The energies of activation for forward and reverse reactions for A 2 + B 2 ⇌ 2 A B are 180 kJ mol − 1 and 200 kJ mol − 1 respectively. The presence of catalyst lowers the activation energy of both

Question

The energies of activation for forward and reverse reactions for $A_{2} + B_{2} ⇌ 2 A B$ are $180 kJ mol^{- 1}$ and $200 kJ mol^{- 1}$ respectively. The presence of catalyst lowers the activation energy of both (forward and reverse) reactions by $100 kJ mol^{- 1}$ . The enthalpy change of the reaction $(A_{2} + B_{2} ⟶ 2 AB)$ in the presence of catalyst will be (in $kJ mol^{- 1}$ )

Question diagram: The energies of activation for forward and reverse reactions

Options:

Answer:

D

Solution:

So, $Δ H_{Reaction} = E_{f} - E_{b}$ $= 80 - 100 = - 20$ Hence, (D) is correct.

TestHub · Accepted Answer

So, Δ H Reaction ​ = E f ​ − E b ​ = 80 − 100 = − 20 Hence, (D) is correct.

Chemistry - Chemical Kinetics Question with Solution | TestHub

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