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WB JEE Medical WB JEE Medical Solved Paper-2015

  • question_answer
    For the reaction,\[{{X}_{2}}{{Y}_{2}}(l)\xrightarrow{{}}2X{{Y}_{2}}(g)\]at \[300\,K,\] the values of \[\Delta U\]and\[\Delta S\] are 2 kcal and \[\text{20}\,\text{cal}\,{{\text{K}}^{-1}}\]respectively. The value of \[\Delta G\]for the reaction is                                     

    A) \[~-3400\,\text{cal}\]         

    B) \[~3400\,\text{cal}\]                  

    C) \[-2800\text{ cal}\]         

    D) \[2000\text{ cal}\]                   

    Correct Answer: C

    Solution :

     For the-reaction\[{{X}_{2}}{{Y}_{4}}(l)\xrightarrow{{}}2X\,{{Y}_{2}}(g),\] \[\Delta {{n}_{g}}=\]number of gaseous products   \[-\]number of gaseous reactants \[\Delta {{n}_{g}}=2-0=0\] Given that, \[\Delta U=2\,kcal\] \[\Delta S=20\,cal{{K}^{-1}}\] From the formula, \[\Delta H=\Delta U+\Delta {{n}_{g}}RT\] Putting the values, we get \[\Delta \Eta =2+\left( \frac{2\times 2\times 300}{1000} \right)\] \[=3.2\,kcal=3.2\,\times {{10}^{3}}\,\text{cal}\] Also, \[\Delta G=\Delta H-T\Delta S\] \[=3.2\times {{10}^{3}}-300\times 20\] \[=3.2\times {{10}^{3}}-6\times {{10}^{3}}=-2800\,cal\]


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