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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2009

  • question_answer
    The following equilibrium establishes on heating 0.2 mole of\[{{H}_{2}}\]and 1.0 mole of sulphur in 1 L vessel at\[90{}^\circ C\]. \[{{H}_{2}}(g)+S(s){{H}_{2}}S(g);\]  \[K=6.8\times {{10}^{-2}}\] The partial pressure of\[{{H}_{2}}S\]in equilibrium state is

    A)  4.20               

    B)  0.42

    C)  0.21               

    D)  0.042

    Correct Answer: B

    Solution :

     \[{{H}_{2}}(g)+S(s){{H}_{2}}S(g)K=6.8\times {{10}^{-2}}\]
    0.2 mol 1 mol 0 mol Initialy
    \[(0.2-x)\] \[x\]mol At equilibrium
    \[K=\frac{[{{H}_{2}}S]}{[{{H}_{2}}]}=\frac{x}{0.2-x}=6.8\times {{10}^{-2}}\] \[\therefore \] \[x=1.273\times {{10}^{-2}}\] \[{{p}_{{{H}_{2}}S}}=\frac{n}{V}\times R\times T\] \[=\frac{1.273\times {{10}^{-2}}\times 0.082\times 363}{1}\] \[=0.38\approx 0.42\]


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