Question

2H2S(g)⇌2H2(g)+S2(g),Kc=1.67×10−7 at 800∘C is carried out at the same temperature with the following initial concentrations: [H2S]=0.100M, [H2]=0.100M, and [S2]=0.00 M. Find the equilibrium concentration of S2. Express the molarity to three significant figures.

Answer 1

Answer:

1.67*10^-7

Explanation:

Kc=[H2]2[S2]/[H2S]2=(.1 M+2x)^2(x)/(0.350M−2x)2≈(0.1M)^2x(0.1M)^2=x  

Answer 2

Answer : The concentration of $S_2$ at equilibrium will be, $1.67\times 10^{-7}M$

Explanation :  Given,

Equilibrium constant = $1.67\times 10^{-7}$

Initial concentration of $H_2S$ = 0.100 M

Initial concentration of $H_2$ = 0.100 M

Initial concentration of $S_2$ = 0.00 M

The balanced equilibrium reaction is,

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

Initial conc.    0.1                0.1          0

At eqm.         (0.1-2x)       (0.1+2x)      x

The expression of equilibrium constant for the reaction will be:

$K_c=\frac{[H_2]^2[S_2]}{[H_2S]^2}$

Now put all the values in this expression, we get :

$1.67\times 10^{-7}=\frac{(0.1+2x)^2\times (x)}{(0.1-2x)^2}$

By solving the term 'x' by quadratic equation, we get two value of 'x'.

$x=1.67\times 10^{-7}M$

Concentration of $S_2$ at equilibrium = $x=1.67\times 10^{-7}M$

Therefore, the concentration of $S_2$ at equilibrium will be, $1.67\times 10^{-7}M$