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NEET Sample Paper NEET Sample Test Paper-89

  • question_answer
    \[{{H}_{2}}S(g)\xrightarrow{{}}H{{S}^{-}}(g)+{{H}^{+}}(g),\]\[\Delta H{}^\circ ={{x}_{1}}\] \[\Delta H_{f}^{{}^\circ }[{{H}_{2}}S(g)={{x}_{2}},\]\[\Delta H_{f}^{{}^\circ }[H(g)]={{x}_{3}}\] hence, \[\Delta H_{f}^{{}^\circ }{{[HS]}^{-}}\] is

    A) \[{{x}_{1}}+{{x}_{2}}-{{x}_{3}}\]    

    B)        \[{{x}_{3}}-{{x}_{1}}-{{x}_{2}}\]

    C) \[{{x}_{1}}-{{x}_{2}}-{{x}_{3}}\]      

    D)        \[{{x}_{3}}-{{x}_{1}}-{{x}_{2}}\]

    Correct Answer: A

    Solution :

    [a] \[\Delta H{}^\circ =\Delta H_{f}^{{}^\circ }\](products)\[-\Delta H_{f}^{{}^\circ }\] (reactants) \[\Delta H{}^\circ =\Delta H_{f}^{{}^\circ }(HS)+\Delta H_{f}^{{}^\circ }(H)-\Delta H_{f}^{{}^\circ }({{H}_{2}}S)\] \[{{x}_{1}}=x\,(?)+{{x}_{3}}-{{x}_{2}}\] \[\therefore \]\[x=({{x}_{1}}+{{x}_{2}}-{{x}_{3}})\]


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