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The energy change accompanying the equil...

The energy change accompanying the equilibrium reaction A `iff` B is `-33.0" kJ mol"^(-1)`. Calculate
(i) Equilibrium constant `K_(c)` for the reaction at 300 K
(ii) Energy of activation for forward and backward reaction `(E_(f)" and "E_(b))` at 300 K. Given that `E_(f)` and `E_(b)` are in the ratio 20 : 31.
Assume that pre-exponential factor is same for forward and backward reaction.

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AI Generated Solution

To solve the problem, we will follow these steps: ### Step 1: Calculate the Equilibrium Constant \( K_c \) Given the energy change for the reaction \( A \iff B \) is \( \Delta E = -33.0 \, \text{kJ mol}^{-1} \), we can use the relationship between the equilibrium constant and the Gibbs free energy change: \[ \Delta G = -RT \ln K_c ...
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