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M olar enthalpy of neutralization

M olar enthalpy of neutralization. Raw data. Calculation heat released. amount of energy released temperature rise = 27.0 o C – 21.0 o C = 6.0 o C heat absorbed by the water = heat released by reaction = mass of water x shc water x temperature rise

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M olar enthalpy of neutralization

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  1. Molar enthalpy of neutralization

  2. Raw data

  3. Calculation heat released amount of energy released • temperature rise = 27.0 oC – 21.0 oC = 6.0oC • heat absorbed by the water = heat released by reaction = mass of water xshcwaterx temperaturerise = 50.0 gx 4.2 J g-1 °C x 6.0 °C = 1260 J

  4. Calculation for 1 mole of acid • moles = concentration x volume = 1.0 mol dm-3x 0.025 dm-3 = 0.025 moles • scaling up to 1 mole of acid gives 1260 / 0.025 = 50400 J Answer: molar enthalpy of neutralization = - 50.4 kJ/mol

  5. Experimental error • Heat loss • Incomplete reaction • We are using the heat capacity of water although there is a solution in the calorimeter

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