Consider the reaction, Cl2 + H2S => 2 HCl + S, which is found to be first order in Cl2. Which step of the proposed mechanism must be slow in order to agree with this rate law? Cl2 => 2 Cl Cl + H2S => HCl + HS Cl + HS => HCl + S
A. only 3
B. only 2
C. 1
D. Either 2 or 3

Respuesta :

Answer:

Second step of the mechanism is slow step.

Explanation:

The given chemical reaction is as follows

[tex]Cl_{2}+H_{2}S\rightarrow 2HCl+S[/tex]

The mechanism of the reaction is as follows.

[tex]Cl_{2}\rightarrow 2Cl[/tex]

[tex]Cl+H_{2}S\rightarrow HCl+HS[/tex]

[tex]Cl+HS\rightarrow HCl+S[/tex]

The rate of the reaction.

[tex]rate=k[Cl_{2}]^\frac{1}{2}[H_{2}S][/tex]

Therefore, Step -2 must be slow.

Second step of the mechanism is slow step.

B. Only 2

Rate of reaction:

It is the speed at which a chemical reaction takes place, defined as proportional to the increase in the concentration of a product per unit time and to the decrease in the concentration of a reactant per unit time.

Chemical reaction:

[tex]Cl_2 + H_2S --> 2 HCl + S[/tex]

The reaction mechanism  is as follows.

[tex]Cl_2-->2Cl\\\\Cl+H_2S--->HCl+HS\\\\Cl+HS-->HCl+S[/tex]

Second step is the slowest step, thus the rate determining step.

Therefore, rate of reaction can be represented as:

[tex]\text{Rate}= k[Cl_2]^{(1/2)}[H_2S][/tex]

Thus, option B is correct.

Find more information about Rate of reaction here:

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