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NikAS [45]
4 years ago
11

Hydrogen sulfide (H2S) is a common and troublesome pollutant in industrial wastewaters. One way to remove H2S is to treat the wa

ter 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?
Chemistry
1 answer:
Kryger [21]4 years ago
5 0

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.

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