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 3.5×10−2 M−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

Explanation:

As the given reaction is as follows.

     $H_{2}S(aq) + Cl_{2}(aq) \rightarrow S(s) + 2H^{+}(aq) + 2Cl^{-}(aq)$

So, according to the balanced equation, it can be seen that rate of formation of $Cl^{-}$ will be twice the rate of disappearance of $H_{2}S$ .

And, it is known that rate of disappearance of reactant will be negative and rate of formation of products will be positive value.

This means that,

Rate of the reaction = -Rate of disappearance of $H_{2}S$

                 = $k[H_{2}S][Cl_{2}]$

                 = $(3.5 \times 10^{-2}) \times (2 \times 10^{-4}) \times (2.8 x 10^{-2})$

                 = $1.96 \times 10^{-7}$ M/s

Therefore, calculate the rate of formation of $Cl^{-}$ as follows.

Rate of formation of $Cl^{-}$ = $2 \times 1.96 \times 10^{-7}$

                                        = $3.92 \times 10^{-7}$ M/s

Thus, we can conclude that the rate of formation of $Cl^{-}$ is $3.92 \times 10^{-7}$ M/s.