The heat of combustion of ethanol into carbon dioxide and water is – 327 kcal at constant pressure. The heat evolved (in cal) at constant volume and 27ᵒC (if all gases behave ideally) is (R = 2 cal mol –1 K –1 ) ________.

Solution

Chemistry ma’am · 3 steps

Count only the gas moles on each side; that Δ n_g is all that separates qₚ from qᵥ.

1. Write the reaction. At 27°C ethanol and water are liquids, so only O₂ and CO₂ are gases: C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l).
2. Change in gas moles. Δ n_g = 2 - 3 = -1. The temperature is T = 27 + 273 = 300 K, and Δ H = -327 kcal = -327000 cal.

   Δ U = Δ H - Δ n_g RT = -327000 - (-1)(2)(300) = -326400 cal

3. So. At constant volume Δ U = -326400 cal, so the heat evolved is 326400 cal. It is a little less than at constant pressure, because at constant pressure the gas volume shrinks and the work done on the system adds to the heat released.

The common slip: 326.4. Leaving the answer in kilocalories gives 326.4, but the question asks for calories, so multiply by 1000 to get 326400.

The idea to keep. To switch between qₚ and qᵥ, use Δ H = Δ U + Δ n_g RT. Count only gaseous moles, and get both terms into the same unit before you combine them.

Answer from the NTA final answer key. Solution by PrepWiser.