Ethyl chloride is prepared by reaction of ethylene with hydrogen chloride as:<br>;<br><br>What is th — Thermodynamics and Thermochemistry Chemistry Question
Question
Ethyl chloride is prepared by reaction of ethylene with hydrogen chloride as:<br>$\text{C}_2\text{H}_4\text{(g)} + \text{HCl(g)} \to \text{C}_2\text{H}_5\text{Cl(g)}$;<br>$\Delta H = -72.3\text{ kJ}$<br>What is the value of $\Delta U$ (in kJ) if 70 g of ethylene and 73 g of HCl are allowed to react?
Answer: D
💡 Solution & Explanation
Moles of $\text{C}_2\text{H}_4 = 70/28 = 2.5$. Moles of $\text{HCl} = 73/36.5 = 2.0$. HCl is limiting. Moles reacted = 2.0. For 1 mole, $\Delta n_g = 1 - 2 = -1$. $\Delta U_m = \Delta H_m - \Delta n_g RT = -72.3 - (-1 \times 8.314 \times 10^{-3} \times 298) = -72.3 + 2.47 = -69.83\text{ kJ/mol}$. For 2.0 moles, $\Delta U = 2 \times (-69.83) = -139.66 \approx -139.6\text{ kJ}$.
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