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sladkih [1.3K]
3 years ago
8

Ethene (c2h4) can be halogenated by this reaction: c2h4( g) + x2( g) ∆ c2h4x2( g) where x2 can be cl2 (green), br2 (brown), or i

2 (purple). examine the three figures representing equilibrium concentrations in this reaction at the same temperature for the three different halogens. rank the equilibrium constants for the three reactions from largest to smallest.

Chemistry
1 answer:
scZoUnD [109]3 years ago
7 0

Answer :The correct order for equilibrium constant = K₁ > K₂ >K₃

Ethene is an alkene and X2 is elemental halogen . The reaction between alkene and halogen is known as Halogenation reaction . This reaction involves electrophilic addition mechanism . In this reaction elemental halogen (X₂ ) produces a elctrophile , X⁺ . This electrophile is attracted by the double bond in alkene and attached to it .

Due to addition of electrophile , it produces carbocation intermediate .On this positive carbon , nucleophile attacks , hence , a resultant 1,2-di halide alkane is produced .( in image )

The reaction between ethene ( C₂H₄) and halogen ( X₂) produces 1,2 - dihalide ethane (C₂H₄X₂) . The image shows reaction between chlorine ,Cl₂ , a halogen (X₂) .

The expression for equilibrium constant of reaction between ethene and halogen is given as:

K = \frac{product}{reactant }

Product = C₂H₄X₂ Reactants = C₂H₄ and X₂

K = \frac{[C_2H_4X_2]}{[C_2H_4] [X_2]}

On applying this concept in figures :

Given : Green = Cl₂ Brown = Br₂ Purple = I₂ Grey = Ethene

Grey + Green = 1,2- dihalide ethane

a) Number of Cl₂(green ) = 2

number of ethane ( grey ) = 2

Number of 1,2-dichloride alkane( grey + green ) = 8

Plugging these number in equilibrium constant expression :

K_1 = \frac{8}{2 * 2 } =  \frac{8}{4} = 2

Hence K₁ for ethene + chlorine = 2

b) Number of Br₂(brown ) = 4

number of ethane ( grey ) = 4

Number of 1,2-dibromide alkane( grey + brown ) = 6

On plugging these values in K expression as :

K_2= \frac{6 }{4 * 4 } = \frac{6}{16} =   0.375

Hence K₂ for ethen + Bromine = 0.375

c) Number of I₂(purple ) = 7

number of ethane ( grey ) = 7

Number of 1,2-diiodide alkane( grey + purple ) = 3

On plugging values in K expression as:

K_3 = \frac{3}{7 * 7 } = \frac{3}{49}   =  0.0612

Hence K₃ for ethene + Iodine = 0.0612

So , the order of equilibrium constant (K) from largest to smallest :

K₃ = 0.0612 K₂ = 0.375 K₁= 2

K₁ > K₂ >K₃

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What characteristic frequencies in the infrared spectrum of your sodium borohydride reduction product will you look for to deter
kkurt [141]

Answer:

A)The characteristic frequency to look out for is  1720-1740 cm-1 (for C=O) for which will disappear in the end product but initially present in the reactant.

B)Characteristic frequency  present in the infrared spectrum will be at a peak of  3300-3400 cm-1 which will be due to O-H stretch.

C)If the product is wet with water there will be no change in the infrared spectrum

Explanation:

The characteristic frequency to look out for is  1720-1740 cm-1 (for C=O) for which will disappear in the end product but initially present in the reactant.

Characteristic frequency  present in the infrared spectrum will be at a peak of 3300-3400 cm-1 which will be due to O-H stretch.

If the product is wet with water there will be no change in the infrared spectrum

5 0
2 years ago
For the following reaction, 22.9 grams of nitrogen monoxide are allowed to react with 5.80 grams of hydrogen gas. nitrogen monox
ololo11 [35]

Answer:

1) Maximun ammount of nitrogen gas: m_{N2}=10.682 g N_2

2) Limiting reagent: NO

3) Ammount of excess reagent: m_{N2}=4.274 g

Explanation:

<u>The reaction </u>

2 NO (g) + 2 H_2 (g) \longrightarrow N_2 (g) + 2 H_2O (g)

Moles of nitrogen monoxide

Molecular weight: M_{NO}=30 g/mol

n_{NO}=\frac{m_{NO}}{M_{NO}}

n_{NO}=\frac{22.9 g}{30 g/mol}=0.763 mol

Moles of hydrogen

Molecular weight: M_{H2}=2 g/mol

n_{H2}=\frac{m_{H2}}{M_{H2}}

n_{H2}=\frac{.5.8 g}{2 g/mol}=2.9 mol

Mol rate of H2 and NO is 1:1 => hydrogen gas is in excess

1) <u>Maximun ammount of nitrogen gas</u> => when all NO reacted

m_{N2}=0.763 mol NO* \frac{1 mol N_2}{2 mol NO}*\frac{28 g N_2}{mol N_2}

m_{N2}=10.682 g N_2

2) <u>Limiting reagent</u>: NO

3) <u>Ammount of excess reagent</u>:

m_{N2}=(2.9 mol - 0.763 mol NO* \frac{1 mol H_2}{1 mol NO})*\frac{2 g H_2}{mol H_2}

m_{N2}=4.274 g

8 0
3 years ago
A sample of hydrated cobalt (II) chloride has a mass of 5.22 g. After heating, it has a mass of 2.85 g. What is the percent by m
Tom [10]
Answer:
mass of water in hydrate = 2.37 grams

Explanation:
The mass of the hydrated cobalt (III) chloride is the summation of the salt and the water it contains.
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Total mass of sample = mass of salt + mass of water

Now, we are given that:
total mass of sample = 5.22 grams
mass of salt = mass of sample after heating = 2.85 grams

Substitute to get the mass of water as follows:
5.22 = mass of water in hydrate + 2.85
mass of water in hydrate = 5.22 - 2.85
mass of water in hydrate = 2.37 grams

Hope this helps :)
5 0
2 years ago
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Answer:

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5 0
2 years ago
What is the molecular formula of a hydrocarbon with m+ = 78?
fiasKO [112]

A molecular formula represents the exact number of atoms present for each element in the compound.  

For carbon atom: \frac{78}{12} = 6\frac{6}{12} ( as molar mass of carbon is 12 g/mol)

Now, 6 carbon atoms are present and rest are hydrogen atoms i.e. 6

Thus, formula becomes C_{6}H_{6} ( C_{x}H_{y})

Now, check for unsaturation:

Degree of unsaturation  = \frac{2x+2-y}{2}

Substitute the value of x and y,

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= \frac{12-4}{2}

= 4 implies one ring and three double bonds.

Thus, formula comes out to be C_{6}H_{6} i.e. benzene ring.



4 0
3 years ago
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