The rate constant for a first order reaction is 4.60 x 10-4 s -1 at 350°C. If the activation energy is 104 kJ mol-1 , calculate the temperature at which its rate constant is 5.28 x 10-2 min-1

Answer 1

Use the Arrhenius equation:

ln (k2/k1) = -Ea/R (1/T2 - 1/T1)

k1 = 4.60x10^-4s^-1

k2 = 5.28x10^-2s^-1 (I'm assuming you mean per second, not per minute as written)

Ea = 104 kJ/mol

R = 8.314 J/molK = 0.008314 kJ/molK

T1 = 350ºC + 273 = 623K

T2 = ?

ln (5.28x10^-2 / 4.60x10^-4) = -104 / 0.008314 (1/T2 - 1/623)

4.74 = -12,498 (1/T2 - 0.001605)

4.74 = -12498/T2 + 20.06

-15.3 = -12498/T2

T2 = 817K = 544ºC

(be sure the check the math)