JEE Main & Advanced JEE Main Paper (Held On 22 April 2013)

  • question_answer The reaction  \[\operatorname{X}\to \operatorname{Y}\] is an exothermic reaction. Activation energy of the reaction for X into \[\operatorname{Y}\] is 150 \[150\operatorname{K}\operatorname{J}{{\operatorname{mol}}^{-1}}\]. Enthalpy of reaction is \[135\operatorname{K}\operatorname{J}{{\operatorname{mol}}^{-1}}\]. The activation energy for the reverse reaction, \[\operatorname{Y}\to \operatorname{X}\] will be:     JEE Main  Online Paper (Held On 22 April 2013)

    A)  \[280\operatorname{kJ}{{\operatorname{mol}}^{-1}}\]                            

    B)  \[285\operatorname{kJ}{{\operatorname{mol}}^{-1}}\]            

    C)  \[270\operatorname{kJ}{{\operatorname{mol}}^{-1}}\]                            

    D)  \[15\operatorname{kJ}{{\operatorname{mol}}^{-1}}\]

    Correct Answer: B

    Solution :

     \[X\xrightarrow{{}}Y;\Delta H=-135\,kJ/mol,\] \[{{E}_{a}}=150\,kJ/mol\] For an exothermic reaction \[{{E}_{a(F.R.)}}=\Delta H+E_{a(B.R)}^{'}\] \[150=-135+E_{a(B.R.)}^{'}\] \[E_{a(B.R)\,}^{'}=285\,kJ/mol\]


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