A reaction mechanism is defined as the sequence of reaction steps that define the pathway from reactants to products. Each step
in a mechanism is an elementary reaction, which describes a single molecular event of usually one or two molecules interacting. The rate law for an overall reaction is the rate law for the slowest step in the mechanism, which is directly related to the stoichiometric coefficients of the reactants.
The exception to this rule occurs when the slowest step contains intermediates. In these cases, the slowest step is usually preceded by an equilibrium step, which can be used to substitute for the intermediates in the overall rate law.
Part A
What is the rate law for the following mechanism in terms of the overall rate constant k?
Step1:Step2:A+BB+C⇌→CD(fast)(slow)
Express your answer in terms of k and the necessary concentrations (e.g., k*[A]^3*[D]).
Part B
Consider the reaction
2X2Y2+Z2⇌2X2Y2Z
which has a rate law of
rate= k[X2Y2][Z2]
Select a possible mechanism for the reaction.
Consider the reaction
which has a rate law of
Select a possible mechanism for the reaction.
Step1:Step2:Step3:Z2→Z+Z (slow)X2Y2+Z→X2Y2Z (fast)X2Y2+Z→X2Y2Z (fast)
Step1:Step2:X2Y2+Z2→X2Y2Z+Z (slow)X2Y2+Z→X2Y2Z (fast)
Step1:Step2:X2Y2+Z2→X2Y2Z2 (slow)X2Y2Z2→X2Y2Z+Z (fast)
D Step1:Step2:2X2Y2⇌X4Y4 (fast)X4Y4+Z2→2X2Y2Z (slow)
E Step1:2X2Y2+Z2→2X2Y2 (slow)
Percent composition by mass is calculated (mass of element within compound)/(mass of compound)*100. The lower the total molar mass of the compound, the greater the percent composition of sulfur. In this case, MgS would be that compound, since Mg has the lowest molar mass of the four elements bonded to S.
<span>we know that each
element has an unique spectra and it can be used to identify the
element. it shows that the energy levels of the electrons and different colors are the result of different wavelengths. hope it helps </span>
Addition of 6 M HCl would form precipitates of all the three cations, since the chlorides of these cations are insoluble: .
Firstly, the solid produced is partially soluble in hot water. Remember that out of all the three solids, lead(II) choride is the most soluble. It would easily completely dissolve in hot water. This is how we separate it from the remaining precipitate. Therefore, we know that we have lead(II) cations present, as the two remaining chlorides are insoluble even at high temperatures.
Secondly, addition of liquid ammonia would form a precipitate with silver: ; Silver hydroxide at higher temperatures decomposes into black silver oxide: .
Thirdly, we also know we have in the mixture, since addition of potassium chromate produces a yellow precipitate: . The latter precipitate is yellow.
= 18.234 grams; Silver hydroxide at higher temperatures decomposes into black silver oxide:
.
in the mixture, since addition of potassium chromate produces a yellow precipitate:
. The latter precipitate is yellow.