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Question

# Ethyl chloride (C2H5Cl) is prepared by reaction of ethylene with hydrogen chloride: C2H4(g)+HCl(g)→C2H5Cl(g)△H=−72.3 kJ/mol What is the value of △U (in kJ), if 98 g of ethylene and 109.5 g of HCl are allowed to react at 300 K?

A
64.81
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B
190.71
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C
209.41
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D
224.38
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Solution

## The correct option is C −209.41△ng=1−2=−1△U=△H−△ngRT=△H+RT=−72.3+8.314×300×10−3=−69.8 kJ/mol Weight of ethylene =98g Mol. wt. of ethylene =28g No. of moles of ethylene =98/28​=3.5 mol Weight of HCl=109.5g Mol. wt. of HCl=36.5g No. of moles of HCl=109.5/36.5​=3 mol HCl is limiting reagent. So, for 3 moles we will get △U=−69.8×3 kJ=−209.42 kJ

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