ΔH from a bomb-calorimeter ΔU

Worked example

Convert ΔcU = −1068 kJ mol⁻¹ into ΔcH at 298 K, given Δn_gas = 0.5.

1. 1. ΔH from ΔU (gas-phase moles): solve for ΔH

   $\Delta H = \Delta U + \Delta n_\mathrm{gas}RT$

   = −1067 kJ mol⁻¹

Answer: ΔH = −1070 kJ mol⁻¹

Method

ΔH = ΔU + Δn_gas RT. The correction is only a few kJ mol⁻¹, so keep 4 significant figures.

Formulas used

• ΔH from ΔU (gas-phase moles)

  $\Delta H = \Delta U + \Delta n_\mathrm{gas}RT$

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