For complete combustion of methanol
CH₃OH(I) + 3/2O₂(g) → CO₂(g) + 2H₂O(I)
the amount of heat produced as measured by bomb calorimeter is 726 kJ mol⁻¹ at 27°C. The enthalpy of combustion for the reaction is -x kJ mol⁻¹, where x is ___________. (Nearest integer)
(Given : R = 8.3 JK⁻¹ mol⁻¹)

Solution

Chemistry ma’am · 4 steps

Bomb calorimeter means constant volume, so 726 kJ is Δ U. Correct it with Δ n_g RT.

1. Read the calorimeter. A bomb calorimeter is sealed, so it measures heat at constant volume. That heat is Δ U. Heat is released, so Δ U = -726 kJ mol⁻¹.
2. Count gas moles. Count only gases. Methanol and water are liquids. That leaves 3/2 mol O₂ on the left and 1 mol CO₂ on the right, so Δ n_g = 1 - 3/2 = -1/2.

   Δ H = Δ U + Δ n_g RT

3. Put in numbers. T = 27 + 273 = 300 K. Then Δ n_g RT = -1/2 × 8.3 × 300 = -1245 J mol⁻¹, which is -1.245 kJ mol⁻¹.
4. So. Δ H = -726 - 1.245 = -727.245 kJ mol⁻¹, so x is 727.

The common slip: 1971. Adding Δ n_g RT = -1245 in joules to -726 in kJ gives -1971, so 1971 instead of 727. Convert 1245 J to 1.245 kJ before you add.

The idea to keep. Bomb calorimeter means constant volume, so it gives Δ U. Use Δ H = Δ U + Δ n_g RT, and count only gaseous moles in Δ n_g. Liquids and solids do not count.

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