Daltons Law of Partial Pressures

1
Daltons Law of Partial Pressures
John Dalton 1766-1844
2
Daltons Law of Partial Pressures

• The total pressure inside a container is equal to
  the partial pressure due to each gas.
• The partial pressure is the contribution by that
  gas.
• PTotal P1 P2 P3

3
Daltons Law of Partial Pressures
The of gases in air Partial pressure (STP)
78.08 N2 593.4 mm Hg 20.95 O2 159.2
mm Hg 0.94 Ar 7.1 mm Hg 0.03
CO2 0.2 mm Hg PAIR PN2 PO2 PAr
PCO2 760 mm Hg Total Pressure 760 mm Hg
4
Daltons Law of Partial Pressures
1. What is the total pressure in a balloon filled
with air if the pressure of the oxygen is 170 mm
Hg and the pressure of nitrogen is 620 mm Hg?
2. In a second balloon the total pressure is
1.3 atm. What is the pressure of oxygen if the
pressure of nitrogen is 720 mm Hg?
5
Daltons Law of Partial Pressures
2 H2O2 (l) ? 2 H2O (g) O2 (g) 0.32 atm
0.16 atm
What is the total pressure in the flask? Ptotal
in gas mixture PA PB Therefore, Ptotal
PH2O PO2 0.48 atm Daltons Law total
P is sum of PARTIAL pressures.
6
Stoichiometry with Gases

• Equal volumes of gas, at the same temperature and
  pressure contain the same number of particles.
• Moles are numbers of particles
• You can treat reactions as if they happen liters
  at a time, as long as you keep the temperature
  and pressure the same.

7
Stoichiometry with Gases

• Convert liters of a gas to moles
• At STP (0ºC and 1 atmosphere pressure)
• At STP 22.4 L of a gas 1 mole

8
Stoichiometry with Gases
How many liters of CH4 at STP are required to
completely react with 11.2 L of O2 ?
CH4 2O2 ? CO2 2H2O 11.2 L O2 x
1 mol 0.5 mol 22.4 L
O2 B 0.25 0.5 C A 0.25 mol CH4 x
22.4 L CH4 5.6 L CH4
1 mol
9
Stoichiometry with Gases

• 1. If 6.45 moles of water are decomposed, how
  many liters of oxygen will be produced at STP?
• 2H2O ?2H2 O2
• 2. How many liters of CO2 at STP will be produced
  from the complete combustion of 23.2 g C4H10 ?
• What volume of oxygen will be required?

10
Stoichiometry and Energy Changes

• Energy is measured in Joules or calories
• Every reaction has an energy change associated
  with it
• Exothermic reactions release energy, usually in
  the form of heat.
• Endothermic reactions absorb energy
• Energy is stored in bonds between atoms

11
Stoichiometry and Energy Changes

• Calorimeter
• Heat given off or absorbed is determined from the
  temperature of a known mass of water.

12
Stoichiometry and Energy Changes
The heat that is released or absorbed in a
chemical reaction is called ?H C O2(g) ?
CO2(g) 393.5 kJ C O2(g) ? CO2(g)
?H -393.5 kJ In thermochemical equations
it is important to indicate the state H2(g) ½
O2 (g) ? H2O(g) ?H -241.8 kJ H2(g)
½ O2 (g) ? H2O(l) ?H -285.8 kJ
13
Stoichiometry and Energy Changes
The heat from the reaction that completely burns
1 mole of a substance C2H4 3 O2 ? 2 CO2 2
H2O C2H6 O2 ? CO2 H2O 2 C2H6 5 O2 ? 2 CO2
6 H2O C2H6 (5/2) O2 ? CO2 3 H2O
14
Stoichiometry and Energy Changes
The ?H for a reaction that produces 1 mol of a
compound from its elements at standard conditions
Standard conditions 25C and 1 atm. Symbol is
?H The standard heat of formation of an element
is 0 This includes the diatomics
15
Stoichiometry and Energy Changes

• There are tables (pg. 190) of heats of formations
  
• The heat of a reaction can be calculated by
  subtracting the heats of formation of the
  reactants from the products

16
Stoichiometry and Energy Changes
CH4(g) 2 O2(g) ? CO2(g) 2 H2O(g) CH4 (g)
-74.86 kJ O2(g) 0 kJ CO2(g) -393.5 kJ
H2O(g) -241.8 kJ ?H -393.5
2(-241.8)--74.86 2 (0) ?H -802.2 kJ
17
Stoichiometry and Energy Changes
If H2(g) 1/2 O2(g) ? H2O(g) ?H -285.5
kJ then H2O(g) ? H2(g) 1/2 O2(g) ?H
285.5 kJ If you turn an equation around, you
change the sign 2 H2O(g) ? 2 H2(g) O2(g)
?H 571.0 kJ If you multiply the equation by a
number, you multiply the heat by that number.
18
Stoichiometry and Solutions

• MolarityA measure of the amount of solute
  dissolved in a certain amount of solvent.
• Concentrated solution has a large amount of
  solute.
• Dilute solution has a small amount of solute
• Sometimes g/l or g/mL of g/100 mL.
• But chemical reactions dont happen in grams

19
(No Transcript)
20
Stoichiometry and Solutions

• Pour in a small amount of solvent,
• then add the solute and dissolve it,
• then fill to final volume.
• M x L moles
• How many moles of NaCl are needed to make 6.0 L
  of a 0.75 M NaCl solution?
• How many grams of CaCl2 are needed to make 625 mL
  of a 2.0 M solution?

21
Stoichiometry and Solutions

• The number of moles of solute in 1 Liter of the
  solution.
• M moles/Liter
• What is the molarity of a solution with 2.0 moles
  of NaCl in 4.0 Liters of solution.
• What is the molarity of a solution with 3.0 moles
  dissolved in 250 mL of solution.

22
Stoichiometry and Solutions

• 10.3 g of NaCl are dissolved in a small amount of
  water then diluted to 250 mL. What is the
  concentration?
• How many grams of sugar are needed to make 125 mL
  of a 0.50 M C6H12O6 solution?

23
Stoichiometry and Solutions

• The number of moles of solute doesnt change if
  you add more solvent.
• The moles before the moles after
• M1 x V1 M2 x V2
• M1 and V1 are the starting concentration and
  volume.
• M2 and V2 are the starting concentration and
  volume.
• Stock solutions are pre-made to known M