2.4 g coal is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atm pressure. The temperature of the calorimeter rises from 298 K to 300 K. The enthalpy change during the combustion of coal is -x kJ mol⁻¹. The value of x is ___________. (Nearest Integer)
(Given : Heat capacity of bomb calorimeter 20.0 kJ K⁻¹. Assume coal to be pure carbon)

Solution

Chemistry ma’am · 4 steps

The calorimeter's temperature rise tells you the heat; divide it by moles of carbon to get per-mole value.

1. Heat released. The calorimeter absorbs the heat of combustion: q = C Δ T = 20.0 kJ K⁻¹ × 2 K = 40 kJ. This much heat is released by burning 2.4 g of coal.
2. Moles of carbon. Coal is taken as pure carbon, so n = 2.4/12 = 0.2 mol.

   Δ U = -(40 kJ)/(0.2 mol) = -200 kJ mol⁻¹

3. From ΔU to ΔH. A bomb calorimeter works at constant volume, so it measures Δ U. For C(s) + O₂(g) → CO₂(g), Δ n_g = 1 - 1 = 0. So Δ H = Δ U + Δ n_g RT = Δ U.
4. So. Δ H = -200 kJ mol⁻¹, so x = 200.

The common slip: 40. Stopping at the heat for 2.4 g gives 40, not 200. The answer is asked per mole, so divide by 0.2 mol of carbon.

The idea to keep. A bomb calorimeter gives Δ U, not Δ H. Convert the heat to a per-mole value, then add Δ n_g RT, which is zero when gas moles do not change.

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