. When a newspaper is left in direct sunlight for a few days, the paper begins to turn yellow. The yellow

Consider the following multistep reaction:

chubhunter [2.5K]

The question is incomplete, here is the complete question:

Consider the following multistep reaction:

C+D⇌CD (fast)

CD+D→CD₂ (slow)

CD₂+D→CD₃ (fast)

C+3D→CD₃

Based on this mechanism, determine the rate law for the overall reaction.

Answer: The rate law for the reaction is $\text{Rate}=k'[C][D]^2$

Explanation:

Rate law is the expression which is used to express the rate of the reaction in terms of the molar concentration of reactants where each term is raised to the power their stoichiometric coefficient respectively from a balanced chemical equation.

In a mechanism of the reaction, the slow step in the mechanism determines the rate of the reaction.

For the given chemical reaction:

$C+3D\rightarrow CD_3$

The intermediate reaction of the mechanism follows:

Step 1: $C+D\rightleftharpoons CD;\text{ (fast)}$

Step 2: $CD+D\rightarrow CD_2;\text{(slow)}$

Step 3: $CD_2+D\rightarrow CD_3;\text{(fast)}$

As, step 2 is the slow step. It is the rate determining step

Rate law for the reaction follows:

$\text{Rate}=k[CD][D]$ ......(1)

As, [CD] is not appearing as a reactant in the overall reaction. So, we apply steady state approximation in it.

Applying steady state approximation for CD from step 1, we get:

$K=\frac{[CD]}{[C][D]}$

$[CD]=K[C][D]$

Putting the value of [CD] in equation 1, we get:

$\text{Rate}=k.K[C][D]^2\\\\\text{Rate}=k'[C][D]^2$

Hence, the rate law for the reaction is $\text{Rate}=k'[C][D]^2$

5 0

3 years ago

What is the correct way to write 1,550,000,000 in scientific notation?

mr_godi [17]

Answer: a

Explanation:

scientific notation is about how many places passed the decimial place you are moving. therefor in order to get 1,550,000,000 you would move the decimial from 1.55 over nine times to the right

4 0

4 years ago

Read 2 more answers

Each of the chemically active Period 2 elements forms stable compounds in which it has bonds to fluorine.(d) Draw Lewis structur

Pavel [41]

Lithium is the first element of period 2 which reacts with fluorine to form LiF ( lithium fluoride ) . it is an inorganic compound . it is also a colorless solid . it is less soluble in water . it is chemically stable because of its comparable molecular mass .

Beryllium is the second element of period 2 which reacts with fluorine to give beryllium difluoride (BeF2) . it is inorganic compound . it is highly soluble in water. it is also a stable compound . it have low melting point .

Boron is the third element of period 2 which reacts with fluorine to form

BF3 (Boron trifluoride ) . it is a inorganic compound . it is colorless and toxic gas forms . it is stable in dry atmosphere but its octet is not satisfied .

Carbon is the 4th element of the period 2 which reacts with fluorine to form carbon tetrafluoride (CF4) . it is not soluble in water . it is a greenhouse gas . it dissolves in oil. it is very stable compound .it forms covalent bond .

Nitrogen is the 5th element of period 2 which reacts with fluorine to form nitrogen trifluoride (NF3) . it is also a inorganic compound . it is colorless and non-flammable . it is a stable gas at room temperature .

Oxygen is the 6th element of period 2 which reacts with fluorine to form oxygen difluoride (OF2) . it is colorless poisonous gas . it is partially stable or relatively stable .

Neon is a noble gas and also a stable element . it is odorless and colorless . so it is nonreactive . so it doesn't form bond with fluorine .

<h3>Learn more about Fluorine here :</h3>

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