Question #f5a83

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Question #f5a83

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Answer 1 · 2017-08-24T17:46:22

$"0.303 atm"$ .

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 $P_(H_2S,eq) = "0.023 atm"$ .

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

$= (0.163 - x) + (x) + (x)$

$= 0.163 + x$

$= "0.163 atm" + "0.140 atm"$

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

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