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KatRina [158]
3 years ago
3

Consider the following elementary steps that make up the mechanism of a certain reaction: 3X→E+F E+M→F+N

Chemistry
1 answer:
romanna [79]3 years ago
6 0

Answer:

Part A- The overall reaction is: 3X + M → 2 F + N.

Part B- The intermediate of this reaction is E.

Part C- Rate law for step 1 = k[X]³ = k*[X]^3.

Part D- Rate law for step 2 = k[E][M] = k*[E]*[M].

Explanation:

<u><em>Part A: What is the overall reaction? Express your answer as a chemical equation.</em></u>

  • We can get the overall reaction by summing the two steps of the reaction.

Step 1: 3X → E + F.

Step 2: E + M → F + N.

The E component in the products side in step 1 will cancel that in the reactants side of step 2.

∴ The overall reaction is: 3X + M → 2 F + N.

<u><em>Part B: Which species is a reaction intermediate? </em></u>

  • We should identify the intermediate firstly to determine the reaction intermediate.

The intermediate is the species that produced in a step of the reaction and consumed in next steps and do not appear in the overall reaction

<em>So, the intermediate of this reaction is</em> E.

<u><em>Part C: What is the rate law for step 1 of this reaction? </em></u>

The rate of the reaction is directly proportional to the concentration of the reactants.

Rate = k[reactants]ˣ.

where, k is the rate constant of the reaction,

x is the no. of moles of the reactants that involved in the reaction mechanism before the rate determining step.

So, The rate law for step 1 of this reaction (neglecting the mechanism) is:

Rate = k[X]³ = k*[X]^3.

<u><em>Part D: What is the rate law for step 2 of this reaction? </em></u>

Also, as in part C:

The rate law for step 2 of this reaction (neglecting the mechanism) is:

Rate = k[E][M] = k*[E]*[M].

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Answer:

2.56 grams of H₂S is needed to produce 18.00g of PbS if the H2S is reacted with an  excess (unlimited) supply of Pb(CH₃COO)₂

Explanation:

The balanced reaction is:

Pb(CH₃COO)₂ + H₂S → 2 CH₃COOH + PbS

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) they react and produce:

  • Pb(CH₃COO)₂: 1 mole
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In this case,  to know how many grams of H₂S are needed to produce 18.00 g of PbS, it is first necessary to know the molar mass of the compounds H₂S and PbS and then to know how much it reacts by stoichiometry. Being:

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So, by stoichiometry they react and are produced:

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Then the following rule of three can be applied: if 239 grams of PbS are produced by stoichiometry from 34 grams of H₂S, 18 grams of PbS from how much mass of H₂S is produced?

mass of H_{2} S=\frac{18 grams of PbS*34 grams of H_{2}S }{239 grams of PbS}

mass of H₂S= 2.56 grams

<u><em>2.56 grams of H₂S is needed to produce 18.00g of PbS if the H2S is reacted with an  excess (unlimited) supply of Pb(CH₃COO)₂</em></u>

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Then you have to multiply the number of kg in a pound (which has units of kg / pound) to get a number in units of Dollars / pound.  This number is 2.2.

 

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