1.3 Reacting Masses and Volumes Reacting Gases Kristin Page IB SL Chemistry

Understandings Avogadro’s law enables the mole ratio of reacting gases to be determined from volumes of the gases. The molar volume of an ideal gas is a constant at specified temperature and pressure. The molar concentration of a solution is determined by the amount of solute and the volume of solution. A standard solution is one of known concentration.

Application & Skills Calculation of reacting volumes of gases using Avogadro’s law. Solution of problems and analysis of graphs involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas. Solution of problems relating to the ideal gas equation. Explanation of the deviation of real gases from ideal behavior at low temperature and high pressure. Obtaining and using experimental values to calculate the molar mass of a gas from the ideal gas equation. Solution of problems involving molar concentration, amount of solute and volume of solution. Use of the experimental method of titration to calculate the concentration of a solution by reference to a standard solution.

Kinetic Theory of Gases Gas particles have high energy are separated by a lot of space Gas particles are move rapidly in straight lines but random directions Gas particles collide with each other and with the container but do not lose energy There is no attractive force between gas particles These hold true for ideal gases

Ideal Gases Model of the behavior of real gases Under “normal” conditions known as standard temperature and pressure (STP) STP: P = 100 kPa and T = 273°K Useful conversions: 100 kPa = 1atm and 273°K = 0°C Gas particles have high energy are separated by a lot of space Gas particles are move rapidly in straight lines but random directions Gas particles collide with each other and with the container but do not lose energy There is no attractive force between gas particles

Absolute Zero When dealing with gases we need to use STP. This means that units of temperature must be in Kelvin (°K) 273°K = 0°C The Kelvin scale starts at absolute zero (0°K) It is not actually possible to reach absolute zero since this would be the point at which particles stopped all motion* * Actually technically not true anymore but for our purposes we will stick with this. If you are curious go to http://www.nature.com/news/quantum-gas-goes-below-absolute- zero-1.12146

Avogadro’s Law Relationship between amount of gas (n) and the volume (m3) At constant temperature & pressure 𝑛∝𝑉 𝑛 1 𝑉 1 = 𝑛 2 𝑉 2 (m3) Demonstration: Balloons, as I add more gas particles, volume increases. When I let gas out, volume decreases This means that volumes can be used directly instead of moles in equations involving gases. Ex: H2(g) + Cl2(g)  2HCl(g)

Avogadro’s Law Problems Consider the following reaction for the synthesis of methanol: CO(g) + 2H2(g)CH3OH(g) (Assume same T & P) What volume of H2 reacts exactly with 2.50dm3 of CO? Mole ratio CO: H2 = 1:2 2.50 𝑑𝑚 3 𝐶𝑂 × 2 𝐻 2 1 𝐶𝑂 = 5.00 dm3 of H2 reacts What volume of CH3OH is produced? Mole ratio CO: CH3OH = 1:1 2.50 𝑑𝑚 3 𝐶𝑂 × 1 𝐶𝐻 3 𝑂𝐻 1 𝐶𝑂 =2.50 𝑑𝑚 3 𝐶𝐻 3 𝑂𝐻 Demonstration: Balloons, as I add more gas particles, volume increases. When I let gas out, volume decreases

Avogadro’s Law Problems If 100 cm3 of oxygen reacts with 30cm3 of methane in the following reaction, how much oxygen will be left over at the end of the reaction? CH4(g) + 2O2(g)CO2(g)+H2O(g) (Assume same T & P) Mole ratio CH4: O2 = 1:2 30 𝑐𝑚 3 𝐶 𝐻 4 × 2 𝑂 2 1 𝐶 𝐻 4 =60 𝑐𝑚 3 𝑂 2 𝑖𝑠 𝑢𝑠𝑒𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 100 𝑐𝑚 3 −60 𝑐𝑚 3 =40 𝑐𝑚 3 𝑂 2 𝑖𝑠 𝑖𝑛 𝑒𝑥𝑐𝑒𝑠𝑠 Demonstration: Balloons, as I add more gas particles, volume increases. When I let gas out, volume decreases

