Question

Consider the following reaction:
2H2S+SO2=S(s)+H2O
A reaction mixture initially containing 0.500M H2S and 0.500M SO2 was found to contain 1.0×10−3M at a certain temperature. A second reaction mixture at the same temperature initially contains 0.255M H2S and 0.320M SO2. Calculate the equilibrium concentration of H2O in the second mixture at this temperature

Answer 1

Is that 1.0 x 10^-3 M H2O or something different. You should read your posts to make sure all of the information is there. I will assume it is (H2O). Have you checked the equation? It isn't balanced. I will balance it. And I wonder if the temperature is high enough so H2O is a gas? I will assume it is. In fact, I will write it as such in the equation.
2H2S + SO2 ==>3S(s) + 2H2O(g)
K = (H2O)^2/(H2S)^2(SO2).
Make an ICE chart.
Initial:
H2S = 0.5 M
SO2 = 0.5 M
H2O = 0

Change:
H2O = +1.0 x 10^-3
SO2 = -0.5 x 10^-3
H2S = -1.0 x 10^-3

Equilibrium:
H2S = 0.5-1.0 x 10^-3
SO2 = 0.5- 0.5) x 10^-3
H2O = 0 + 1.0 x 10^-3
Plug those equilibrium concns into the Keq expression I wrote above to find Keq.

Then prepare another ICE chart for the new mixture which will be
initial:
H2S = 0.255 M
SO2 = 0.320 M
H2O = 0
Change:
H2O = +2x
SO2 = -x
H2S = -2x

Equilibrium:
H2O = 2x
SO2 = 0.320- x
H2S = 0.255-2x
Plug those values into the Keq expression and using Keq you calculated before, determine the value of x. That times 2 will be the equilibrium value for H2O. Post your work if you get stuck.

Answer 2

write an expression for ka hor the dissociation of HBr, when HBr is dissociated, it loses its H+