Dear friend your answer is
                 Â
H2SÂ Â Â
==>Â Â Â Â Â
H2Â Â Â Â
+Â Â Â Â Â S
initial....       0.161
atm.     Â
0...          Â
0
change.....    Â
-2p.....         Â
p....        p
equil.....      Â
0.161-2p..      Â
p....         Â
p
Kp = pH2*pS/pH2S
Substitute into the Kp expression from the ICE chart above and
solve for p through quadratic equation
0.846 = pH2*pS/pH2S = (p)(p)/(0.161-p)
Solve for p = 0.1905 so
pH2 = 0.1905
pS = 0.1905
1 mol H2S gas decomposes to 2 mols that the pressure should
increase ( 2X 0.161) = 0.322
pH2S = 0.322 - 0.1905 = 0.1315
Ptotal = 0.1905 + 0.1905+ 0.1315 =
0.5125atm Â
Thank you, all the best.
