How long would it take for HSO 3 - to reach half its initial concentration? The

Question

How long would it take for HSO3- to reach half its initial concentration?

The oxidation of HSO_3^- by O_2 in aqueous solution is a reaction of importance to the processes of acid rain formation and flue gas desulfurization. R.E. Connick, Y.-X. Zhang, S. Lee, R. Adamic, and P. Chieng (Inorg. Chem. 34, 4543 (1995)) report that the reaction 2 HSO_3^- + O_2 rightarrow 2 SO_4^2-+ 2H^+ follows the rate law v = k[HSO_3^-]^2[H^+]^2. Given a pH of 5.6 and an oxygen molar concentration of 2.4 times 10^-4 mol dm^-3 (both presumed constant), an initial HSO_3^- molar concentration of 5 times 10^-5 mol dm^-3, and rate constant of 3.6times10^6 dm^9 mol^-3 s^-1, what is the initial rate of reaction?

Explanation / Answer

2 HSO3(-) + O2 --> 2 SO4(-2) + 2 H(+)

Given rate law :-

v = k[HSO3(-)]^2[H(+)]^2

pH of 5.6

oxygen molar concentration = 2.4 * 10^-4 mol dm^-3

initial HSO3(-) molar concentration = 5 * 10^-5 mol dm^-3

rate constant = k = 3.6 * 10^6 dm^9 mol^-3 s^-1,

pH = - log [H+]

[H+] = 10 ^ (- pH)

[H+] = 2.512 * 10 ^-6

Therefore initial rate ;-

v = k[HSO3(-)]^2[H(+)]^2

   = 3.6 * 10^6 * 5 * 10^-5 * 2.512 * 10 ^-6

   = 4.5216 * 10^-4

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