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For the equilibrium
2H2S(g) ⇋ 2H2(g) + S2(g) Kc = 9 .0X 10-8 at 700°C
the initial concentrations of the three gases are 0.300 M H2S, 0.300 M H2, and 0. 1 50 M S2' Determine the equilibrium concentrations of the gases.



Sagot :

Lublanaidn

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]

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