Consider the following reaction:
2H2S(g)+SO2(g)⇌3S(s)+2H2O(g)
A reaction mixture initially containing 0.510M H2S and 0.510M SO2 was found to contain 1.2×10−3M H2O at a certain temperature. A second reaction mixture at the same temperature initially contains [H2S]= 0.255M and [SO2]= 0.325M .


Find the concentration of H2O in the second mixture.

Usually with this kind of things you seperate your equation in 3 steps

Initial state (I)
Reaction (R)
Equilibrium (E)

------2H2S(g)+SO2(g)⇌3S(s)+2H2O(g)
I------(0.510) ---(0.510)------0 -------0
R------(-2x) ----(-x) -------(3x)------(2x)
E ---(0.510-2x)--(0.510-x)---(3x)----(2x)

You know that 2x = 1.2x10^-3 at the final state for H20. Also you know that your limiting factor is the component on the left side of the equaiton that will react completely before the others (H2S in this case because it reacts following 2x)

from there you have that k (equilibrium constant) = (2x)/[(0.510-2x)(0.510-x))] where x can be found with 2x = 1.2x10^-3

i'll let you continue from here

edit : formatting fucked up on my table
look here to see what i tried to do : http://chemwiki.ucdavis.edu/Physical_Chemistry/Chemical_Equilibrium/Calculating_an_Equilibrium_Concentration_from_the_Equilibrium_Constant

This post was edited by LeB on Nov 14 2013 08:41pm

Doesn't work or I don't understand it.

calculate the equilibrium constant from the first reaction: (dont forget stoichiometry)

kc = (1.2*10^-3)^2 / (0.510^2)*0.510 = 0.0001591812

rearrange the equation with the new values to find the concentration of H2O:

kc * (0.255^2)*0.325 = [H2O]^2

square root

=0.001834

i think ^^

take a read: http://www.chemguide.co.uk/physical/equilibria/kc.html

Kc is basically a constant,I get it now.

Your calculation weren't correct but it worked.

well if you understand it can i get the 50fg



also can u tell me where i went wrong lol, havn't done equilibrium calculations in 5+ years

This post was edited by ArnoldChlamydia on Nov 14 2013 10:24pm

nvm lol i used initial concentrations not equilibrium

The equilibrium constant is equal to the ratio of the concentration (written with brackets) of the products over the reactants

K = [Prod]/[React]

Making the ICE table allows you to see where the moles are being distributed as the reaction undergoes.

Initial is the amount that you start with
Change is where the reaction is going
the numbers in front of the x's denote the ratios given to you in the original problem

Equilibrium is the addition of the Initial and C columns

======2H2S(g) + SO2(g) => 3S(s) +2H2O(g)

I-------------0.510------ 0.510----------0------------0
C-------------(-2x)---------(-x)----------(+3x)--------(+2x)
E-------(0.510-2x)-----(0.510-x)-------+3x----------+2x

K= [Prod]/[Reactant]

Normally here we'd plugin all our x's and be given a K and find the x

But the problem tells you theres 1.2*10^-3 M H2O. This is the concentration and so

2x = 1.2*10^-3
x = 6*10^-4

x here represents the concentration of something at equilibrium. problems often stop here and would ask things like (what is the pH of chemical y at equilibrium?)


for the second part of the problem, i believe you
find the K of the first eqn

second part you have K and concentrations of H2S and SO2 and then you can calculate concentration of H2O

isn't there math websites with walkthrough's?