Haryana PMT Haryana PMT Solved Paper-2004

  • question_answer
    According to equation,\[{{C}_{6}}{{H}_{6}}(l)+\frac{15}{2}{{O}_{2}}(g)\xrightarrow{{}}3{{H}_{2}}O(l)+6C{{O}_{2}}(g)\]\[\Delta H=-3264.4kJ/\,mol\]. The energy          evolved when \[7.8g\]of benzene is burnt in air will be:

    A)  \[3.264\text{ }kJ/mol\]

    B)  \[32.64\text{ }kJ/mol\]

    C)  \[326.4\text{ }kJ/mol\]

    D)  \[163.22\text{ }kJ/mol\]

    Correct Answer: C

    Solution :

                    \[\underset{(6\times 12+6\times 1=78)}{\mathop{{{C}_{6}}{{H}_{6}}(l)+\frac{15}{2}{{O}_{2}}(g)}}\,\to 3{{H}_{2}}O(l)+6C{{O}_{2}}(g)\] \[\because \] The energy evolved when 78 g. benzene is burnt \[=3264.4\text{ }kJ/mol\] \[\therefore \]The energy evolved when 7.8g benzene is burnt                                 \[=\frac{3264.4\times 7.8}{78}=326.44kJ/mol\]


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