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For a given reaction, energy of activation for forward reaction (Eaf) is80 \mathrm{~kJ} \mathrm{mol-1.} Enthalpy change ,\Delta \mathrm{H}=-40 \mathrm{~kJ} \mathrm{~mol}^{-1} or the reaction. A catalyst lowers Eaf  to  20 \mathrm{kJmol}^{-1}  The ratio of energy of activation for reverse reaction before and after addition of catalyst is:

Option: 1

1


Option: 2

0.5


Option: 3

1.5


Option: 4

2.0


Answers (1)

best_answer

Enthalpy change,  \mathrm{\Delta H=E_f-E_b \Rightarrow-40=80-E_b }
                                \mathrm{E_b=120 \mathrm{~kJ} / \mathrm{mal} }
The catalyst lowers the \mathrm{E_{a f} } to  \mathrm{20 \mathrm{~kJ} / \mathrm{mol} }  for forward reaction then                                                              \mathrm{E_{\mathrm{af}}^{\prime}=20 \mathrm{~kJ} / \mathrm{mol} }
As we know that a catalyst decreases the activation energy by equal amount in both direction, the decrease in both the directions is by  \mathrm{60 \mathrm{~kJ} / \mathrm{mol} }
                                 \mathrm{ E_b^{\prime}=(120-60)=60 \mathrm{~kJ} / \mathrm{mol} } 
\mathrm{\frac{E_{\mathrm{b}}}{E_{\mathrm{b}}^{\prime}}=\frac{120}{60}=2.0 }.

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