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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

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
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.


Continued
( 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


Endothermic or exothermic

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?


3 ways to determine enthalpy change h of a reaction
3 Ways to determine Enthalpy change (∆H) of a reaction

  • Calorimetry

  • Hess’s Law

  • Standard Enthalpies of Formation


Calorimetry
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)


Calorimetry example
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



Hess s law
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.


Hess s law example
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


Standard enthalpies of formation
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


Multiple choice
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


B 2 h 6 6h 2 0 6h 2 2h 3 bo 3
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



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