The oxidation of nitric oxide (nitrogen monoxide. NO) has an overall reaction wr

Question

The oxidation of nitric oxide (nitrogen monoxide. NO) has an overall reaction written 2NO(g) + O_2(g) rightarrow 2NO_2(g) (a) The gas-phase reaction goes through the following elementary step: NO + NO N_2O_2 N_2O_2 NO + NO N_2 O_2 O_2 2NO_2 Use the steady stale approximation to derive the rate law for NOj formation to phase. (b) A metal catalyst such as Pt can catalyze the oxidation of NO. The metal catalyzed NO oxidation goes through a Langmuir-Hinshelwood mechanism consists of the following elementary steps: BO + NO^- K_NO y_2O_2 + O* K_O_2 NO + O* NO_2(g) + 2* Derive the rate law based on the Langmuir-Hinshelwood mechanism. (e) Compare the rate laws obtained for gas-phase and surface-catalyzed NO oxidations and discuss the difference in the context of reaction mechanism. The exchange current density and the electron transfer coefficient for the reaction 2H^+ + 2z H_2(g) on nickel at 25 degree C are 6.3 times 10^-6 A mc^-2 and 0.42, respectively. a) Determine at what current density to obtain an over potential of 0.25 V according to the Butler-Volmer equation. b) What will be the overpotential at the same current density according to the Tafel equation?

Explanation / Answer

For the given second order reaction,

1/[A] = 1/[Ao] +kt

k ia rate constant

1/0.02 = 1/0.06 + k x 1

k = 33.33 M-1.s-1

half-life t1/2,

t1/2 = 1/k[Ao]

       = 1/33.33 x 0.06

       = 0.50 h-1

(a) Activation energy for,

A <==> B

Using Arrhenius equation,

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

ln(2) = Ea/8.314[1/298 - 1/523]

Ea = 3.992 kJ/mol

so activation energy for reverse reaction = 3.992 kJ/mol

(b) Gas phase reaction for formation of NO2

d[NO2]/dt = 2k2[N2O2][O2]

for intermediate [N2O2],

applying steady state approximation

rate of formation = rate of decomposition

k1[NO]^2 = k-1[N2O2] + k2[N2O2][O2]

[N2O2] = k1[NO]^2/(k1+k2[O2])

So rate of formation of [NO2],

d[NO2]/dt = 2k1k2[NO]^2.[O2]/(k1 + k2[O2])

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