3..thermodynamics Q3. Ethyl chloride (C2H5Cl) is prepared by reaction of ethylene with hydrogen chloride : C2H4(g) + HCl(g) → C 2 H 5 CL ( g ) ∆ HH = – 72 . 3 kJ What is the value of ∆ E( in kJ ), if 70 g of ethylene and 73 g of HCL are allowed to react at 300 K? (A) –69.8 (B) - 180.75 (C) – 174.5 (D) – 139.6 Share with your friends Share 27 Vartika Jain answered this Dear Student, ∆H=∆E+P∆Vand, P∆V=∆nRTSo, ∆H=∆E+∆nRTFor the reaction, C2H4(g) + HCl(g) → C2H5Cl(g)1 mol i.e. 28 g of C2H4 reacts with 1 mol i.e. 36.5 g of HClSo, 70 g of C2H4 will react with 36.528×70=91.25 g of HClBut we have only 73 g of HCl, so HCl is the limiting reagent. Now, 36.5 g of HCl = 28 g of C2H4 So, 73 g of HCl =2836.5×73=56 g of C2H4So, Moles of C2H4=5628=2 molMoles of HCl =7336.5=2 molNow, 1 mol of C2H4 gives 1 mol of C2H5ClSo, 2 mol of C2H4 will give 2 mol of C2H5ClHence, ∆n = 2-(2+2)=2-4=-2 molR=8.314 JK-1T=300 K∆H=-72.3 ×103 JHence, ∆E=∆H-∆nRT =-72.3 ×103-(-2)(8.314)(300) =-72300+4988.4 =-67311.6 J =-67.311 kJSo, the correct answer is (a) -45 View Full Answer
