Question

At elevated temperatures, in the absense of a catalyst, nitrous oxide decomposes by a first order proces according the the equation: 2N₂O (g) --> 2N₂ (g) + O₂ (g). From an experiment at 430 degrees Celsius, k is found to be 3.8 x 10^-5 s^-1; at 700 degrees Celsius, k is found to be 1.0 s^-1.

a. Using the two-point version of the linearized Arrhenius equation, please find the activation energy (kJ/mol) for the decomposition of N₂O (g).

b. Given that the decomposition is first order, please write the rate law.

c. Using your rate law information, please write a mechanism that proceeds in three elementary steps, with the first being rate-determining, the third step being the fastest, and the intermediate N₂O₂ being formed in the second step. Please write the rate law next to each step. Please write the overall reaction underneath the three steps and make sure it all "adds up."

d. On a grid, to scale, please sketch the potential energy vs. reaction progress profile for the first order decompositon of N₂O (g). Please clearly label the graph axes, the reactants and products, E_a, and ΔE_reaction.

Answer 1

From Arhenius equation, ln K= lnKo-Ea/RT

Where K =Rate constant, Ko= Frequency factor, T= Temperature in K, R= 8.314 J/mole.K, Ea= activation energy.

At two different temperatures, the equation can be written as

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

K2= 1/s and K1= 3.8*10-5/s, T1= 430deg.c= 430+273=703K, T2= 700+273=973K

Ln(1/3.8*10-5)= (Ea/R)*(1/703-1/973)

Ea= 214375 J/mole=214.375 KJ/mole

b) The rate law (r) is = K[N2O]

c) N2O--->N2+1/2O2 ( step-1), slowest step( rate determining step)

N2O+1/2O2<-<àN2O2 ( step-2),

N2O2------> N2+O2 fast and equilibrium

d)