Ethyl chloride is prepared by reaction of ethylene with hydrogen chloride as:
C₂H₄(g) + HCl(g) → C₂H₅Cl(g); ΔH = -72.3 kJ
What is the value of ΔU (in kJ) if 70 g of ethylene and 73 g of HCl are allowed to react? (Assume 298 K.)

Question

Ethyl chloride is prepared by reaction of ethylene with hydrogen chloride as:

C₂H₄(g) + HCl(g) → C₂H₅Cl(g); ΔH = -72.3 kJ

What is the value of ΔU (in kJ) if 70 g of ethylene and 73 g of HCl are allowed to react? (Assume 298 K.)

TestHub · Accepted Answer

Moles: C 2 H 4 = 70/28 = 2.5 mol, HCl = 73/36.5 = 2 mol, so HCl is limiting and 2 mol of reaction occurs. Per mole of reaction Δn g = 1 - 2 = -1, so ΔU = ΔH - Δn g RT = -72.3 + (8.314 × 298)/1000 = -72.3 + 2.48 = -69.82 kJ. For 2 mol: ΔU = 2 × (-69.82) = -139.6 kJ. This is option (d).

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