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JJ25 & KID SWAGG

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.)
- 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 )

- 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: 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

- Calorimetry
- Hess’s Law
- Standard Enthalpies of Formation

Calorimetry:

- Measurement of heeet flow
- Coffee cup Calorimeter
C x M x ∆T=q

- Coffee cup Calorimeter

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

- 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 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

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 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

- 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

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

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

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