2. Sample Papers
  3. NEET

NEET Sample Paper NEET Sample Test Paper-83

  • question_answer
    What will be the value of equilibrium constant for the reaction? \[\operatorname{AgCl}(s)\,\,\xrightarrow{{}}\,\,A{{g}^{+}}\,(aq)+C{{l}^{-}}\,(aq)\]\[E{{{}^\circ }_{AgCl\,(s)/Ag(s),\,C{{l}^{-}}}}\,\,=\,\,0.22\,V\]and      \[E{{{}^\circ }_{[A{{g}^{+}}\left( aq \right)/Ag\left( s \right)]}}\,\,=\,\,0.80\,V\]

    A)  - 22.57            

    B)  \[1.6\,\,\times \,\,1{{0}^{-}}^{10}\]

    C)  \[1.6\,\,\times \,\,1{{0}^{10}}\]                        

    D)  \[1.6\,\,\times \,\,1{{0}^{9}}\]

    Correct Answer: C

    Solution :

    Half cell reactions are (i) \[\operatorname{AgC}l \xrightarrow{{}}\,A{{g}^{+}}\left( aq \right) + C{{l}^{-}}\,\left( aq \right)\,\,+ E{}^\circ  = 0.22\,V\] (ii) \[A{{g}^{+}}\,+e\,\,\xrightarrow{{}}\,\,Ag,\,\,E{}^\circ =\,\,0.80\,V\] Subtracting Eqs. (i) and (ii) \[\operatorname{AgC}l \xrightarrow{{}}\,\,A{{g}^{+}}+ C{{l}^{-}}\] \[E_{cell}^{{}^\circ } = 0.22 - 0.80 = - 0.58 V\] We know that \[\Delta G = -nFE{}^\circ \] \[-\,RT\,\,In\,k\,\,=\,\,-nFE{}^\circ \] In \[k=-\frac{nFE{}^\circ }{RT}\] \[=-\frac{2\times 96500\times -\,0.58}{8.314\,\,\times \,\,300}\] \[k=-\,22.57\] \[k=\,\,Antilog\,\,\left( \frac{-22.57}{2.303} \right)\,\,=\,\,1.6\times 1{{0}^{-10}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner