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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=-72.3 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 300KA. `-64.81`B. `-190.71`C. `-209.41`D. `-229.38` |
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Answer» Correct Answer - 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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