Chemistry

Problem

For a reaction with activation energy

E_a = 75\,\text{kJ mol}^{-1}

, the fraction of molecules with kinetic energy equal to or greater than

E_a

is proportional to

e^{-E_a/(RT)}

, where

R = 8.314\,\text{J mol}^{-1}\text{K}^{-1}

. If this fraction is 0.040 at 510 K, determine the temperature (in K) at which the fraction will be 0.080. Report your answer to the nearest kelvin.