BCECE Engineering BCECE Engineering Solved Paper-2012

  • question_answer
    Change in enthalpy for the reaction, \[2{{H}_{2}}{{O}_{2}}(l)\xrightarrow{{}}2{{H}_{2}}O(l)+{{O}_{2}}(g)\] if heat of formation of \[{{H}_{2}}{{O}_{2}}(l)\]and\[{{H}_{2}}O(l)\] are -188 and -286 kJ/mol respectively is

    A) \[-196\text{ }kJ/mol\]  

    B)         \[+196\text{ }kJ/mol\]

    C)        \[+948\text{ }kJ/mol\]  

    D)         \[-948\text{ }kJ/mol\]

    Correct Answer: A

    Solution :

    \[2{{H}_{2}}{{O}_{2}}(l)\xrightarrow{{}}2{{H}_{2}}O(l)+{{O}_{2}}(g);\Delta H=?\] \[\Delta H=[(2\times \Delta {{\Eta }_{f}}\,of\,{{H}_{2}}O(l)+(\Delta {{\Eta }_{f}}of\,{{O}_{2}})]\] \[-(2\times \Delta {{H}_{f}}\,\text{of}\,{{H}_{2}}{{O}_{2}}(l))]\]                 \[=[(2\times -286)+(0)-(2\times -188)]\] \[=[-572+376]\] \[=-196\,kJ/mol\]


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