Question

A 0.15 mol sample of H2S is placed in a 10 L reaction vessel and heated to 1132◦C. At equilibrium, 0.03 mol H2is present. Calculate the value of Kc for the reaction2 H2S(g)⇀↽2 H2(g) + S2(g)at 1132◦C

Answer 1

Answer: The value of the equilibrium constant is 0.00009.

Explanation:

Initial moles of  $H_2S$ = 0.15 mole

Volume of container = 10 L

Initial concentration of $H_2S=\frac{moles}{volume}=\frac{0.15moles}{10L}=0.015M$  

Moles of $H_2$ at equilibrium= 0.03 mole

equilibrium concentration of $H_2=\frac{moles}{volume}=\frac{0.03moles}{10L}=0.003M$ [/tex]

The given balanced equilibrium reaction is,

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

Initial conc.           0.015 M                     0     0

 At eqm. conc.    (0.015-2x) M           (2x) M   (x) M

The expression for equilibrium constant for this reaction will be,

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

$K_c=\frac{(2x)^2\times x}{(0.015-2x)^2}$

we are given : 2x= 0.003 M

x= 0.0015 M

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

$K_c=\frac{(0.003)^2\times 0.0015}{(0.015-0.003)^2}$

$K_c=0.00009$

Thus the value of the equilibrium constant is 0.00009.