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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/molWhat 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 −139.6Mass of ethylene =70gMolar mass of ethylene =28g∴ No. of moles of ethylene =7028=2.5 molesMass of HCl=73gMolecular mass of HCl=36.5gNo. of moles of HCl=7336.5=2 molesSince 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/molGiven ΔH is for 1 mole of product.For 2 moles,ΔH=2×(−72300)=144600KJ/molΔng=nP−nR=(2−(2+2))=−2As we know that,ΔH=ΔU+ΔngRT⇒ΔU=ΔH−ΔngRTΔU=−144600−(−2×8.314×300)=−139.61KJ/mol.

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