products/reactants
(1.2)2/(0.3) (0.9)3 = 6.58 or 6.6 mol-2 dm6
2/3 of 9 is 6 not 3??? same for 2/3 of 27 it is 18 not 9??
(edited 8 years ago)
Original post by bobwibbles
2/3 of 9 is 6 not 3??? same for 2/3 of 27 it is 18 not 6??
For that question it says if 2/3 dissociates meaning only 1/3 would be left at equilibrium
Original post by bobwibbles
2/3 of 9 is 6 not 3??? same for 2/3 of 27 it is 18 not 9??
N2
3H2
2NH3
Initial
9
27
0
Change
2/3 of 9= 6
2/3 of 27= 18
Equilibrium
9-6= 3
27-18= 9
9+3= 12
Conc: n/v
3/10= 0.3
9/10= 0.9
12/10= 1.2
i tried to simplify it by putting it in an ICE table... you can never go wrong with ICE tables.... i hope this makes it easier, let me know how it goes
(edited 8 years ago)
Original post by KaylaB
For that question it says if 2/3 dissociates meaning only 1/3 would be left at equilibrium
Oh yeh I know that.. Its just to me it looks like it was quoted that 2/3 of 9 is 3... Yes if 2/3 of 9 moles dissociates then 9 -6 = 3 moles left at equillibrium.. I understand now...
Trying to learn how to do these. Couldnt at AS... Im getting it kinda... Where to go for more questions and answers and even notes for this?
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