Question

a.)A certain reaction has an activation energy of 25.10 kJ/mol. At what Kelvin temperature will the reaction proceed 7.00 times faster than it did at 289 K?
b.A certain reaction has an enthalpy of ΔH = 39 kJ and an activation energy of Ea = 51 kJ. What is the activation energy of the reverse reaction?
c.)At a given temperature, the elementary reaction A<=> B in the forward direction is the first order in A with a rate constant of 3.40 × 10-2 s–1. The reverse reaction is first order in B and the rate constant is 5.80 × 10-2 s–1.What is the value of the equilibrium constant for the reaction A<=>B at this temperature. What is the value of the equilibrium constant for the reaction B<=> A at this temperature.
d.) consider reaction mechanism: step 1: A<=> B+C equilibrium
step 2: C+D-> E slow
overall: A+D-> B+E and determine the rate law overall. (wrong answer would be =k[A][D])
Why?

Answer 1

To do part a) use arrhenius equation:

ln (K1/K2) = Ea/R(1/T2 - 1/T1)

And we want a reaction 7 times faster than the first, and the temperature that it reached. In other words, T2 when K2 = 7K1 so:

ln (K1/7K1) = 25100/8.3144(1/T2-1/289)

ln(1/7) = 3018.86(1/T2 - 0.00346)

-1.9459/3018.86 = 1/T2 - 0.00346

-0.000645 = 1/T2 - 0.00346

1/T2 = 0.00346 - 0.000645

T2 = 1/0.002815

T2 = 355.23 K

For PArt b) In a reversible reaction, the difference between the activation energy of the forward and reverse rxns. is equal to the heat of the rxn.

delta H = AE of fwd - AE of rvrs

AE rvrs = AEfwd - H

AE rvrs = 51 - 39 = 12 kJ

Post part c and d in another question thread.

Hope this helps