Molar Volume According to Avogadro’s law, the volume of 1 mole of and ideal gas at STP is constant, this is known as the Molar Volume Molar volume = 22.7 dm3 mol-1 Note: 1 dm3 = 1L =1000cm3 Calculate the number of moles in 250.cm3 of O2 at STP 250 𝑐𝑚 3 × 1 𝑑𝑚 3 1000𝑐𝑚 3 × 1 𝑚𝑜𝑙 22.7 𝑑𝑚 3 =0.0110 𝑚𝑜𝑙 Calculate the volume of 0.135 mol CO2 at STP 0.135 𝑚𝑜𝑙 𝐶𝑂 2 × 22.7 𝑑𝑚 3 1 𝑚𝑜𝑙 =3.06 𝑑𝑚 3 Demonstration: Balloons, as I add more gas particles, volume increases. When I let gas out, volume decreases

Boyle’s Law Relationship between pressure and volume At constant temperature, the volume of a fixed mass of an ideal gas is inversely proportional to its pressure 𝑃∝ 1 𝑉 Gases in smaller volume will have more collisions with the container and so higher pressure This can also be written 𝑉 1 𝑃 1 = 𝑉 2 𝑃 2 where 𝑉 1 𝑎𝑛𝑑 𝑃 1 =𝑖𝑛𝑖𝑡𝑖𝑎𝑙 and 𝑉 2 𝑎𝑛𝑑 𝑃 2 =𝑓𝑖𝑛𝑎𝑙 Pressure chamber (vacuum) – inflate a marshmallow, balloon, bags of chips (decrease pressure increase volume) http://www.uccs.edu/vgcl/gas-laws/experiment-1-boyles-law.html

Boyle’s Law Example 𝑃 2 = 𝑉 1 𝑃 1 𝑉 2 𝑃 2 = 1.50∙101 0.462 =328𝑘𝑃𝑎 To make an air horn, 1.50 dm3 of air at 101 kPa are compressed into a can with a volume of 0.462 dm3. Assuming a constant temperature, what is the pressure on the compressed air? 𝑃 1 =101𝑘𝑃𝑎, 𝑉 1 =1.50 𝑑𝑚 3 , 𝑃 2 =? kPa, 𝑉 2 =0.462 𝑑𝑚 3 , 𝑉 1 𝑃 1 = 𝑉 2 𝑃 2 𝑃 2 = 𝑉 1 𝑃 1 𝑉 2 𝑃 2 = 1.50∙101 0.462 =328𝑘𝑃𝑎 Pressure chamber (vacuum) – inflate a marshmallow, balloon, bags of chips (decrease pressure increase volume)

Charles’s Law Relationship between volume and temperature At constant pressure the volume of a fixed mass of ideal gas is directly proportional to its temperature (K) 𝑉∝𝑇 or 𝑉 1 𝑇 1 = 𝑉 2 𝑇 2 As temperature increase particles move more rapidly (higher kinetic energy) and collide with surface more causing an increase in volume Temp. must be in Kelvin! Balloon in boiling water bath, balloon in freezer http://www.uccs.edu/vgcl/gas-laws/experiment-2-charles-law.html

Charles’s Law Example On hot days, you may have noticed that potato chip bags seem to “inflate”, even though they have not been opened. If I have a 250mL bag at a temperature of 19°C, and I leave it in my car which has a temperature of 45°C, what will the new volume of the bag be? Temp. must be in Kelvin! 𝑇 1 =19+273=292𝐾 𝑇 2 =45+273=318 𝐾 𝑉 1 𝑇 1 = 𝑉 2 𝑇 2 𝑽 𝟐 = 𝑽 𝟏 𝑻 𝟐 𝑻 𝟏 𝑉 2 = 250𝑚𝐿∙318𝐾 292𝐾 =272 𝑚𝐿 Balloon in boiling water bath, balloon in freezer

