Measuring and Comparing the Volumes of Reacting Gases Early 1800s, Gay-Lussac studied gas volume relationships with chemical reaction between H and O Observed 2 L H can react with 1 L O to form 2 L of water vapor at constant temperature and pressure Hydrogen gas + oxygen gas water vapor 2 L 1 L 2 L 2 volumes1 volume 2 volumes
Reaction shows 2:1:2 relationship between volumes of reactants and product Ratio applies to any proportions (mL, L, cm 3 ) Gay-Lussac also noticed ratios by volume between other reactions of gases Hydrogen gas + chlorine gas hydrogen chloride gas 1 L 1 L 2 L
Law of Combining Volumes of Gases 1808 Gay-Lussac summarized results in Gay-Lussac’s law of combining volumes of gases at constant temperature and pressure, the volumes of gaseous reactants and products can be expressed as ratios of small whole numbers
Avogadro’s Law Important point of Dalton’s atomic theory: atoms are indivisible Dalton also thought particles of gaseous elements exist in form of single atoms Believed one atom of one element always combines with one atom of another element to form single particle of product
Gay-Lussac’s results presented problem for Dalton’s theory Ex. Reactions like formation of water Hydrogen gas + oxygen gas water vapor 2 L 1 L 2 L Seems that oxygen involved would have to divide into two parts
1811 Avogadro found way to explain Gay- Lussac’s simple ratios of combining volumes without violating Dalton’s idea of indivisible atoms Rejected Dalton’s idea that reactant elements are always in monatomic form when they combine to form products Reasoned these molecules could contain more than 1 atom
Avogadro’s law equal volumes of gases at the same temperature and pressure contain equal numbers of molecules At the same temp and pressure, volume of any given gas varies directly with the number of molecules
1 mol CO2 at STP = 22.4 L 1 mol O2 at STP = 22.4 L 1 mol H2 at STP = 22.4 L
Consider reaction of H and Cl to produce HCl According to Avogadro’s law, equal volumes of H and Cl contain same number of molecules b/c he rejected Dalton’s theory that elements are always monatomic, he concluded H and Cl components must each consist of 2 or more atoms joined together
Simplest assumption was that H and Cl molecules had 2 atoms each Leads to following balanced equation: H 2 (g) + Cl 2 (g) → 2HCl(g) 1 volume 1 volume 2 volumes 1 molecule 1 molecule 2 molecules If the simplest formula for hydrogen chloride, HCl indicates molecule contains 1 H and 1 Cl Then the simplest formulas for hydrogen and chlorine must be H 2 and Cl 2
Avogadro’s law also indicates that gas volume is directly proportional to the amount of gas, at given temp and pressure V = kn k = constant n = amount of gas in moles
Molar Volume of Gases Remember 1 mol of substance contains 6.022 x 10 23 According to Avogadro’s law, 1 mol of any gas occupies same volume as 1 mol of any other gas at same temperature and pressure, even though masses are different Standard molar volume of gas volume occupied by 1 mol of gas at STP = 22.4 L
Knowing volume of gas, you can use 1mol/22.4 L as conversion factor Can find number of moles Can find mass Can also use molar volume to find volume if you have number of moles or mass
Sample Problem A chemical reaction produces 0.0680 mol of oxygen gas. What volume in liters is occupied by this gas sample at STP?
1. Analyze Given: moles of O2 = 0.0680 mol Unknown: volume of O2 in liters at STP
2. Plan moles of O2 → liters of O2 at STP The standard molar volume can be used to find the volume of a known molar amount of a gas at STP
Practice Problems At STP, what is the volume of 7.08 mol of nitrogen gas? 159 L N 2 A sample of hydrogen gas occupies 14.1 L at STP. How many moles of the gas are present? 0.629 mol H 2 At STP, a sample of neon gas occupies 550. cm 3. How many moles of neon gas does this represent? 0.0246 mol Ne
Sample Problem A chemical reaction produced 98.0 mL of sulfur dioxide gas, SO 2, at STP. What was the mass (in grams) of the gas produced?
1. Analyze Given: volume of SO 2 at STP = 98.0 mL Unknown: mass of SO 2 in grams
2. Plan liters of SO 2 at STP→moles of SO 2 →grams of SO 2
Practice Problems What is the mass of 1.33 × 10 4 mL of oxygen gas at STP? 19.0 g O 2 What is the volume of 77.0 g of nitrogen dioxide gas at STP? 37.5 L NO 2 At STP, 3 L of chlorine is produced during a chemical reaction. What is the mass of this gas? 9 g Cl 2
A gas sample can be characterized by 4 quantities 1. Pressure 2. Volume 3. Temperature 4. Number of moles
Number of moles present always affects at least one of the other 3 quantities Collision rate per unit area of container wall depends on number of moles Increase moles, increase collision rate, increase pressure
Pressure, volume, temperature, moles are all interrelated A mathematical relationship exists to describe behavior of gas for any combination of these conditions Ideal gas law mathematical relationship among pressure, volume, temperature, and number of moles of a gas
Derivation of Ideal Gas Law Derived by combining the other gas laws Boyle’s law: at constant temp, the volume of a given mass of gas is inversely proportional to the pressure.
