Answer:
0.00193 mol/L
Explanation:
Given that:
numbers of moles of H₂S = 0.59 moles
Volume = 3.0-L
Equilibrium constant [tex]K_c[/tex] = 9.30 × 10⁻⁸
The equation for the reaction is given as :
2H₂S ⇄ 2H₂(g) + S₂(g)
The initial concentration of H₂S = [tex]\frac{number of moles of H_2S }{Volume}[/tex]
The initial concentration of H₂S = [tex]\frac{0.59 moles}{3.0}[/tex]
= 0.1966 mol/L
The ICE table is shown be as :
2H₂S ⇄ 2H₂(g) + S₂(g)
Initial 0.9166 0 0
Change -2 x +2 x + x
Equilibrium (0.9166 - 2x) 2x x
[tex]K_c= \frac{[H_2]^2[S_2]}{[H_2S]^2}[/tex]
[tex]9.38*10^{-8}= \frac{[2x]^2[x]}{[0.1966-2x]^2}[/tex]
[tex]9.38*10^{-8}= \frac{[4x]^3}{[0.1966]^2}[/tex]
(since 2x < 0.1966 if solved through quadratic equation)
[tex]4x^3= {9.30*10^-8 * 0.1966^2}[/tex]
[tex]x^3= \frac{9.30*10^-8 * 0.1966^2}{4}[/tex]
[tex]x^3= 8.9865*10^{-10[/tex]
[tex]x=\sqrt[3]{8.9865*10^{-10}}[/tex]
[tex]x= 9.65*10^{-4}[/tex]
The equilibrium concentration for H₂(g) = 2x
∴ [tex]2*9.65*10^{-4[/tex]
= 0.00193 mol/L
Thus, the equilibrium concentration of H₂(g) at 700°C = 0.00193 mol/L
