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Chemical Kinetics

Question
CBSEENCH12006323

The energy of activation for a reaction is 100 kJ mol–1. The presence of a catalyst lowers the enregy of activation by 75%. What will be the effect on rate of reaction at 20% C, other things being equal.

Solution

According to the Arrhenius equation 

K = Ae–Ea/R
In absence of catalyst k1 = Ae–100/RT
In presence of catalyst k2 = Ae–25/RT

           k1k2 = e-100/RTe-25/RT=e-75/RT 

or       2.303 log10k1k2 = (75/RT)

or         2.303 log10 k1k2 = 758.314×10-3×298

                  k1k2 = 2.35 × 1013


Since Rate k[ ]
n at any temperature for a reaction n and concentration of reactants are same and temperature changes.

r2r1 = k2k1 = 2.35 × 1013