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The forward and backward activation energy of exothermic reaction A→B is 80 and 280 \mathrm{~kJ} \mathrm{~mol}^{-1} respectively. The enthalpy change (in kJ/mol) of the forward reaction, A→B will be:

Option: 1

-200


Option: 2

-360


Option: 3

-280


Option: 4

-80


Answers (1)

The given reaction is an exothermic reaction
For exothermic reaction, \mathrm{\Delta \mathrm{H}=-\mathrm{ve}}

      

\mathrm{E_{a_p}=} Activation energy of the forward reaction
\mathrm{E_{a_B}=} Activation energy of the backward reaction
Enthalpy change can be expressed also in terms of activation energy,
\mathrm{ \Delta H=E_{a_F}-E_{a_B} \\ }

\mathrm{ \Delta H=E_f-E_b \\ }

\mathrm{ \Delta H=80-280 \\ }

\mathrm{ \Delta H=-200 \mathrm{~kJ} / \mathrm{mol} }

Posted by

Ramraj Saini

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