Question

Hydrogen sulfide (H2S) is a common and troublesome pollutant in industrial wastewaters. One way to remove H2S is to treat the water with chlorine, in which case the following reaction occurs: H2S(aq)+Cl2(aq)?S(s)+2H+(aq)+2Cl?(aq) The rate of this reaction is first order in each reactant. The rate constant for the disappearance of H2S at 28?C is 3.5×10?2M?1s?1.

If at a given time the concentration of H2S is 2.0×10^?4 M and that of Cl2 is 2.8×10^?2 M , what is the rate of formation of Cl??

Answer 1

H₂S(aq) + Cl₂ (aq)    S (s) + 2H⁺ (aq) + 2Cl^-(aq)

From the balanced equation,we can see that rate of formation of Cl^- will be twice the rate of disappearance of H₂S

[Note that rate of disappearance of reactant will be negative and rate of formation of products will be positive value]

Rate of the reaction = -Rate of disappearance of H₂S = k[H₂S][Cl₂] = (3.5 x 10^-2) x (2 x 10^-4) x (2.8 x 10^-2) = 1.96 x 10^-7 M/s

Rate of formation of Cl^- = 2 x 1.96 x 10^-7 = 3.92 x 10^-7 M/s