The value of Kc for rxn below 2 H2S(g) <----> 2H2(g) + S2 (g) find value of [H2S] ?

Question

The value of Kc for rxn below

2 H2S(g) <----> 2H2(g) + S2 (g)

is 2.2 x 10^-4 at 1400 K. A sample of [H2S] = 5.00 M is heated to 1400 K in a sealed vessel. After chemical equilibrium, find value of [H2S] No H2 or S2 was present original sample.

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Answer 1 · 2018-04-11T02:59:57

The equilibrium concentration of #H_2S# is #4.78M#

#K_c=2.2xx10^-4=([H_2]^2[S_2])/([H_2S]^2)#

Let

#n_(H_2S)^0# = the initial concentration of #H_2S=5M#

Assume that #x# molar concentration of #H_2S# decomposes at equilibrium. At equilibrium, the concentration of #H_2S# will be #n_(H_2S)^0-x#.

This will produce #x# molar concentration of #H_2# and #x/2# molar concentration of #S_2#. Thus at equilibrium we will have

#K_c=(x^2(x/2))/((n_(H_2S)^0-x)^2)#

Because the equilibrium constant is small, we will make the approximation that #n_(H_2S)^0-x~~n_(H_2S)^0#. This allows us to write

#K_c~~(x^3)/(2(n_(H_2S)^0)^2)#

Solving for #x# we have

#x=[2K_c(n_(H_2S)^0)^2]^(1/3)=[2*2.2xx10^-4(5)^2]^(1/3)~~0.22#

This would give the final concentration of #H_2S# as

#n_(H_2S)^0-x=5-0.22=4.78M#

As a check of our approximation we can calculate the equilibrium constant as

#(0.22^2(0.22/2))/((4.78)^2)~~2.3xx10^-4#

which is less than a 5% error from the given value.