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OLEGan [10]
3 years ago
14

Consider the following elementary steps that make up the mechanism of a certain reaction: 3A→B+C B+2D→C+F Part A What is the ove

rall reaction? Express your answer as a chemical equation. View Available Hint(s) nothing Part B Which species is a reaction intermediate? View Available Hint(s) Which species is a reaction intermediate? A B C D F Part C What is the rate law for step 1 of this reaction? Express your answer in standard MasteringChemistry notation. For example, if the rate law is k[A][C]3 type k*[A]*[C]^3. View Available Hint(s) Rate = Part D What is the rate law for step 2 of this reaction? Express your answer in standard MasteringChemistry notation. For example, if the rate law is k[A][C]3 type k*[A]*[C]^3. View Available Hint(s) Rate = Provide Feedback
Chemistry
1 answer:
saveliy_v [14]3 years ago
7 0

Answer:

A.    3A + 2D → 2C + F

B.    B is an intermediate

C.   rate = k[A]³

 D.  rate = k`[D]²

Explanation:

Part A

The overall reaction is the sum of all the elementary steps:

       3A→B+C

+

       B+2D→C+F

<u>                                         </u>

         3A + 2D → 2C + F

Part B

A reaction intermediate is one that participates in the elementary steps, but it is not a product. Notice B is produced in the first step and consumed in the second, therefore it is an intermediate species.

Part C

The rate law is  the product of the reaction constant, k, times the concentration of the reactants raised to the power of their coefficients in the balanced reaction assuming that it is the slow step in the reaction mechanism.

rate = k[A]³

Part D

Again as in part C:

rate = k [D]²  ( B is not included since as we saw it is an intermediate )

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3 years ago
Mendeleev organized the elements:
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Answer: (3)

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4 0
4 years ago
A chemist titrates 130.0mL of a 0.4248 M lidocaine (C14H21NONH) solution with 0.4429 M HBr solution at 25 degree C . Calculate t
jeka57 [31]

Answer:

pH = 3.36

Explanation:

Lidocaine is a weak base to be titrated with the strong acid HBr, therefore at equivalence point we wil have the protonated lidocaine weak conjugate acid of lidocaine which will drive the pH.

Thus to solve the question we will need to calculate the concentration of this weak acid at equivalence point.

Molarity = mol /V ∴ mol = V x M

mol lidocaine = (130 mL/1000 mL/L) x 0.4248 mol/L = 0.0552 mol

The volume of 0.4429 M HBr required to neutralize this 0.0552 mol is

0.0552 mol x  (1L / 0.4429mol) = 0.125 L

Total volume at equivalence is  initial volume lidocaine + volume HBr added

0 .130 L +0.125 L = 0.255L

and the concentration of protonated lidocaine at the end of the titration will be

0.0552 mol / 0.255 L = 0.22M

Now to calculate the pH we setup our customary ICE table for  weak acids for the equilibria:

protonated lidocaine + H₂O   ⇆  lidocaine + H₃O⁺

                      protonated lidocaine          lidocaine        H₃O⁺

Initial(M)               0.22                                       0                  0

Change                   -x                                      +x                 +x

Equilibrium          0.22 - x                                  x                    x

We know for this equilibrium

Ka = [Lidocaine] [H₃O⁺] / [protonaded Lidocaine] =  x² / ( 0.22 - x )

The Ka can be calculated from the given pKb for lidocaine

Kb = antilog( - 7.94 ) = 1.15 x 10⁻⁸

Ka = Kw / Kb = 10⁻¹⁴ / 1.15 x 10⁻⁸  = 8.71 x 10⁻⁷

Since Ka is very small we can make the approximation 0.22  - x  ≈ 0.22

and solve for x. The pH  will then  be the negative log of this value.

8.71 x 10⁻⁷  = x² / 0.22 ⇒ x = √(/ 8.71 X 10⁻⁷ x 0.22) = 4.38 x 10⁻⁴

( Indeed our approximation checks since 4.38 x 10⁻⁴ is just 0.2 % of 0.22 )

pH = - log ( 4.4x 10⁻⁴) = 3.36

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