C(s) + (g) -> (g) + (g). Given δ H_c (kcal/mol): (g) = -68.0; (g) = -68.0; C(s) = -94.0. Select the — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

100 mol C. (a) 100 * 94 = 9400 kcal (correct). (b) 100 mol $H_2$ + 100 mol $CO$: 6800 + 6800 = 13,600 kcal (correct). (c) 1200 g equimolar $H_2$/$CO$ mixture: 30n = 1200, n = 40 mol each; heat = 40*68 + 40*68 = 5440 kcal (correct). (d) 13,600 ≠ 9400 (incorrect).