Rate Constant Question for Chemistry?

Question

If a reaction carried out at 25 degrees C has a rate constant of 3.15*10^-5 1/M*s and an activation energy of 36.6 kJ/mol, what is the temperature in degrees celcius when the rate constant is 4.80*10^-4 1/M*s?

Answer 1

Let us start from Arrhenius equation:
k = A*exp(-E/RT), where k is the rate constant, E is the activation energy, R is the gas constant, and T is the temperature in kelven.
Let us write k as K1 and k2, and T as T1 and T2, correspondingly:
k1 = A*exp(-E/RT1),
k2 = A*exp(-E/RT2),
ln(k2) - ln(k1) = (E/R)*(1/T1 - 1/T2)
This is basically a Clausius-Clapeyron relation
Put the values in, we have:
ln(4.80*10^-4) - ln(3.15*10^-5) = (36.6*10^3/8.314)*(1/298 - 1/T)
Solve for T we get T = 365(K)
Thus the requested temperature in degrees celcius is 92C.