Question

Hydrogen sulfide decomposes according to the following reaction, for which Kc=9.30E-8 at 700 degrees Celsius. 2 H2S(g) --> 2 H2(g) + S2(g) If 0.29 moles of H2S is placed in a 3.0-L container, What is the equilibrium concentration of H2(g) at 700 degrees Celsius?

Answer 1

Answer: The equilibrium concentration of $H_2(g)$ at 700 degrees Celsius is 0.0012 M

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_c$

Moles of  $H_2S$ = 0.29 mole

Volume of solution = 3.0 L

Initial concentration of $H_2S$ = $\frac{0.29mol}{3.0L}=0.097M$

The given balanced equilibrium reaction is,

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

         Initial conc.         0.097 M           0M          0M

At eqm. conc.    (0.097-2x) M            (2x) M   (x) M

The expression for $K_c$ is written as:

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

$K_c=\frac{(2x)^2\times x}{(0.097-2x)^2}$

$9.30\times 10^{-8}=\frac{(2x)^2\times x}{(0.097-2x)^2}$

$x=0.00060$

Equilibrium concentration of $[H_2]$= 2x= $2\times 0.00060=0.0012M$