contestada

2H2S (g) <-> 2H2 (g) + S2 (g)
Kc = 1.67 *10 ^-7 at 800 C
A 0.500 L reaction vessel initially contains 0.0125 mol of H2S at 800 C. Find the equilibrium concentrations of H2 and S2.

Respuesta :

Edufirst
Chemical reaction:

2H2S (g) ⇄ 2H2(g) + S2(g)

The equilibrium constant is given by: Kc = [H2]^2 * [S2] / [H2S]^2 = 1.67 * 10^ -7


The initial concentration is 0.0125 mol / 0.500 L = 0.0250 M

Make a table showing the initial concentrations, the change and the final concentrations of each species


                  2H2S (g)     ⇄    2H2(g) + S2(g)
start           0.0250M                   0            0

change          - 2x                   +2x       + x

end            0.0250 - 2x             2x          x


Kc = (2x)^2 (x) / (0.0250 - 2x)^2


Kc = 4x^3 / (0.0250 - 2x)^2


To solve that equation in an easy way you can assume that 2x is << 0.0250, which leads to

Kc = 4x^3 / (0.0250)^2 = 1.67 * 10^ -7

=> x^3 = 1.67 * 10^ -7 * 0.0250 / 4 = 2.6  * 10 ^-11


=> x = 2.97 * 10^ -4 M

With this you can check that your assumption that x << 0.0250 is good and continue.

From the table you know that the concentrations at equilibrium are:

[H2] = 2x = 2 * 2.97 * 10 ^ -4 M = 5.94 * 10 ^ -4

[S2] = x = 2.97 * 10^ -4 M

Otras preguntas