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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=12=1U=HngRT=H+RT=72.3+8.314×300×103=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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