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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?


  1. −174.5

  2. −139.6

  3. −180.75

  4. −69.8


Solution

The correct option is B

−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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