Question #f5a83

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Question #f5a83

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Answer 1 · 2017-08-24T17:46:22

$0.303 atm$ $"0.303 atm"$ .

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

$K_{P} = \frac{P_{H_{2}, e q} P_{S, e q}}{P_{H_{2} S, e q}} = 0.833$ $K_P = (P_(H_2,eq) P_(S,eq))/(P_(H_2S,eq)) = 0.833$

All the coefficients were $1$ $1$ so there are no exponents to worry about. We know that $P_{H_{2} S, i} = 0.163 atm$ $P_(H_2S,i) = "0.163 atm"$ . An ICE table gives:

$H_{2} S (g) ⇌ H_{2} (g) + S (g)$ $"H"_2"S"(g) " "rightleftharpoons" " "H"_2(g) + "S"(g)$

$I 0.163 00$ $"I"" "0.163" "" "" "" "0" "" "" "0$
$C - x + x + x$ $"C"" "-x" "" "" "" "+x" "" "+x$
$E 0.163 - x x x$ $"E"" "0.163 - x" "" "x" "" "" "x$

We define $x$ $x$ as the partial pressure of $H_{2} (g)$ $"H"_2(g)$ or $S (g)$ $"S"(g)$ at equilibrium, i.e. $P_{H_{2}, e q} = P_{S, e q}$ $P_(H_2,eq) = P_(S,eq)$ .

This becomes:

$0.833 = \frac{x^{2}}{0.163 - x}$ $0.833 = x^2/(0.163 - x)$

The $K_{P}$ $K_P$ is not small, so we'll have to solve this quadratic equation in full.

$0.833 (0.163 - x) - x^{2} = 0$ $0.833(0.163 - x) - x^2 = 0$

You should get:

$x \equiv P_{H_{2}, e q} = P_{S, e q} = 0.140 atm$ $x -= P_(H_2,eq) = P_(S,eq) = "0.140 atm"$

This says that $H_{2} S$ $"H"_2"S"$ decomposes almost completely, as $P_{H_{2} S, e q} = 0.023 atm$ $P_(H_2S,eq) = "0.023 atm"$ .

$P_{tot} = P_{1} + P_{2} + . . .$ $color(blue)(P_("tot")) = P_1 + P_2 + . . .$

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

$= 0.163 + x$ $= 0.163 + x$

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

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

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