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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2003

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

    A)  \[-2730kJ\]

    B)  \[-462kJ\]

    C)  \[-1365kJ\]

    D)  \[+2730kJ\]

    Correct Answer: B

    Solution :

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


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