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Question

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

C2H4(g)+HCl(g)C2H5Cl(g) ΔH=72.3kJ. What is the value of ΔE (in KJ), if 70g of ethylene and 73g of HCl are allowed to react at 300K?


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


No. of moles C2H4=7028=2.5, No. of moles of HCl (limiting Reagent) = 7336.5=2

ΔH=ΔE+ΔngRT;

72.3=ΔE+(1×8.314×300)1000

ΔE=69.80; for two moles ΔE=69.80×2

139.6kJ


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