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EAMCET Medical EAMCET Medical Solved Paper-1996

  • question_answer
    The heat of combustion of methane at 298 K is expressed by: \[C{{H}_{4}}(g)+2{{O}_{2}}(g)\to C{{O}_{2}}(g)+2{{H}_{2}}O(l)\]\[\Delta H=890.2\,\text{kJ}\text{.}\] The magnitude of AE of the reaction at this temperature is:

    A)  infinity                                

    B)  less than \[\Delta H\]

    C)  equal to \[\Delta H\]    

    D)  greater than \[\Delta H\]

    Correct Answer: D

    Solution :

                     \[C{{H}_{4}}(g)+2{{O}_{2}}(g)\xrightarrow{{}}\] \[C{{O}_{2}}(g)+2{{H}_{2}}O(l)-890.2\,kJ\] \[\Delta H=\Delta E+\Delta nRT\] \[\Delta n=\] moles of gaseous products - moles of gaseous reactants \[=1-3=-2\] So, for the above reaction \[\Delta \Epsilon \] value is greater than \[\Delta H.\]


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