The energy of activation for a reaction is $100 k J m o l^{- 1}$ . Presence of a catalyst lowers the energy of activation by 75%. The ratio of final rate to initial rate is:

2.34 \times 10^{- 13}

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4.34 \times 10^{- 13}

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1.34 \times 10^{- 13}

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None of these

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Solution

The correct option is D $2.34 \times 10^{- 13}$
$∵ K = A e^{- E_{a} / R T}$
In absence of catalyst $K_{1} = A e^{- 100 / R T}$
In presence of catalyst $K_{2} = A e^{- 25 / R T}$
$∴ \frac{K_{1}}{K_{2}} = \frac{e^{- 100 / R T}}{e^{- 25 / R T}} = e^{- 75 / R T}$
or $2.303 l o g_{10} (K_{2} / K_{1}) = (75 / R T) (∵ E_{A}$ in kJ and thus $R = 8.314 \times 10^{- 3} k J$ )
or $2.303 l o g_{10} \frac{K_{2}}{K_{1}} = \frac{75}{8.314 \times 10^{- 3} \times 293}$
$∴ (K_{2} / K_{1}) = 2.34 \times 10^{- 13}$
Since, $r a t e = K [R e a c t a n t]^{n}$ at any temperature for a reaction. n and concentration of reactants are same and temperature changes.
$∴ (r_{2} / r_{1}) = (K_{2} / K_{1}) = 2.34 \times 10^{- 13}$

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

Assertion:

A catalyst enhances the rate of a reaction.

Reason:

The energy of activation of the reaction is lowered in the presence of a catalyst.

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The energy of activation for a reaction is 100kJmol−1. Presence of a catalyst lowers the energy of activation by 75%. The ratio of final rate to initial rate is:

The energy of activation for a reaction is $100 k J m o l^{- 1}$ . Presence of a catalyst lowers the energy of activation by 75%. The ratio of final rate to initial rate is: