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")#