# Question #f5a83

##### 1 Answer

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

#"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

This becomes:

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

The

#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

#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")#