[candiishop]

1) In which one of the following reactions will an increase in total pressure cause an increase in the concentration of the products?

 a) H2(g) + Cl2(g) <---> 2HCl(g)

 b) 2SO2(g) + O2(g) <--> 2SO3(g)

 c) CaCO3(s) <--> CaO(s) + CO2(g)

 d) N2O4(g) <--> 2NO2(g)

 e) CO2(g) + H2(g) <--> CO(g) + H2O(g) [/b]

Is the answer D? Base on the the higher no of products over reactants?


2) When a chemical system is at equilibrium,

a) the concentrations of the reactants are equal to the concentrations of the products.
 
b) the concentrations of the reactants and products have reached constant values.
 
c) the forward and reverse reactions have stopped.
 
d) the reaction quotient, Q, has reached a maximum.
 
e) all of the above

I got D


3) A 7.30 L container had 64.50 moles of N2O4 added to it. When equilibrium was established, the concentration of N2O4 was found to be 2.35M. What is Kc for the following reaction
     N2O4(g) <---> 2NO2(g)  

a) 60.86 M
 
b) 82.34 M
 
c) 71.60 M
 
d) 5.520 M
 
e) 6.348 M
 
f) 4.692 M

I got that answer that doesn't matches the above.

M= n/v = 64.50mol/ 7.30L = 8.836M


4) For the equilibrium
H2(g) + CO2(g) <--->  H2O(g) + CO(g)

Kc = 6.00 at 1166K. If each of the four species was initially present at a concentration of 4.000M, calculate the equilibrium concentration of the CO(g) at this temperature.

a) 1.681
 
b) 5.681
 
c) 6.249
 
d) 2.319

e) 6.553

                      H2(g) + CO2(g) <--->  H2O(g) + CO(g)
Initial               4.000      4.000            4.000      4.000
change              -x            -x                 +x           +x
Equilibrium       4.000-x     4.000-x        4.000+x     4.000+x

Kc= [H2O] [CO2] / [H2] [CO2]
6 = (4.000 + x) ^2 / (4.000 - x) ^2

Any help would be greatly appreciated =)

[Will]

For Q1, if you increase the pressure, the system will try and reduce the pressure by shifting to the side with less moles of gas. There is only one of those equilibria which have less moles of gas on the right hand side and unfortunately it isn't D!

For Q2, I would have put a different answer (B) because when a system has reached equilibrium then the concentrations are constant, as the forward reaction and backward reaction occur at the same rate.

Q3 is quite long! You have to work out the moles of N₂O₄ that have become NO₂, by multiplying the concentration by the volume which gives 17.155 moles, so 47.345 moles of N₂O₄ have become NO₂. Then you multiply by 2, to give you the moles of NO₂ (1:2 ratio), which is 94.69.
Then find the concentration of this by dividing by 7.3, which gives 12.97123288...
And if you've written out the equation for K_c, then its [NO₂]² divided by [N₂O₄].
So you square the 12.97... to give 168.2528823...
Then as the equation says, you divide by the concentration of N₂O₄, which is 2.35, so you end up with 71.59697121...
Hence I think the answer is C. If there is a shorter way then somone please say!

You are on the right track for Q4! Just expand out the brackets and then solve to find x!

[candiishop]

Oh... thanku so much. So the answer to question 1 is B right?

[candiishop]

Also with question 3, how u got  17.155 moles, how did u get that? I seem to get 8.836M from 64.50mol/ 7.30L

[Will]

Oh... thanku so much. So the answer to question 1 is B right?

You're welcome , and yep, B is right.

Also with question 3, how u got 17.155 moles, how did u get that? I seem to get 8.836M from 64.50mol/ 7.30L

You do 2.35 (Concentration of N₂O₄ at equilibrium) multiplied by 7.3 (Volume) which will give you the moles of N₂O₄ at equilibrium (17.155), so the number of moles of N₂O₄ that have dissociated is 60-17.55, which is 42.845, and from that you can work out the concentration of NO₂.
Dividing 64.5 by 7.3 will give you the concentration of N₂O₄ at the beginning of the 'reaction', not when it has reached equilibrium. They give the concentration of N₂O₄ when the equilibrium has been reached, you only need to find the concentration of NO₂ when equilibrium has been reached. Hope that makes sense!

[edwinksl]

For Q1, if you increase the pressure, the system will try and reduce the pressure by shifting to the side with less moles of gas. There is only one of those equilibria which have less moles of gas on the right hand side and unfortunately it isn't D!

For Q2, I would have put a different answer (B) because when a system has reached equilibrium then the concentrations are constant, as the forward reaction and backward reaction occur at the same rate.

Q3 is quite long! You have to work out the moles of N₂O₄ that have become NO₂, by multiplying the concentration by the volume which gives 17.155 moles, so 47.345 moles of N₂O₄ have become NO₂. Then you multiply by 2, to give you the moles of NO₂ (1:2 ratio), which is 94.69.
Then find the concentration of this by dividing by 7.3, which gives 12.97123288...
And if you've written out the equation for K_c, then its [NO₂]² divided by [N₂O₄].
So you square the 12.97... to give 168.2528823...
Then as the equation says, you divide by the concentration of N₂O₄, which is 2.35, so you end up with 71.59697121...
Hence I think the answer is C. If there is a shorter way then somone please say!

You are on the right track for Q4! Just expand out the brackets and then solve to find x!

I got B for Qn.4