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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2010

  • question_answer
    Calculate the standard enthalpy change (in kJ \[mo{{l}^{-1}}\]) for the reaction \[{{H}_{2}}(g)+{{O}_{2}}(g)\xrightarrow{{}}{{H}_{2}}{{O}_{2}}(g)\] Given that bond enthalpies of\[HH,\text{ O}=O,\]\[OH\]and\[OO\](in kJ\[mo{{l}^{-1}}\]) are respectively 438, 498, 464 and 138.

    A)  \[-130\]                              

    B)  \[-\,65\]

    C)  + 130                   

    D)         \[-\,334\]

    E)  + 334

    Correct Answer: A

    Solution :

    \[{{H}_{2}}(g)+{{O}_{2}}(g)\xrightarrow{{}}{{H}_{2}}{{O}_{2}}(g)\] \[\Delta {{H}_{reaction}}=B{{E}_{reac\tan ts}}-B{{E}_{products}}\] \[=[BE(H-H)+BE(O=O)]\] \[-2BE(O-H)+BE(O-O)]\] \[=[438+498]-[2\times 464+138]\] \[=936-1066=-130\text{ }kJ\text{ }mo{{l}^{-1}}\]


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