Question

Question #f5a83

Answer 1

`"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")`