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Enthalpy and Calorimetry

Enthalpy and Calorimetry. Chapter 5 part 2. Enthalpy. H is heat under constant pressure or H=q P H=E+PV And therefore Δ H= Δ E+P Δ V Δ H=H final -H initial. Example:.

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Enthalpy and Calorimetry

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  1. Enthalpy and Calorimetry Chapter 5 part 2

  2. Enthalpy • H is heat under constant pressure or • H=qP • H=E+PV • And therefore • ΔH= ΔE+P ΔV • ΔH=Hfinal-Hinitial

  3. Example: • When 1 mole of methane is burned at a constant pressure. 890kJ of energy is released as heat. Calculate the ΔH for a process in which a 5.8g sample of methane is burned at constant pressure.

  4. Answer • ΔH= -890kJ/mol • CH4 is methane, MM = 16 g/mol • 5.8g/16g mol-1 = 0.36 mol • 0.36 mol CH4 x -890kJ/mol CH4 = -320kJ

  5. Calorimetry • The science of measuring heat. • The device used to measure heat changes associated with chemical reaction is a calorimeter

  6. Heat capacity of an object is the heat absorbed by the change in temperature or C=(heat absorbed)/(ΔT) Specific heat capacity is the heat capacity per gram of a substance or Csp=C/gram Molar heat capacity is the heat capacity per mole of a substance or Cmol=C/mole Heat Capacity

  7. Constant pressure calorimeter Calculations (per gram) Csp* mass*ΔT= ΔH Remember: the specific heat and the mass are of the same object. The measurement is of the surrounding water, not the actual system so the sign is reversed. Calorimeters

  8. Example: • When 1.00 L of 1.00M Ba(NO3)2 solution at 25.0 °C is mixed with 1.00 L of 1.00 M Na2SO4 solution at 25 °C in a calorimeter, the white solid BaSO4 forms and the temperature of the mixture increases to 28.1 °C . Assuming that the calorimeter absorbs only a negligible amount of heat and the specific heat of the solution is 4.18J/°Cg and the density of the solution is 1.00g/mL, calculate the enthalpy change per mole of BaSO4 formed.

  9. Ba2+ (aq) + SO4 2- (aq)→BaSO4 (s) • Mass of solution =2 liters = 2000 grams • Csp= 4.18J/°Cg • ΔT=28.1-25.0 =3.1 °C • qsurroundings = 2.6 x104J • qsystem= -2.6 x104J

  10. Bomb Calorimeter • This is also known as a constant volume calorimeter. • Since it is under constant volume, but there is a change in temperature, the pressure is not constant.

  11. Note: Since there is no change in volume, then no work is done. The constant volume calorimeter measures the system directly. The specific heat is of the calorimeter itself. Bomb Calorimeter

  12. Example • In comparing potential fuels a bomb calorimeter with a specific heat of 11.3kJ/ °C was employed. When a 1.50 g sample of methane was burned with an excess of oxygen in the calorimeter, the temperature increased by 7.3 °C . When a 1.15 g sample of hydrogen gas was burned with an excess of oxygen, the temperature increase was 14.3 °C. Calculate the energy of combustion per gram for these two fuels.

  13. Methane: released energy from 1.5 gram. =(11.3 kJ/g °C)(7.3 °C) =83kJ Per gram 83kj/1.5g=55kJ/g Hydrogen: released energy from 1.15 g. =(11.3 kJ/g °C)(14.3 °C) =162 kJ Per gram 162kJ/1.15 g= 141 kJ/g

  14. Hess’s Law • Since enthalpy is a state function, then • In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or a series of steps.

  15. Hess’s Law

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