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Question

Chemistry Question on Chemical Kinetics

From the above figure, the activation energy for the reverse reaction would be

A

−120KJ mol−1-120 KJ \, mol^{-1}

B

+152KJ mol−1+152 KJ \, mol^{-1}

C

+120KJ mol−1+120 KJ \, mol^{-1}

D

+1760KJ mol−1+1760 KJ \, mol^{-1}

Answer

+1760KJ mol−1+1760 KJ \, mol^{-1}

Explanation

Solution

Given ΔH=−120 KJ mol−1;Ea=1640KJ mol−1\Delta H=-120 \, KJ \, mol^{-1}; E_{a}=1640 KJ \, mol^{-1} For reverse reaction, ΔH=+120 KJ mol−1\Delta H=+120 \, KJ \, mol^{-1} ∴ \therefore\, Activation energy for the reverse reaction =1640+120=+1760 KJ mol−1=1640+120=+1760 \, KJ \, mol^{-1}