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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

`-229.38`

Text Solution

Verified by Experts

The correct Answer is:
C
`Deltan_(g)=1-2=-1`
`DeltaE=DeltaH-Deltan_(g)RT -DeltaH-RT =-72.3 + 8.314 xx 300 xx 10^(-3)=-69.806 KJ//"mole"`
so for 3 mole we will get `DeltaE=-69.806 xx 3 KJ//"mol" =209.42 KJ//"mole"`
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