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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=-72.3kJ//mol`
What is the value of `DeltaE (` in `kJ)`, if `98g` of ethylene and `109.5g` of `HCl` are allowed to react at `300K`
A. `-64.81`
B. `-190.71`
C. `-209.41`
D. `-224.38`

1 Answer

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Correct Answer - 3
`Deltan_(g)=1-2=-1`
`DeltaE=DeltaH-Deltan_(g)RT=DeltaH-RT=-72.3+8.314xx300xx10^(-3)`
`=-69.806kJ //mol e`
so for 3 mole we will get `DeltaE=-69.806xx3kJ //mol e=209.42kJ//mol e`

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