Answer:
Equilibrium concentrations of the gases are
[tex]H_2S=0.596M[/tex]
[tex]H_2=0.004 M[/tex]
[tex]S_2=0.002 M[/tex]
Explanation:
We are given that for the equilibrium
[tex]2H_2S\rightleftharpoons 2H_2(g)+S_2(g)[/tex]
[tex]k_c=9.0\times 10^{-8}[/tex]
Temperature, [tex]T=700^{\circ}C[/tex]
Initial concentration of
[tex]H_2S=0.30M[/tex]
[tex]H_2=0.30 M[/tex]
[tex]S_2=0.150 M[/tex]
We have to find the equilibrium concentration of gases.
After certain time
2x number of moles of reactant reduced and form product
Concentration of
[tex]H_2S=0.30+2x[/tex]
[tex]H_2=0.30-2x[/tex]
[tex]S_2=0.150-x[/tex]
At equilibrium
Equilibrium constant
[tex]K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}[/tex]
Substitute the values
[tex]9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}[/tex]
[tex]9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}[/tex]
[tex]9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}[/tex]
By solving we get
[tex]x\approx 0.148[/tex]
Now, equilibrium concentration of gases
[tex]H_2S=0.30+2(0.148)=0.596M[/tex]
[tex]H_2=0.30-2(0.148)=0.004 M[/tex]
[tex]S_2=0.150-0.148=0.002 M[/tex]
