Ethyl chloride is prepared by reaction of ethylene with hydrogen chloride as: (g) + (g) → C2H5Cl(g); — Thermodynamics and Thermochemistry Chemistry Question
Question
Ethyl chloride is prepared by reaction of ethylene with hydrogen chloride as: $C_2H_4$(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?
💡 Solution & Explanation
Let's determine the limiting reactant:<br>- Moles of ethylene ($C_2H_4$, M = 28 g/mol) = 70 g / 28 = 2.5 mol.<br>- Moles of $HCl$ (M = 36.5 g/mol) = 73 g / 36.5 = 2.0 mol.<br>Since $HCl$ is the limiting reactant, exactly 2.0 moles of $C_2H_4$ react with 2.0 moles of $HCl$ to form 2.0 moles of C2H5Cl(g).<br>Therefore, ΔH = 2.0 × (-72.3 kJ) = -144.6 kJ.<br>The gaseous mole change is Δng = n_products(g) - n_reactants(g) = 1 - (1 + 1) = -1 per mole of reaction. For 2.0 moles of reaction, Δng = -2.0.<br>Using ΔU = ΔH - Δng × R × T (assuming standard temperature 298.15 K):<br>ΔU = -144.6 kJ - (-2.0 mol × 8.314 × 10^-3 kJ/(mol K) × 298.15 K) = -144.6 + 4.96 = -139.64 kJ.