Answer :
Answer : The total pressure in the container at equilibrium is, 0.4667 atm
Solution : Given,
Initial pressure of [tex]H_2S[/tex] = 0.259 atm
Equilibrium constant, [tex]K_p[/tex] = 0.841
The given equilibrium reaction is,
[tex]H_2S(g)\rightleftharpoons H_2(g)+S(g)[/tex]
Initially 0.259 0 0
At equilibrium (0.259 - x) x x
Let the partial pressure of [tex]H_2[/tex] and [tex]S[/tex] will be, 'x'
The expression of [tex]K_p[/tex] will be,
[tex]K_p=\frac{(p_{H_2})(p_{S})}{p_{H_2S}}[/tex]
Now put all the values of partial pressure, we get
[tex]0.841=\frac{(x)\times (x)}{(0.259-x)}[/tex]
By solving the term x, we get
[tex]x=0.2077atm[/tex]
The partial pressure of [tex]H_2[/tex] and [tex]S[/tex] = x = 0.2077 atm
Total pressure in the container at equilibrium = [tex]0.259-x+x+x=0.259+x=0.259+0.2077=0.4667atm[/tex]
Therefore, the total pressure in the container at equilibrium is, 0.4667 atm