2. Solved Papers
  3. BCECE Medical

BCECE Medical BCECE Medical Solved Papers-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 kJ/mol

    B)  +196 kJ/mol

    C)  +948 kJ/mol  

    D)  -948 kJ/mol

    Correct Answer: A

    Solution :

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


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