In an exothermic reaction $A \to B$ , the activation energy of reverse reaction is twice that of forward reaction. If enthalpy of the reaction is -80 kJ $m o l^{- 1}$ , the activation energy of the reverse reaction is :

$80.0 k J m o l^{- 1}$

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$60.0 k J m o l^{- 1}$

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$40.0 k J m o l^{- 1}$

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$160.0 k J m o l^{- 1}$

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Solution

The correct option is D
$160.0 k J m o l^{- 1}$

$E_{a} =$ Activation energy for forward reaction, $E_{a}^{'}$ that for reverse reaction. Then $E_{a} - E_{a}^{'} = Δ H$
Or, $E_{a} - 2 E_{a} = - 80, E_{a} = 80 k J, E_{a}^{'} = 2 \times 80 k J$

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In an exothermic reaction A→B, the activation energy of reverse reaction is twice that of forward reaction. If enthalpy of the reaction is -80 kJ mol−1, the activation energy of the reverse reaction is :

The correct option is D 160.0kJ mol−1 Ea= Activation energy for forward reaction, E′a that for reverse reaction. Then Ea−E′a=ΔH Or, Ea−2Ea=−80,Ea=80 kJ,E′a=2×80 kJ

In an exothermic reaction $A \to B$ , the activation energy of reverse reaction is twice that of forward reaction. If enthalpy of the reaction is -80 kJ $m o l^{- 1}$ , the activation energy of the reverse reaction is :

The correct option is D
$160.0 k J m o l^{- 1}$

$E_{a} =$ Activation energy for forward reaction, $E_{a}^{'}$ that for reverse reaction. Then $E_{a} - E_{a}^{'} = Δ H$
Or, $E_{a} - 2 E_{a} = - 80, E_{a} = 80 k J, E_{a}^{'} = 2 \times 80 k J$