Question #f5a83

1 Answer
Truong-Son N.
Aug 24, 2017

#"0.303 atm"#.

Given the reaction, write the mass action expression using the partial pressures of each gas:

#K_P = (P_(H_2,eq) P_(S,eq))/(P_(H_2S,eq)) = 0.833#

All the coefficients were #1# so there are no exponents to worry about. We know that #P_(H_2S,i) = "0.163 atm"#. An ICE table gives:

#"H"_2"S"(g) " "rightleftharpoons" " "H"_2(g) + "S"(g)#

#"I"" "0.163" "" "" "" "0" "" "" "0#
#"C"" "-x" "" "" "" "+x" "" "+x#
#"E"" "0.163 - x" "" "x" "" "" "x#

We define #x# as the partial pressure of #"H"_2(g)# or #"S"(g)# at equilibrium, i.e. #P_(H_2,eq) = P_(S,eq)#.

This becomes:

#0.833 = x^2/(0.163 - x)#

The #K_P# is not small, so we'll have to solve this quadratic equation in full.

#0.833(0.163 - x) - x^2 = 0#

You should get:

#x -= P_(H_2,eq) = P_(S,eq) = "0.140 atm"#

This says that #"H"_2"S"# decomposes almost completely, as #P_(H_2S,eq) = "0.023 atm"#.

#color(blue)(P_("tot")) = P_1 + P_2 + . . . #

#= (0.163 - x) + (x) + (x)#

#= 0.163 + x#

#= "0.163 atm" + "0.140 atm"#

#=# #color(blue)("0.303 atm")#

Related questions
See all questions in Dynamic Equilibrium
Impact of this question
1897 views around the world
You can reuse this answer
Creative Commons License