Chemical equilibrium

  • Thread starter Anna_Kwong
  • Start date
  • #1

Homework Statement


It is a chem equilibrium question, involving the ICE table.

The following reaction:
S2(g) + 2H2 (g) ←→ 2SH2(g)
takes place at high temperatures. If the initial amount of H2 and SH2 is 1.75 g and 1.95 g, respectively, and at equilibrium the amount of S2(g) is found to be 0.00188 atm, what is the value of the equilibrium constant? Please give your answer to three significant figures. Given: T = 1670 K and Reaction Volume = 0.500 L. (Hint: you will need to use the ideal gas law).


Homework Equations



pv=nRT, ICE table

The Attempt at a Solution


What I did was find the the mols of H2 and SH2 since we are given n=m/M. So I have those 2 for "I" initial.
Then I found the equilibrium mols of S2, by plugging into the formula pv=nRt, solving for n=pv/Rt and I get 2.00E-5.

Essentially, I know K = [products]/[reactants]. But I only know the "I" of H2 and SH2, and I only know the "E" of S2.

Your help is greatly appreciated!
 

Answers and Replies

  • #2
39
0
K = P(product)/ P(reactant) when everything is in gas state.
 
  • #3
how do I figure out the "C" in change? or is that even necessary..
 
  • #4
remember when dealing with gases, use partial pressure. when dealing with solutions, use concentration.
 
  • #5
ok so, kp= pSH2^2/ pH2^2 x pS2 , i know the value of pS2, which is 0.00188atm, how do I find the other 2 values needed to solve for the kp value? Thanks
 
  • #6
My kp value is 0.580, can anyone confirm this please and thanks
 
  • #7
Borek
Mentor
28,600
3,078
Use stoichiometry of the reaction.

Kc vs Kp - it depends on what the question asks. When dealing with gases it is more likely that the answer expected is Kp, but it is also perfectly possible to calculate Kc and it will be a valid answer as well. Whether it will be accepted by your teacher... that's another story.

Seems to me like your Kp value is much too low. Show details of your work.
 
  • #8
H2 :1.75g / 2.02g/mol = 0.866mol , use pv=nRT , p = nRT/v = (0.866mol)(0.082)(1670)/(0.5) = 237atm
SH2: 1.95g/34.06g/mol = 0.0572mol, use pv=nRT, p = nRT/v = (0.0572mol)(0.082)(1670)/(0.5)=15.7 atm
S2: 0.00188atm

kp= pSH2^2/pS2 x pH2^2 = (15.7)^2/((237)^2 x (0.00188)) = 2.33, eekk guess my calculations for the first run was incorrect, however, is the work done correctly?
 
  • #9
Borek
Mentor
28,600
3,078
Results looks much better now, although you have not accounted for the fact that your final mixture doesn't contain 1.75 g H2 nor 1.95 g of H2S - the first was produced together with S2, the latter was decomposed. You should use reaction stoichiometry for that.

Not that it will change much, but you should check it to be sure.
 

Related Threads on Chemical equilibrium

  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
902
  • Last Post
Replies
5
Views
6K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
12
Views
7K
Replies
0
Views
8K
  • Last Post
Replies
5
Views
2K
Top