The heat of combustion of solid benzoic acid at constant volume is -321.30 kJ at 27° C. The heat of combustion at constant pressure is (-321.30-x R) kJ, the value of x is __________.

Solution

Chemistry ma’am · 3 steps

Count only gas moles on each side: Δ n_g RT is all that separates Δ H from Δ U.

1. Balanced equation. C₆H₅COOH(s) + 15/2 O₂(g) → 7CO₂(g) + 3H₂O(l). At 27°C, water forms as a liquid, and benzoic acid is a solid.
2. Change in gas moles. Only gases count. The products have 7 mol of CO₂ and the reactants have 7.5 mol of O₂. So Δ n_g = 7 - 7.5 = -0.5.

   Δ H = Δ U + Δ n_g RT = -321.30 + (-0.5) × R × 300 = -321.30 - 150R

3. So. With T = 300 K, the heat of combustion at constant pressure is (-321.30 - 150R) kJ, so x = 150.

The common slip: 13.5. Putting T = 27 instead of 300 K gives x = 13.5, not 150. Convert Celsius to kelvin first.

The idea to keep. To go from Δ U to Δ H, use Δ H = Δ U + Δ n_g RT. Leave out solids and liquids, and always put T in kelvin.

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