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Chem10 Topic 01 - ThermochemistryPowerPoint Presentation

Chem10 Topic 01 - Thermochemistry

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Stoichi what?

- Stoikheion (greek)= element
- Metron (greek)= to measure
- Stoichiometry: the calculation of amounts of substances involved in chemical reactions
- This will be Topic 01 in IB Chemistry

Moles

- Moles (latin): heap or pile
- 1 Mole (mol) = 6.02 x 1023 representative particles (atoms or molecules)
- This is called Avogadro’s number

Molar Mass

- The atomic mass of one mole of an element or molecule expressed in grams
- Round to 2 decimal places, your periodic table should already have this!

Calculating Molar Mass

- Step One: Find the molar mass (atomic mass) of each element on the PT
- round to 2 decimal places
- Unit is grams (g) per mole (mol)

- Step Two: Multiply the number of atoms of the element by the molar mass
- Remember – the subscript tells you how many atoms you have of each element

- Step Three: Add up the masses of all elements in the compound

Units for Molar Mass

- H2O is 18.02 g/mol
- Stated as “grams per mol”
- Meaning 18.02 grams = 1 mole
- OR 1 mole = 18.02 grams

- Use like common conversions
- How many meters in 1.000 km?
- Use the conversion of 1000 m = 1.000 km

- How many meters in 1.000 km?

(Molar Volume of a gas 22.4)

Moles to Molecules

(use Avogadro’s Number)

Moles to Mass

(use Molar Mass)

Simple Conversions:

- Mole / Mass Conversions -

Use the Molar Mass of a substance to convert from Moles to Mass and Mass to Moles

80. g CuSO4

1 mol CuSO4

Mass to Moles

= 0.50 mol CuSO4

159.5 g CuSO4

0.50 mol CuSO4

159.5 g CuSO4

Moles to Mass

= 80. g CuSO4

1 mol CuSO4

Simple Conversions:

- Mole / Molecule Conversions -

Use Avogadro’s Number : 6.022 x 1023 molecules (mc) in one mole of the substance

2 mol CuSO4

6.022x1023 (mc) CuSO4

Moles to (mc)

= 1.2x1024 (mc) CuSO4

1 mol CuSO4

1.2x1024 (mc) CuSO4

1 mol CuSO4

(mc) to Moles

= 2 mol CuSO4

6.022x1023 (mc) CuSO4

Let’s use it, but first…

- You needed to calculate q (heat change) for NaOHand Na2S2O3. Lets use NaOH as our example:
- qNaOH cannot be found directly, but qH2O can be
- qNaOH = - (qH2O) (this is the conservation of energy)
- qH2O = msΔT
- m = 50.00 g
- s = 4.184 J/goc
- ΔT =20.00 oc

- qH2O = (50.00 g)(4.184 J/goc)(20.00oc)
- qH2O = 4,184 J
- qNaOH = -4,184 J

- Now, how many grams of NaOH were used to produce this amount of heat? (for my data it was 1.24 g NaOH)

First Law of Thermodynamics!

Energy lost = Energy gained

q = -q

Calculate Molar Heat of Solution

- For a 7 on the Heat Change Lab you were asked to calculate the molar heat change
- In my reaction example, with 1.24 g NaOHI produced a value of qNaOH= -4,184 J

- So, lets write this in units of Joules per gram
- qNaOH =

- You need to find Joules per mole (molar heats of solution) so that you can relate to the heat production of the other salt.
- NaOH(s) Na+(aq) + OH-(aq) qNaOH = -134 kJ/mol
- Na2S2O3(s) 2 Na+(aq) + S2O32-(aq) qNa2S2O3 = + or - ??

Enthalpy: Thermochemical Equations

NaOH(s)

Is DH negative or positive?

System gives off heat

Exothermic

DH < 0

Na+(aq) + OH-(aq)

134 kJ are released for every 1 mole of sodium hydroxide that is dissolved into water.

- NaOH(s) Na+(aq) + OH-(aq) qNaOH= -134 kJ/mol
- ΔH = -134 kJ/mol

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