Question

A certain reaction has an activation energy of 67.72 kJ/mol. At what Kelvin temperature will the reaction proceed 4.00 times faster than it did at 353 K?

Answer 1

Rate of a reaction is:\[kA^n*B^m.....\] with each uppercase letter being the concentration of each reactant. Since we are dealing with only temperature, we only have to worry about k, which is dependent upon temperature. Using the Arrhenius equation: \[K = Ae^x , x= -E/RT, \] with K as the rate constant, E as the activation energy, R as the gas constant, and T the temperature. We want a reaction that proceeds 4 times faster than at 353 K, or having a rate constant K 4 times the original. Thus, we want a temperature that satisfies this. Using T* = (ln 4) *T we get \[K* = Ae^(-\ln(1/4)x = 4e^-x= 4K\] since e^-(ln(1/4)= e^ln4= 4. Thus, the temperature needed in terms of Degrees Kelvin is \[353 \ln 4 \], or approximately 489.362.