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THERMOCHEMISTRY

THERMOCHEMISTRY. JJ25 & KID SWAGG. Definition:. Branch of chemistry that studies the relationship between the chemical action and the amount of heeeet absorbed or generated. Thermochemistry Terminology. Heeet (not to be confused with temp.)

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THERMOCHEMISTRY

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  1. THERMOCHEMISTRY JJ25 & KID SWAGG

  2. Definition: Branch of chemistry that studies the relationship between the chemical action and the amount of heeeet absorbed or generated.

  3. Thermochemistry Terminology • Heeet (not to be confused with temp.) • Transfer of energy to an object of higher energy to an object of lower energy. • Symbolized by ( q ) • Heeet is stoichiometric which means more stuff in reaction more heeet involved.

  4. ( Continued ) • System is the concentrated object • Surroundings are everything around the system • Enthalpy • Total heeet of a system • Symbolized by ( H ) • State function: Only initial and final conditions matter not how you get thurr

  5. Endothermic: Requires input of heeet from surrounding for reaction to take place The system feels to cool to touch ∆H > 0 Exothermic: Releases heeet into the surroundings as the process occurs The system feels to hot to touch ∆H < 0 Endothermic or Exothermic?

  6. 3 Ways to determine Enthalpy change (∆H) of a reaction • Calorimetry • Hess’s Law • Standard Enthalpies of Formation

  7. Calorimetry: • Measurement of heeet flow • Coffee cup Calorimeter C x M x ∆T=q Specific heeet capacity ( j/g°C) Heeet Change in Temp. ( C° ) Mass ( g ) • “ BOMB “ Calorimeter q= C x ∆T Heeet Change in Temp. ( C° ) Specific heeet capacity ( j/g°C)

  8. Calorimetry Example • H+(aq) + OH-(aq) → H2O(l) • The temperature of 110 g of water rises from 25.0°C to 26.2°C when 0.10 mol of H+ is reacted with 0.10 mol of OH-. • Calculate q of the water • Calculate ∆H

  9. Hess's Law

  10. Hess’s law • Hess’s Law states that the heat of a whole reaction is equivalent to the sum of it’s steps. • For example: C + O2 CO2 This occurs as 2 steps C + ½O2  CO H = – 110.5 kJ CO + ½O2  CO2 H = – 283.0 kJ C + CO + O2  CO + CO2 H = – 393.5 kJ I.e. C + O2  CO2 H = – 393.5 kJ Hess’s law allows us to add equations. We add all reactants, products, & H values.

  11. Hess’s law: Example We may need to manipulate equations further: 2Fe + 1.5O2  Fe2O3 H=?, given Fe2O3 + 3CO  2Fe + 3CO2 H= – 26.74 kJ CO + ½O2  CO2 H= –282.96 kJ 1: Align equations based on reactants/products. 2: Multiply based on final reaction. 3: Add equations. 2Fe + 3CO2  Fe2O3 + 3CO H= + 26.74 kJ CO + ½O2  CO2 H= –282.96 kJ 3CO + 1.5O2  3CO2 H= –848.88 kJ 2Fe + 1.5O2  Fe2O3 H= –822.14 kJ

  12. Standard Enthalpies of Formation • Standard conditions: Most stable form of the substance • 1atm and 25°C ( 298K ) • Standard Enthalpy, ∆H°, is enthalpy measured when everything is measured in standard state

  13. Multiple Choice • 1.) Which of the following is NOT a characteristic of an exothermic reaction? • Reaction feels warm • System gains energy • Enthalpy change of reaction is negative

  14. B2H6 + 6H20  6H2 + 2H3BO3 • 2.) ∆H=? KJ/mol • -3604 KJ/mol • -772 KJ/mol • 3604 KJ/mol • 772 KJ/mol B2H6: ∆H°= +36 KJ/mol H20: ∆H°= -242 KJ/mol H3OBO3: ∆H°= -1094 KJ/mol

  15. 3.) Which of the ∆H’s is exothermic? • 563 • 0 • -375 • 989

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