A sample of n -octane (1.14 g) was completely burnt in excess of oxygen in a bomb calorimeter, whose heat capacity is 5 kJ K⁻¹. As a result of combustion reaction, the temperature of the calorimeter is increased by 5 K . The magnitude of the heat of combustion of octane at constant volume is__________ kJ mol⁻¹ (nearest integer).

Solution

Chemistry ma’am · 3 steps

The calorimeter tells you the heat; divide it by the moles of octane burnt to get it per mole.

1. Heat released. The heat given out by the reaction warms the calorimeter. So q = C × Δ T = 5 kJ K⁻¹ × 5 K = 25 kJ.
2. Moles of octane. n-Octane is C₈H₁₈, with molar mass 8 × 12 + 18 × 1 = 114 g mol⁻¹. So 1.14 g is 1.14/114 = 0.01 mol.

   |Δ_c U| = q/n = (25 kJ)/(0.01 mol) = 2500 kJ mol⁻¹

3. So. A bomb calorimeter works at constant volume, so this heat is Δ U. Its magnitude is 2500 kJ mol⁻¹.

The common slip: 25. Stopping at C × Δ T gives 25, which is the heat for 0.01 mol only, not per mole.

The idea to keep. A bomb calorimeter gives q_V = Δ U, from C × Δ T. Always convert that to per mole by dividing by the moles of the substance burnt.

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