Electron transfer from an aqueous H^+/H_2 couple at a nickel, Ni, electrode foll
ID: 1061132 • Letter: E
Question
Electron transfer from an aqueous H^+/H_2 couple at a nickel, Ni, electrode follows the Butler-Volmer equation (j = j_0{e^(1 - alpha)^fn - e^-alpha fn), where f = F/RT and represents overpotential) with an exchange current density of 6.3 times 10^-6 A cm^-2 and a transfer coefficient of 0.58 at 25 degree C. Determine the current density for an overpotential of 0.50 V at this temperature. The decomposition of N_2O_5(g) follows a first order kinetics. The half-life of the reaction at 25 degree C was measured as 10.000 s. (a) calculated the rate constant of the reaction; (b) calculate the needed to reduce the concentrate of N_2O_5 to 1/s of the initial value. The gas-phase oxidation of nitric oxide (nitrogen monoxide, NO) has an overall reaction written as: 2 NO(g) + O_2(g) rightarrow 2 NO_2(g). The reaction has an overall third order with a rate law rate = k_r[NO]^2[O_2]. The reaction goes through the following elementary steps: Derive the above rate law for NO_2 formation using the steady-state approximation; (b) Assuming all the rate constants obeys Arrhenius equation, drive a relationship that relates the "apparent" activation energy in k_r to the activation energies of the elementary steps.Explanation / Answer
II. for a first order reaction,
rate constant k,
k = ln(2)/t1/2
with,
t1/2 = 10,000 s
we get,
k = ln(2)/10,000 = 6.93 x 10^-5 s-1
(b) For first order kinetics,
time required t,
t = (ln[Ao] - ln[A])/k
So to reduce the concentration to half the time required would be,
t = (ln(100) - ln(50))/6.93 x 10^-5
= 10002.124 s
III. From the given set of equations
(a) d[NO2]/dt = 2kb[N2O2][O2]
Applying steady state rule for [N2O2]
ka[NO]^2 = ka'[N2O2] + 2kb[N2O2][O2]
[N2O2] = ka[NO]^2/(ka' + 2kb[O2])
feeding in the rate equation,
d[NO2]/dt = 2kbka[NO]^2.[O2]/(ka' + 2kb[O2])
Is the final equation for formation of NO2
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