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The activation energy of H(2)+I(2) hAr...

The activation energy of
`H_(2)+I_(2) hArr 2HI(g)` in equilibrium for the forward reaction is `167 kJ mol^(-1)` whereas for the reverse reaction is `180 kJ mol^(-1)`. The presence of catalyst lowers the activation energy by `80 kJ mol^(-1)`. Assuming that the reactions are made at `27^(@)C` and the frequency factor for forwatd and backward reactions are `4xx10^(-4)` and `2xx10^(-3)` respectively, calculate `K_(c)`.

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To solve the problem, we will follow these steps: ### Step 1: Determine the new activation energies for both reactions The activation energy for the forward reaction is given as 167 kJ/mol. The presence of a catalyst lowers this by 80 kJ/mol. \[ \text{New Activation Energy for Forward Reaction (E_a1)} = 167 \, \text{kJ/mol} - 80 \, \text{kJ/mol} = 87 \, \text{kJ/mol} \] ...
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