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Calorimetry

Calorimetry. 1st Law of Thermodynamics. 1 St Law: Energy cannot be created or destroyed, but simply converted from one form to another When a system absorbs energy, the surroundings release it When a system releases energy, the surroundings absorb it

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Calorimetry

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  1. Calorimetry

  2. 1st Law of Thermodynamics 1St Law: Energy cannot be created or destroyed, but simply converted from one form to another • When a system absorbs energy, the surroundings release it • When a system releases energy, the surroundings absorb it ∆Etotal = ∆Esystem + ∆Esurroundings = 0 ∆Esystem = -∆Esurroundings

  3. Enthalpy • To express thermochemical changes, chemists need a variable that is not affected by the conditions under which the process occurs • Enthalpy, H, H = E + PV • It is not possible to measure H, but it is possible to measure changes in enthalpy, ∆H

  4. Enthalpy & 1st Law ∆Esystem = -∆Esurroundings ∆Hsystem = -Qsurroundings • If heat enters the system (endothermic), ∆H will be positive • If heat leaves the system (exothermic), ∆H will be negative

  5. Calorimetry • A calorimeter is a device used to measure heat released or absorbed by a process • Calorimetry is using a calorimeter to study energy changes associated with processes • The calorimeter isolates the system from its external environment.

  6. Calorimetry • The 2 stacked Styrofoam cups isolate the system from the external environment • The reaction takes place in the inner cup with a known mass of water (or dilute solution) • A thermometer is used to measure heat absorbed/released to the surroundings (water) • Limitations: • Does not work for reactions involving gases • Does not work for high temperature reactions

  7. Bomb Calorimeter

  8. Assumptions in Calorimetry • No heat is transferred between the calorimeter and the outside environment • Any heat absorbed or released by the calorimeter itself is negligible • A dilute aqueous solution will have a density and specific heat capacity equal to water

  9. Calorimetry & The 1st Law ∆Hsystem = - Qsurroundings n∆Hsystem = - Qsurroundings n∆Hsystem = - mc∆T | n∆Hsystem | = | mc∆T |

  10. Sample Problem 1 A pellet of potassium hydroxide with a mass of 0.648 g is dissolved in 40.0 mL of water in a styrofoam calorimeter. The temperature increases from 22.6°C to 27.8°C. What is the molar enthalpy of solution for potassium hydroxide? Heat was released (exothermic) Therefore ∆H = -75.3 kJ/mol

  11. Sample Problem 2 50.00mL of a 0.300M cupric sulfate solution reacts with 50.00mL of a 1.00M sodium hydroxide solution. The initial temperature of both solutions is 21.40°C. After mixing the solutions in a styrofoam calorimeter, the highest temperature recorded is 24.60°C. Determine the molar enthalpy change of reaction. Heat was released (exothermic) Therefore ∆H = -89.4 kJ/mol

  12. Practice • p. 310-311 #4-10 • p. 312 #1-5

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