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Question

Ethyl chloride (C2H5Cl), is prepared by reaction of ethylene with hydrogen chloride :

C2H4(g)+HCl(g)C2H5Cl(g);H=72.3kJ/mol
What is the value of U (in kJ), if 70 g of ethylene and 73 g of HCL are allowed to react at 300 K?

A
69.8
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B
180.75
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C
174.5
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D
139.6
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Solution

The correct option is D 139.6
Mass of ethylene =70g

Molar mass of ethylene =28g

No. of moles of ethylene =7028=2.5 moles

Mass of HCl=73g

Molecular mass of HCl=36.5g

No. of moles of HCl=7336.5=2 moles

Since no. of moles of HCl are less than that of ethylene.

Hence HCl will be limiting reagent.

C2H4(g)+HCl(g)C2H5Cl(g)

From the reaction-

1 mole of HCl reacts with 1 mole of C2H4 and gives 1 moles of C2H5Cl.

2 moles of HCl will react with 2 moles of C2H4 and will give 2 mole of C2H5Cl.

Given that ΔH=72.3KJ/mol=72300KJ/mol

Given ΔH is for 1 mole of product.

For 2 moles,

ΔH=2×(72300)=144600KJ/mol

Δng=nPnR=(2(2+2))=2
As we know that,

ΔH=ΔU+ΔngRT

ΔU=ΔHΔngRT

ΔU=144600(2×8.314×300)=139.61KJ/mol.



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