Given the rate constant to be 2.3E-4 s^-1 AT 40 degrees C and activation energy to be 62 kj/mol, at what temperature will the rate constant be 3.8E-2?

Question

Answer choices:

831 K

None of the above

398 K

None of the above

125 K

Answer 1

Use the Arrhenius equation:

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

k1 = 2.3x10^-4 s^-1

k2 = 3.8x10^-2 s^-1

Ea = 62 kJ/mol

R = 8.314 J/Kmol = 0.008314 kJ/Kmol (change units to be consistent with those for Ea)

T1 = 40ºC + 273 = 313K

T2 = ?

ln (3.8x10^-2 / 2.3x10^-4 ) = -62 / 0.008314 (1/T2 - 1/313)

5.11 = -7457 (1/T2 - 0.00319)

5.11 = -7457/T2 + 23.8

-18.7 = -7457/T2

T2 = 399K

T2 = 399K - 273 = 126ºC