Ethyl chloride ( C 2 H 5 Cl ) , is prepared by reaction of ethylene with hydrogen chloride: C 2 H 4 ( g ) + HCl ( g ) ⟶ C 2 H 5 Cl ( g ) Δ H = − 72.3 kJ What is the value of Δ E (in kJ ),

Question

Ethyl chloride ( C 2 ​ H 5 ​ Cl ) , is prepared by reaction of ethylene with hydrogen chloride: C 2 ​ H 4 ​ ( g ) + HCl ( g ) ⟶ C 2 ​ H 5 ​ Cl ( g ) Δ H = − 72.3 kJ What is the value of Δ E (in kJ ), if 98 g of ethylene and 109.5 g of HCl are allowed to react at 300 K .

TestHub · Accepted Answer

No. of mole of C 2 ​ H 4 ​ = 28 98 ​ = 3.5 , No. of mole of HCl ( Limiting Reagent ) = 36.5 109.5 ​ = 3 Δ H = Δ E + Δ n g ​ RT ; − 72.3 = Δ E + ( − 1 × 8.314 × 300 ) /1000 Δ E = − 69.80 ; for three mole Δ E = − 69.80 × 3 ⇒ − 209.41 kJ / mol

Chemistry - Thermochemistry Question with Solution | TestHub

Options:

Doubts & Discussion

JEE Main (2028)