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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2003

  • question_answer
    The enthalpies of formation of \[A{{l}_{2}}{{O}_{3}}\] and \[C{{r}_{2}}{{O}_{3}}\] are \[-1596\text{ }kJ\] and \[-1134kJ\] respectively. \[\Delta H\] for the reaction, \[2Al+C{{r}_{2}}{{O}_{3}}\xrightarrow{{}}2Cr+A{{l}_{2}}{{O}_{3}}\] is:

    A)  -2730 kJ                              

    B)  -462 kJ

    C)  -1365 kJ                              

    D)  +2730 kJ

    Correct Answer: B

    Solution :

    According to question                 \[2Al+\frac{3}{2}{{O}_{2}}\xrightarrow{{}}A{{l}_{2}}{{O}_{3}}\]                   ... (i)                 \[\Delta H=-1596\,\,kJ\]                 \[2Cr+\frac{3}{2}{{O}_{2}}\xrightarrow{{}}C{{r}_{2}}{{O}_{3}}\]                  ... (ii)                 \[\Delta H=-1134\,\,kJ\] Reversing eq. (ii) and adding both reactions, we get                 \[2Al+C{{r}_{2}}{{O}_{3}}\xrightarrow{{}}2Cr+A{{l}_{2}}{{O}_{3}}\]                                           \[\Delta H=-462\,\,kJ\]


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