Respuesta :
Answer: [tex]K_c=\frac{(2\times 0.032)^2\times (0.032)}{(0.15-2\times 0.032)^2}[/tex]
Explanation:
Moles of [tex]H_2S[/tex] = [tex]\frac{\text {given mass}}{\text {molar mass}}=\frac{3.75g}{34g/mol}=0.11mol[/tex]
Volume of solution = 0.75 L
Initial concentration of [tex]H_2S[/tex] = [tex]\frac{moles}{Volume}=\frac{0.11mol}{0.75L}=0.15M[/tex]
Equilibrium concentration of [tex]S_2[/tex] = [tex]\frac{moles}{Volume}=\frac{2.42\times 10^{-2}mol}{0.75L}=0.032M[/tex]
The given balanced equilibrium reaction is,
[tex]2H_2S(g)\rightleftharpoons 2H_2(g)+S_2(g)[/tex]
Initial conc. 0.15 M 0 0
At eqm. conc. (0.15-2x) 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]
Now put all the given values in this expression, we get :
[tex]K_c=\frac{(2x)^2\times (x)}{(0.15-2x)^2}[/tex]
Given : x = 0.032 M
[tex]K_c=\frac{(2\times 0.032)^2\times (0.032)}{(0.15-2\times 0.032)^2}[/tex]
[tex]K_c=0.017[/tex]
