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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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How many formula units are there in 212 grams of mgCl2
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Formula units are there in 212 grams of MgCl₂ are 830.56

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Here given data is

MgCl₂ = 212 grams

1 mole of magnesium chloride has mass = 95.211 gram and contains 6.022×10²³formula units of magnesium chloride

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1 year ago
An isotonic solution has: a)The same pH as blood b) More ions per unit volume than blood c) More water per unit volume than blo
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Answer:

d) osmotic pressure equal to that of blood

Explanation:

Answer is not a) because pH is concentration of positive hydrogen in solution but it not note all solutes. Answer is not b) because like answer a), it not note solutes without charge and, additionally it not note the pressure equilibrium when solvent flows through a membrane . Answer is not c) because it can be more or less water (solvent).

8 0
3 years ago
30.5 g of sodium metal reacts with chlorine gas to produce sodium chloride. How much (in grams) chlorine gas must react with thi
Lena [83]

Answer:

Mass of chlorine = 47.22 g

Explanation:

Given data:

Mass of sodium = 30.5 g

Mass of chlorine= ?

Solution:

Chemical equation:

 2Na + Cl₂      →      2NaCl  

Number of moles of Na:

Number of moles = mass/molar mass

Number of moles = 30.5g/ 23 g/mol

Number of moles = 1.33 mol

Now we will compare the moles of Cl ₂ with Na from balance chemical equation.

                    Na            :              Cl ₂

                     2              :               1

                     1.33          :            1/2×1.33 = 0.665 mol

Mass of chlorine gas:

Mass = number of moles × molar mass

Mass = 0.665 mol × 71 g/mol

Mass = 47.22 g

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3 years ago
Problem PageQuestion Gaseous ethane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water . If of w
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Answer:

<h2> = ( 1.08 / 2.2 ) 100% = 49%</h2>

Explanation:

Balanced Equation: 2CH₃CH₃(g) + 5O₂(g) → 2CO₂(g) + 6H₂O(g)

Calculate moles of CH₃CH₃ and O₂

1.2 ₃₃ ( 1 ₃₃/ 30.0694 ₃₃ ) = 0.040 ₃₃

8.6 ₂ ( 1 2/ 31.998 ₂ ) = 0.27 ₃₃

Find limiting reagent  0.040 ₃₃ ( 5 ₂/ 2 ₃₃ ) = 0.10 ₂

CH₃CH₃ is the limiting Reagent

CH₃CH₃ (L.R.) O₂ CO₂ H₂O

Initial (mol) 0.040 0.27 0 0

Change (mol) -2x=-0 -5x= -0.10  +2x=+0.040 +6x=+0.12

Final (mol) 0 0.117 0.040 0.12

0.040 − 2 = 0 = 0.020

Determine percent yield

0.12 ₂ ( 18.0148 ₂ /1 ₂ ) = 2.2 ₂  

= ( 1.08 / 2.2 ) 100% = 49%

5 0
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