Question

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.

Answer 1

Equilibrium concentrations of the gases are

`H_2S=0.596M`

`H_2=0.004 M`

`S_2=0.002 M`

Step-by-step explanation:

We are given that for the equilibrium

`2H_2S\rightleftharpoons 2H_2(g)+S_2(g)`

`k_c=9.0* 10^(-8)`

Temperature,
`T=700^(\circ)C`

Initial concentration of

`H_2S=0.30M`

`H_2=0.30 M`

`S_2=0.150 M`

We have to find the equilibrium concentration of gases.

After certain time

2x number of moles of reactant reduced and form product

Concentration of

`H_2S=0.30+2x`

`H_2=0.30-2x`

`S_2=0.150-x`

At equilibrium

Equilibrium constant

`K_c=(product)/(Reactant)=([H_2]^2[S_2])/([H_2S]^2)`

Substitute the values

`9* 10^(-8)=((0.30-2x)^2(0.150-x))/((0.30+2x)^2)`

`9* 10^(-8)=((0.30-2x)^2(0.150-x))/((0.30+2x)^2)`

`9* 10^(-8)=((0.30-2x)^2(0.150-x))/((0.30+2x)^2)`

By solving we get

`x\approx 0.148`

Now, equilibrium concentration of gases

`H_2S=0.30+2(0.148)=0.596M`

`H_2=0.30-2(0.148)=0.004 M`

`S_2=0.150-0.148=0.002 M`