A catalyst lowers the activation energy of a reaction from 20 kJ mole–1 to  10 kJ mole–1. The temperature at which the uncatalysed reaction will have the same rate as that of the catalysed at 27°C is

Q:  A catalyst lowers the activation energy of a reaction from 20 kJ mole–1 to  10 kJ mole–1. The temperature at which the uncatalysed reaction will have the same rate as that of the catalysed at 27°C is

(A) –123°C

(B) 327°C

(C) 327°C

(D) + 23°C

Solution: $\large \frac{E_a’}{T_1} = \frac{E_a}{T_2}$

$\large \frac{10}{300} = \frac{20}{T_2}$

T2 = 600 K = 327° C

Ans: (B)