Sticking with the topic of sulfides, consider the kinetics of the following elem

Question

Sticking with the topic of sulfides, consider the kinetics of the following elementary reaction: H_2 S implies HS + H^+ with rate constants k_f and k_b for the forward and back reactions, respectively. Write the overall rate equation for this reaction. If k_f = 10^3 s^-1, and K_a1 = 10^-7 M, what are the value and units of k_b? What are the "orders" of k_f and k_b? And what is the "order" of the overall rate equation? If the pH=7, what are the concentrations of H_2 S and HS^- if the [S]r = 10^-4 M? What are the forward and reverse exchange rates at this pH and Sr concentration at equilibrium? if a large amount of "marsh gas" (methane; CH_4 were to bubble out of the sediments, and thereby strip much of the gaseous H_2 S out of the water, how long would the sulfide system take to return to equilibrium? if the pH of this water were to drop to 5, what would happen to the rates of the forward and back reactions? And what would happen to the relaxation time of the sulfide system?

Explanation / Answer

(a) Rate equation is

rate = k * [H2S]

(b): Given kf = 103 s-1

Ka1 = 10-7 M

ka1 = kf / kb

=> kb = kf / Ka1

=> kb = 103 s-1 / 10-7 M = 1010 M-1s-1 (answer)

Since the unit of kf is s-1, kf is 1st order reaction.

Since the unit of kb is M-1s-1, kb is second order reaction.

Hence the overall rate is 1st order.

(c): Given the pH = 7

=> -log[H+] = 7

=> [H+] = 10-7 M

Hence [H+] = [HS-] = 10-7 (answer)

Also Ka1 = 10-7 = [HS-] x [H+] / [H2S]

=> 10-7 = (10-7 M x 10-7 M) / [H2S]

=> [H2S] = (10-7 M x 10-7 M) / 10-7 = 10-7 M (answer)

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