0.3 g of ethane undergoes combustion at 27° C in a bomb calorimeter. The temperature of calorimeter system (including the water) is found to rise by 0.5° C. The heat evolved during combustion of ethane at constant pressure is ____________ kJ mol ⁻¹. (Nearest integer)
[Given : The heat capacity of the calorimeter system is 20 kJ K⁻¹, R=8.3 JK⁻¹ mol⁻¹.
Assume ideal gas behaviour.
Atomic mass of C and H are 12 and 1 g mol⁻¹ respectively]

Solution

Chemistry ma’am · 4 steps

A bomb calorimeter gives you Δ U, not Δ H. Correct it with Δ n_g RT.

1. Heat from calorimeter. Heat released = C × Δ T = 20 kJ K⁻¹ × 0.5 K = 10 kJ. A bomb calorimeter is sealed, so this heat is measured at constant volume.
2. Per mole. The molar mass of C₂H₆ is 30 g mol⁻¹, so 0.3 g is 0.01 mol. That gives Δ U = -10/0.01 = -1000 kJ mol⁻¹.
3. Count gas moles. For C₂H₆(g) + 7/2 O₂(g) → 2CO₂(g) + 3H₂O(l), water is a liquid at 300 K. So Δ n_g = 2 - (1 + 3.5) = -2.5.

   Δ H = Δ U + Δ n_g RT = -1000 + (-2.5)(8.3 × 10⁻³)(300) = -1006.2 kJ mol⁻¹

4. So. The heat evolved at constant pressure is about 1006.2 kJ mol⁻¹, which rounds to 1006.

The common slip: 7225. Adding Δ n_g RT = -6225 in joules directly to -1000 in kJ gives 7225. Convert RT to kJ first, so the correction is only -6.225 kJ mol⁻¹.

The idea to keep. Bomb calorimeter means constant volume, so it measures Δ U. To get Δ H, add Δ n_g RT, counting only gaseous moles and keeping R in kJ.

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