Respuesta :
Answer: The value of the equilibrium constant is 0.00009.
Explanation:
Initial moles of [tex]H_2S[/tex] = 0.15 mole
Volume of container = 10 L
Initial concentration of [tex]H_2S=\frac{moles}{volume}=\frac{0.15moles}{10L}=0.015M[/tex]
Moles of [tex]H_2[/tex] at equilibrium= 0.03 mole
equilibrium concentration of [tex]H_2=\frac{moles}{volume}=\frac{0.03moles}{10L}=0.003M[/tex] [/tex]
The given balanced equilibrium reaction is,
[tex]2H_2S(g)\rightleftharpoons 2H_2(g)+S_2(g)[/tex]
Initial conc. 0.015 M 0 0
At eqm. conc. (0.015-2x) M (2x) M (x) M
The expression for equilibrium constant for this reaction will be,
[tex]K_c=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}[/tex]
[tex]K_c=\frac{(2x)^2\times x}{(0.015-2x)^2}[/tex]
we are given : 2x= 0.003 M
x= 0.0015 M
Now put all the given values in this expression, we get :
[tex]K_c=\frac{(0.003)^2\times 0.0015}{(0.015-0.003)^2}[/tex]
[tex]K_c=0.00009[/tex]
Thus the value of the equilibrium constant is 0.00009.
