The experimental rate law for the reaction 2NO_2 (g) + O_3 (g) rightarrow N_2O_5

Question

The experimental rate law for the reaction 2NO_2 (g) + O_3 (g) rightarrow N_2O_5 (g) + O_2 (g) has been determined to be: Rate = k[NO_Z]^Z[O_3]. Which of the following mechanisms is consistent with this rate law? NO_2 + O_3 NOs (fast equilibrium) NOs + NOs rightarrow N_2O_5 + 5/2O_2 (slow) NO_2 + NO_2 N_2O_4 (fast equilibrium) N_2O_4 + O_3 rightarrow N_2O_5 + O_2 (slow) NO_2 + O_3 rightarrow NO_3 + O_2 (slow) NO_3 + NO_2 rightarrow N_2O_5 (fast) NO_2 + NO_2 rightarrow N_2O_2 + O_2 (slow) N_2O_2 + O_3 rightarrow N_2O_5 (fast) None of the above mechanisms are consistent with this rate law. For questions 29-30 consider a class of biological catalysts called enzymes. Enzymes are complex biological molecules that help to speed up complex chemical processes in the body. The specificity of an enzyme comes from its unique 3-dimensional structure as shown in the image to the right. Almost every metabolic process in the body requires an enzyme in order to occur at rates that are fast enough to allow the body to function at our normal physiological temperature of 37 degree C. If the activation energy for a certain biological reaction is 50.0 kJ/mol in a healthy human body, by what factor would the rate of the reaction increase if you have a fever of 40 degree C. 2.0 times 10^4 2.0 1.2 1.0 0.83 The enzyme carbonic anhydrase catalyzes the hydration of carbon dioxide m your blood according to the equilibrium shown below: CO_2(g) + H_2O(l) box H_2CO_3(aq) box HCO_3^-1(aq) In the absence of the carbonic anhydrase enzyme and at normal physiological temperature [37 degree C] the rate constant for this reaction is 0.1 s^-1. By how many kilojoules does carbonic anhydrase lower the activation energy for this reaction? 581 kJ 43.1 kJ 25.7 kJ 6.48 kJ 4.83 kJ.

Explanation / Answer

If rate is:

R = K[NO2]^2 [O3]

then...

we must consider REACTANTS of the SLOW rate, since those are the one that determine the overall reaction

so

a. cant be since no O3 reactant

b) cant be since N2O4 is not included

c) cant be since N2O has no 2 as stoichiometric coefficeint, so no ^2 will be found

d) can't be since no O3

therefore, there is NONE of mechanism above consistent

q29

Ea = 50 kJ/mol at 37C

then

factor:

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

ln(xK1/K1) = 50000/8.314 * (1/(37+273) - 1/(273+40))

x = exp(0.18594) = 1.20434

this will be 1.20434 faster

since K2 = xK1 and x = 1.20434

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