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=

Question

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=

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`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - 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"`

4 views Answered 2023-02-05 15:29