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NEET Sample Paper NEET Sample Test Paper-12

  • question_answer
    Heat of combustion\[\Delta {{H}^{o}}\]for\[C(s),{{H}_{2}}(g)\]and \[C{{H}_{4}}(g)\]are -94, -68 and -213 kcal/mol. Then \[\Delta {{H }^{o}}\] for \[C(s)+2{{H}_{2}}(g)\Delta C{{H}_{4}}(g)\]is

    A)  -17 kcal

    B)  -111 kcal

    C)  -170 kcal

    D)  -85 kcal

    Correct Answer: A

    Solution :

     For reaction, \[C(s)+2{{H}_{2}}(g)\xrightarrow{{}}C{{H}_{4}}(g),\]   \[\Delta {{\Eta }^{o}}=?\] \[\Delta {{H}^{o}}=-[(\Delta \Eta _{c}^{o}\,of\,C{{H}_{4}})-(\Delta \Eta _{c}^{o}\,of\,C+2\,\times \,\Delta \Eta _{c}^{o}\,of\,{{H}_{2}})]\]\[C+{{O}_{2}}\xrightarrow{{}}C{{O}_{2}};\Delta \Eta =-94\,kcal\]                  (i) \[2{{H}_{2}}+{{O}_{2}}\xrightarrow{{}}2{{H}_{2}}O;\Delta H=-68\times 2\,kcal\]           (ii) \[C{{H}_{4}}+2{{O}_{2}}\xrightarrow{{}}\] \[C{{O}_{2}}+2{{H}_{2}}O;\Delta \Eta =-213\,kcal\]                (ii) Eq. (i)+(ii) -Eq (iii) \[=-[(-213)-(-94+2\times -68)]\,kcal/mol\] \[=-[-213+230]=-17\,kcal/mol\]


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