The evaporation of a 120-nm film of n-pentane from a single crystal of aluminum

Question

The evaporation of a 120-nm film of n-pentane from a single crystal of aluminum oxide is zero order with a rate constant of 1.92 times 10^13 molecules/cm^2 middot s at 120 K. a. If the initial surface coverage is 8.9 times 10^16 molecules/cm^2 how long will it take for one-half of the film to evaporate? b. What fraction of the film is left after 10 s? Assume the same initial coverage as in part a. The kinetics of this reaction were studied as a function of temperature. (The reaction is first order in each reactant and second order overall.) C_2H_5Br(aq) + OH^- (aq) rightarrow C_2H_5OH(l) + Br^-(aq)

Explanation / Answer

98.

a) for a zero-order reaction,

[A]t = [A]o - kt

with,

[A] = 8.9 x 10^16 molecules/cm^2

[A]o = 8.9 x 10^16/2 = 4.45 x 10^16 molecules/cm^2

k = 1.92 x 10^13 molecules/cm^2.s

we get,

time t required for initial concentration to reduce to half by evaporation,

t = (8.9 x 10^16 - 4.45 x 10^16)/1.92 x 10^13 = 2317.71 s

b) fraction of film [A]t left after t = 10 s

would be,

[A]t = 8.9 x 10^16 - 1.92 x 10^13 x 10

      = 8.881 x 10^16

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