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Enthalpy

Enthalpy. In an exothermic reaction: e nergy flows out of the system . The flow of heat is negative (-) for the system. In an endothermic reaction: e nergy flows into the system . The flow of heat is positive (+) for the system. Enthalpy.

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Enthalpy

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  1. Enthalpy In an exothermic reaction: energy flows out of the system. The flow of heat is negative (-) for the system. In an endothermic reaction: energy flows into the system. The flow of heat is positive (+) for the system.

  2. Enthalpy Chemists like to know exactly how much energy is produced or absorbed by a given reaction. To make that process more convenient, the special energy function, enthalpy, was invented..

  3. Enthalpy Enthalpy is designated by H. The change in enthalpy (ΔH) for a reaction is equal to the energy that flows as heat. That is (ΔH) = heat

  4. Problem When 1 mol of sulfur dioxide reacts with excess oxygen to form sulfur trioxide, 198.2 kJ of energy is released as heat. Calculate ΔH for a process in which a 12.8g sample of sulfur dioxide reacts with excess oxygen. ΔH = -198.2 kJ/mol SO2 12.8g sample of SO2 (molar mass = 64.07g) is burned

  5. Problem When 1 mol of sulfur dioxide reacts with excess oxygen to form sulfur trioxide, 198.2 kJ of energy is released as heat. Calculate ΔH for a process in which a 12.8g sample of sulfur dioxide reacts with excess oxygen. 12.8g SO2 x 1 mol SO2/64.07g SO2 = 0.200 mol SO2 and 0.200 mol SO2 x -198.2kJ/molSO2 = -39.6kJ = ΔH

  6. Problem When 1 mol of hydrogen gas reacts with excess oxygen to form water, 241.8 kJ of energy is released as heat. Calculate ΔH for a process in which a 15.0g sample of hydrogen gas reacts with excess oxygen. ΔH = -241.8 kJ/mol H2 15.0g sample of H2 (molar mass = 2.016g) is burned

  7. Problem When 1 mol of hydrogen gas reacts with excess oxygen to form water, 241.8 kJ of energy is released as heat. Calculate ΔH for a process in which a 15.0g sample of hydrogen gas reacts with excess oxygen. 15.0g H2 x 1 mol H2/2.016g H2 = 7.44 mol H2 and 7.44 mol H2 x 241.8kJ/molH2 = -1800kJ = ΔH

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