Gay-Lussac’s Law (1778-1850) 1800’s studied gases using hot air balloons When volume is constant, pressure of the gas is directly proportional to temperature 𝑝∝𝑇 or 𝑝 1 𝑇 1 = 𝑝 2 𝑇 2 Pressure in Nm-2 (Pa) Temp in K Demonstrations: Crushing Can, Egg in a Bottle Large Erlenmeyer flask, small amount of water, heat on hot plate. Balloon on top – balloon inflates, put flask in ice water bath, balloon inflates inside the jar http://www.uccs.edu/vgcl/gas-laws/experiment-3-gay-lussacs-law.html

Combined Gas Law All 3 gas laws can be combined into one equation known as the combined gas law For a fixed amount of gas 𝒑 𝟏 𝑽 𝟏 𝑻 𝟏 = 𝒑 𝟐 𝑽 𝟐 𝑻 𝟐 This equation can technically be used for ANY gas law problem since the constant variable will cancel out. Note: all units must be the same on both sides of the equation and temperature must be in Kelvin!

Combined Gas Law Example If the volume of an ideal gas collected at 0°C and 100 kPa is 50.0cm3, what would be the volume at 60°C and 108 kPa? 𝑇 1 =0+273=273𝐾, 𝑝 1 =100𝑘𝑃𝑎, 𝑉 1 =50.0 𝑐𝑚 3 , 𝑇 2 =60+273=333𝐾, 𝑝 2 =108 𝑘𝑃𝑎, 𝑉 2 =? 𝒑 𝟏 𝑽 𝟏 𝑻 𝟏 = 𝒑 𝟐 𝑽 𝟐 𝑻 𝟐 so 𝑽 𝟐 = 𝒑 𝟏 𝑽 𝟏 𝑻 𝟐 𝑻 𝟏 𝒑 𝟐 𝑽 𝟐 = 𝟏𝟎𝟎∙𝟓𝟎.𝟎∙𝟑𝟑𝟑 𝟐𝟕𝟑∙𝟏𝟎𝟖 =𝟓𝟔.𝟓 𝒄𝒎 𝟑

Ideal Gas Equation 𝒑𝑽=𝒏𝑹𝑻 Now we can add the gas equations with Avogadro’s Law to get the ideal gas equation 𝒑𝑽=𝒏𝑹𝑻 R = gas constant = 8.31 J K-1mol-1 (Data Book p.2) Must use the following units: p (Pa) V (m3) T (K) Conversion Factors 1 Pa = 1 J m-3 1 dm3 = 1 x 10-3 m3 (Data book)

Ideal Gas Equation Practice An ideal gas occupies 590 cm3 at 120°C and 202 kPa. What amount of gas (in moles) is present? 𝒑𝑽=𝒏𝑹𝑻 𝑝=202 𝑘𝑃𝑎∙ 1000 𝑃𝑎 1 𝑘𝑃𝑎 =202,000 𝑃𝑎 𝑉=590 𝑐𝑚 3 ∙ 0.001 𝑚 3 1000 𝑐𝑚 3 =5.9× 10 −4 𝑚 3 R = 8.31 J K-1mol-1 𝑇=120+273=393 𝐾 𝑛= 202,000𝑃𝑎×(5.9× 10 −4 ) 𝑚 3 8.31𝐽 𝐾 −1 𝑚𝑜𝑙 −1 × 393𝐾 =0.036 𝑚𝑜𝑙

Ideal Gas Equation Practice A gas has a density of 1.24 g dm-3 at 0°C and 1.00 x 105 Pa. Calculate its molar mass. 𝑺𝑻𝑷 𝑮𝒊𝒗𝒆𝒏 𝑀𝑜𝑙𝑎𝑟 𝑉𝑜𝑙𝑢𝑚𝑒=22.7 𝑑𝑚 3 𝑚𝑜𝑙 −1 1.24 𝑔 𝑑𝑚 3 ∙ 22.7 𝑑𝑚 3 𝑚𝑜𝑙 = 28.1𝑔 𝑚𝑜𝑙