Question

A catalyst lowers the activation energy of a reaction from 20 kJ $mo{l}^{-1}to10kJmo{l}^{-1}.$ The temperature at which the uncatalyzed reaction will have the same rate as that of the catalyzed at ${27}^{\circ }C$ is:

• A. $-{123}^{\circ }C$
• B. ${327}^{\circ }C$
• C. ${32.7}^{\circ }C$
• D. $+{23}^{\circ }C$

Answer 1

The correct option is B ${327}^{\circ }C$

As we know,
$\frac{E^{\prime }}{T_1}=\frac{E_a}{T_2}$
$\frac{10}{300}=\frac{20}{T_2}$
$\therefore T_2$ = 600 K = ${327}^{\circ }C$