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Ethyl chloride (C(2)H(5)Cl) , is prepare...

Ethyl chloride `(C_(2)H_(5)Cl)` , is prepared by reaction of ethylene with hydrogen chloride:
`C_(2)H_(4)(g) + HCl(g) rarr C_(2)H_(5)Cl(g) " " DeltaH=-72.3 KJ//"mol"`
What is the value of `DeltaE ("in KJ")`, if 98 g fo ethylene and 109.5 g if HCl are allowed to react at 300K

A

`-64.81`

B

`-190.71`

C

`-209.41`

D

`-224.38`

Text Solution

Verified by Experts

The correct Answer is:
C
No. of mole of `C_(2)H_(4)=98/28=3.5," "` No. of mole of `HCl` (Limiting Reagent)`=109.5/36.5=3`
`DeltaH=DeltaE+Deltan_(g) RT," "-72.3=DeltaE+(-1xx8.314xx300)//1000`
`DeltaE=-69.80,` for three mole `DeltaE=-69.80xx3" "rArr -209.41" kJ/mol"`
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