The activation energy for the gas phase isomerization of dimethyl citraconate is 105 kJ. The rate constant at 685 K is 9.67x10^-4. The rate constant will be at 731 K.

Question

Please give steps, Include how to solve for ln(k2/k1)

Answer 1

Mary

Now that you have provided the other value of k, we can proceed using the Arrhenius equation:

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

k1 = 9.67x10^-4

k2 = ?

Ea = 105 kJ

R = 8.314 J/Kmol = 0.008314 kJ/Kmol (converted to kJ to be consistent with units of Ea)

T1 = 685K

T2 = 731K

ln(k2/9.67x10^-4) = - 105/0.008314 (1/731 - 1/685)

ln(k2/9.67x10^-4) = - 12629 (0.001370 - 0.001460)

ln(k2/9.67x10^-4) = - 12629 x -9x10^-5

ln(k2/9.67x10^-4) = +1.14

Now take the natural log of both sides to get

ln k2 - ln 9.67x10^-4 = ln 1.14

ln k2 = ln 1.14 + ln 9.67x10^-4

ln k2 = 0.131 -6.94

ln k2 = -6.81

Take anti ln of -6.81 to get k2

k2 = 1.10x10^-3

Does this make sense? k2 is the rate constant at 731K, so it should be greater than the rate constant k1 at 685K because as temperature increases, rate increases. So, the answer does make sense.

Answer 2

So -- rate(T) = k(arbitrary)exp(-.delta.G(act)/RT)

You should be able to take ln of both sides of this equation, surely?

ln(k2/k1) = (-.delta.G(act)/R)*(1/T2 - 1/T1)

Remember that ratioing on an exponent, you add / subtract the multiplied / divided portions of the exponent argument. If you're rusty on that, please review it, you'll need to use that info going forward (or backwards, if it's an equilibrium!!). Incidentally, you will note that the activation energy is different in the forward and reverse directions, in general -- that implies that approach to equilibrium is at different speeds from all reactants vs. all products. But, reactions still get there eventually!

Also remember, T is in K.

-- Cheers, --Mr. d.