In a study of the decomposition of hydrogen iodide on a gold surface at 150 °C H

Question

In a study of the decomposition of hydrogen iodide on a gold surface at 150 °C HI(g)- H2(g) + ½ 12(g) the following data were obtained: [HI], M seconds 8.45x102 782 4.23×10-2 913 0.338 0.169 522 Hint: It is not necessary to graph these data. The observed half life for this reaction when the starting concentration is 0.338 M is s and when the starting concentration is 0.169 M is The average rate of disappearance of HI from t = 0 s to t = 522 s is Ms The average rate of disappearance of H1 from t = 522 s to t = 782 s is Ms Based on these data, the rate constant for this order reaction is

Explanation / Answer

(1)

Half life time is the time required to convert half of the reactant in to product.

SO,

Half life time when the intial concentration is 0.338 M = 522 s

And

Half life time when the initial concentration is 0.169 M = 782 - 522 = 260 s

(2)

Rate of disappearance of HI = - d[HI] / dt = - ( 0. 169 - 0.338 ) / 522 = 0.000324 M.s-1

(3)

Rate of disappearance of HI = - ( 0.0845 - 0.169 ) / ( 782 - 522 ) = 0.000325 M.s-1

(4)

Since the rate of reaction is independent on the initial concentration of reactant,

the reaction follows zero order reaction mechanism.

Integrated rate consrant for zero order reaction is,

[A] = [A]0 - kt

0.169 = 0.338 - k (522)

k = 0.000324 M.s-1

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