1. Tardigrade
  2. Question
  3. Chemistry
  4. At 90° C, the following equilibrium is established H2(g)+S(s)H2S(g); Kc=6.8× 10-2 If 0.20 mole hydrogen and 1.0 mole of sulphur are heated to 90° C in a 1.0 L vessel, what will be the partial pressure of H2S at equilibrium?

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Q. At $ 90{}^\circ C, $ the following equilibrium is established $ {{H}_{2}}(g)+S(s){{H}_{2}}S(g); $ $ {{K}_{c}}=6.8\times {{10}^{-2}} $ If 0.20 mole hydrogen and 1.0 mole of sulphur are heated to $ 90{}^\circ C $ in a 1.0 L vessel, what will be the partial pressure of $ {{H}_{2}}S $ at equilibrium?

JamiaJamia 2013

Solution:

$ {{H}_{2}}(g)+S(s){{H}_{2}}S(g); $ $ {{K}_{c}}=6.8\times {{10}^{-2}} $ Initial 0.20 0 Atequiation $ (0.20-x)~ $ $ x $ $ {{K}_{c}}=6.8\times {{10}^{-2}}=\frac{[{{H}_{2}}S]}{{{H}_{2}}} $ $ 6.8\times {{10}^{-2}}=\frac{x}{0.20-x} $ $ x=1.273\times {{10}^{-2}}mol\,{{L}^{-1}} $ $ {{p}_{{{H}_{2}}S}}=\left( \frac{n}{v} \right)RT $ $ =(1.273\times {{10}^{-2}})\times 0.0821\times 363 $ $ =0.38\text{ }atm $