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Question

# Ethyl chloride (C2H5Cl), is prepared by reaction of ethylene with hydrogen chloride:C2H4 + HCl(g)→ C2H5Cl(g);△H=−72.3 kJWhat is the value of △E (in kJ), if 70 g of ethylene and 73g of HCl are allowed to react at 300 K.

A
-69.8
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B
-180.75
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C
-174.5
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D
-139.6
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Solution

## The correct option is D -139.6Solution:-Weight of ethylene =70gMol. wt. of ethylene =28gNo. of moles of ethylene =7028=2.5 molWeight of HCl=73gMol. wt. of HCl=36.5gNo. of moles of HCl=7336.5=2 molHCl is limiting reagent.Now,C2H4(g)+HCl(g)⟶C2H5Cl(g)ΔH=−72.3kJFrom the above reaction,1 mole of HCl gives 1 mole of C2H5Cl.Therefore,2 moles of HCl gives 2 moles of C2H5Cl.Again from the above reaction,Δng=nP−nR=1−(1+1)=−1As we know that,ΔH=ΔE+ΔngRTat 300K,−72.3=ΔE+(−1)×8.34×10−3×300ΔE=−72.3+2.5=−69.8kJNow,For 1 mole of C2H5Cl-ΔE=−69.8kJTherefore,For 2 mole of C2H5Cl-ΔE=2×(−69.8)=−139.6kJ

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