Question

I'm having some trouble with this question. will post my thought process in a bit!
For the thermal decomposition of hydrogen sulfide, H2S(g) ↔ 2H2(g) + S2(g), K = 2.2x10-4 at 1,400 K. A sample of gas in which [H2S] = 6.00 M is heated to the decomposition point in a sealed vessel. After equilibrium is reached, what is [H2S]? You can assume that no H2 or S2 was present in the original sample.
A. 3.22 M
B. 5.75 M
C. 4.45 M
D. 1.77 M
E. 5.10 M

Answer 1

First I balanced the equation: 2H2S(g) ↔ 2H2(g) + S2(g)
and use the ICE table, which I would get 2.2*10^-4=(x)*(2x^2)/(6-2x)^2
solving for x, I would get 0.153, plugging it back into the equation is 6-2(0.153)=5.69M
but no choices are that concentration. the closest is B 5.75, but I think this question is harder than I think it is.

Answer 2

\[K = 2.2E-4 = \frac{ [H2(g)]^2[S2(g)] }{ [H2S(g)] } = \frac{ [2x]^2[x] }{ [6-2x]^2 } = \frac{ [4x^2][x] }{ [6-2x]^2 } = \frac{ [4x^3] }{ [6-2x]^2 }\] is what i have so far

Answer 3

gosh this equation thingy takes forever to do

Answer 4

I also tried using the free energy side of it. deltaG=-RTln(K)
free energy is always 0 at equilibrium
so 0=-8.314*1400*ln((x)*(2x^2)/(6-2x)^2)
which solving for x would be 1.587, plugging that into the ice table; it would be 6-2(1.587)=2.83M which again is none of the choices

Answer 5

Thanks for staying with me, you can use just the easier way to enter equations if you want to :)

Answer 6

wait

Answer 7

how did you solve for x? I've having a hard time doing that lol

Answer 8

i tried square rooting both sides and got 0.014 = \[\sqrt{x}\]/6-2x

Answer 9

ops i mean,
\[0.014 =\frac{ 2x \sqrt{x} }{ 6-2x }\]

Answer 10

ya i was going to say that the equation in the problem wasn't balanced lol

Answer 11

But @premedhelp , x can't be 0.153 b/c if you plug back 0.153 into the equilibrium expression, you don't get the same equilibrium constant of 2.2E-4

Answer 12

thats what I was thinking, but doesnt the constant change when you heat it up to the decomposition point? i think thats what im confused about

Answer 13

and neither can it be 1.587 b/c it doesn't give you 2.2E-4

Answer 14

Oh snaps, yes you're definitely right. The eq constant's value changes with temperature, but we're not told the final temperature so we can't find the new eq constant

Answer 15

my other question is if 6M is even the initial concentration of H2S though

Answer 16

yep, has been bugging me an hour

Answer 17

ya it has to be. it can't be the final eq concentration b/c that would be the answer then lol

Answer 18

Look, i think we have our equilibrium expressions set up correctly. I'm thinking that where we're going wrong is actually solving for x

Answer 19

that's all we have to do, solve for x properly

Answer 20

http://www.wolframalpha.com/input/?i=2.2*10^-4%3D%28x%29*%282x^2%29%2F%286-2x%29^2

here could be our reference, just cant get step by step

Answer 21

no
you typed it up wrong

Answer 22

it should be (x)(2x)^2 or (x)(4x^2) or just 4x^3

Answer 23

YAY after we type it in correctly, we'll solve this problem correctly YES!

Answer 24

and viola!, x = 0.122
http://www.wolframalpha.com/input/?i=2.2*10%5E-4%3D%28x%29*%284x%5E2%29%2F%286-2x%29%5E2

Answer 25

and just to check, we can plug x = 0.122 into the equilibrium expression and we DO get 2.2E-4 :)

Answer 26

Now we can solve for the equilibrium concentration of H2S(g), which is represented by 6-2x

Answer 27

which is 5.75M!

Answer 28

so we can substitute our new x of 0.122 into 6-2x and we get 5.756

Answer 29

but i would still like to know how we can solve for x manually

Answer 30

YAY! WE FRIGGING DID IT LOL

Answer 31

high five :)

Answer 32

;)

Answer 33

ya me too :(

Answer 34

stay on for a bit, and ill try to figure it out :)

Answer 35

but damn...that wolf ram alpha tho... lol i've never used that before. That program is cool

Answer 36

haha message me it, i'm going to eat right now :) Good luck

Answer 37

i know right! alrighty, eat up! :)

Answer 38

I need some...."food for thought" haha

Answer 39

haha, you're right! If you want to make a friend, I'll add you on fb?

Answer 40

ya sure

Answer 41

i've sent you my fb link to ur message inbox