For a reversible reaction A B, the H_forward reaction = 20 ~kJ ~mol^-1. The activation energy of the uncatalysed forward reaction is 300 ~kJ ~mol^-1. When the reaction is catalysed keeping the reactant concentration same, the rate of the catalysed fo

For a reversible reaction A leftharpoons B the Delta Hforwar: null (Chemistry)

Using the Arrhenius equation:

k = Ae^{-E_a/RT}

where

k

is the rate constant,

A

is the pre-exponential factor,

E_a

is the activation energy,

R

is the gas constant, and

T

is the temperature in Kelvin.

The equation is used for both the uncatalyzed forward reaction and the catalyzed backward reaction. By setting the two equations equal to each other and cancelling out the pre-exponential factor, we get:

e^{\frac{300 \times 10^3}{600 \times R}} = e^{\frac{-E_a}{300 \times R}}

Simplifying the equation, we get:

\frac{300 \times 10^3}{600 \times R} = \frac{E_a}{300 \times R}

Solving for

E_a

, we get:

E_a = \frac{10^3}{2}\times 300 = 150 \times 10^3\ \mathrm{J~mol^{-1}} = 150\ \mathrm{kJ~mol^{-1}}

This gives us the activation energy for the uncatalyzed forward reaction.

To find the activation energy for the catalyzed backward reaction, we use the relationship:

E_{\text{rev,catalysed}} = E_{\text{fwd,uncat}} - \Delta H_{\text{forward reaction}}

Substituting the given values, we get:

E_{\text{rev,catalysed}} = 150\ \mathrm{kJ~mol^{-1}} - 20\ \mathrm{kJ~mol^{-1}} = 130\ \mathrm{kJ~mol^{-1}}

Therefore, the activation energy for the catalyzed backward reaction is

\boxed{130\ \mathrm{kJ~mol^{-1}}}

Explanation

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