Charles’s law: At constant pressure, volume of given mass of gas is directly proportional to Kelvin temperature V α T Avogadro’s law: at constant temp and pressure, volume of given mass of gas is directly proportional to the number of moles V α n
Volume is proportional to pressure, temp and moles in each equation Combine the 3: Can change proportion to equality by adding constant, this time R
This equation says the volume of a gas varies directly with the number of moles of gas and its Kelvin temperature Volume also varies inversely with pressure
Ideal gas law combines Boyle’s, Charles’s, Gay-Lussac’s, and Avogadro’s laws Ex: PV = nRT n and T are constant, nRT is constant b/c R is also constant This makes PV = constant which is Boyle’s law
The Ideal Gas Constant Ideal gas constant R Value depends on units for volume, pressure, temp Unit of RValue of R Unit of PUnit of VUnit of TUnit of n 62.4mmHgLKMol 0.0821AtmLKMol 8.314Pam3m3 KMol 8.314kPaLKMol
Sample Problem What is the pressure in atmospheres exerted by a 0.500 mol sample of nitrogen gas in a 10.0 L container at 298 K?
1. Analyze Given: V of N2 = 10.0 L n of N2 = 0.500 mol T of N2 = 298 K Unknown: P of N2 in atm
2. Plan n,V,T → P The gas sample undergoes no change in conditions Therefore, the ideal gas law can be rearranged and used to find the pressure as follows
Practice Problems What pressure, in atmospheres, is exerted by 0.325 mol of hydrogen gas in a 4.08 L container at 35°C? 2.01 atm A gas sample occupies 8.77 L at 20°C.What is the pressure, in atmospheres, given that there are 1.45 mol of gas in the sample? 3.98 atm
What is the volume, in liters, of 0.250 mol of oxygen gas at 20.0°C and 0.974 atm pressure? 6.17 L O 2 A sample that contains 4.38 mol of a gas at 250 K has a pressure of 0.857 atm. What is the volume? 105 L How many liters are occupied by 0.909 mol of nitrogen at 125°C and 0.901 atm pressure? 33.0 L N 2
What mass of chlorine gas, Cl 2, in grams, is contained in a 10.0 L tank at 27°C and 3.50 atm of pressure? 101 g Cl 2 How many grams of carbon dioxide gas are there in a 45.1 L container at 34°C and 1.04 atm? 81.9 g CO 2 What is the mass, in grams, of oxygen gas in a 12.5 L container at 45°C and 7.22 atm? 111 g O 2
A sample of carbon dioxide with a mass of 0.30 g was placed in a 250 mL container at 400. K. What is the pressure exerted by the gas? 0.90 atm
Finding Molar Mass or Density from Ideal Gas Law If P, V, T and mass are known you can calculate number of moles (n) in sample Can calculate molar mass (g/mol) Equation shows relationship between density, P, T, molar mass Mass divided by molar mass gives moles Substitute m/M for n in equation PV=nRT
Density (D) = mass (m) per unit volume (V) D = m/V
Sample Problem At 28°C and 0.974 atm, 1.00 L of gas has a mass of 5.16 g. What is the molar mass of this gas?
1. Analyze Given: P of gas = 0.974 atm V of gas = 1.00 L T of gas = 28°C + 273 = 301 K m of gas = 5.16 g Unknown: M of gas in g/mol
2. Plan P, V, T, m → M You can use the rearranged ideal gas law provided earlier to find the answer
Practice Problems What is the molar mass of a gas if 0.427 g of the gas occupies a volume of 125 mL at 20.0°C and 0.980 atm? 83.8 g/mol What is the density of a sample of ammonia gas, NH 3, if the pressure is 0.928 atm and the temperature is 63.0°C? 0.572 g/L NH 3
The density of a gas was found to be 2.0 g/L at 1.50 atm and 27°C. What is the molar mass of the gas? 33 g/mol What is the density of argon gas,Ar, at a pressure of 551 torr and a temperature of 25°C? 1.18 g/L Ar
You can apply gas laws to calculate stoichiometry of reactions involving gases Coefficients in balanced equations represent mole AND volume ratios
Volume-Volume Calculations Propane, C 3 H 8, is a gas that is sometimes used as a fuel for cooking and heating. The complete combustion of propane occurs according to the following equation. C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(g) (a) What will be the volume, in liters, of oxygen required for the complete combustion of 0.350 L of propane? (b) What will be the volume of carbon dioxide produced in the reaction? Assume that all volume measurements are made at the same temperature and pressure.
1. Analyze Given: balanced chemical equation V of propane = 0.350 L Unknown: a. V of O 2 in L; b. V of CO 2 in L
2. Plan a. V of C 3 H 8 → V of O 2 ; b. V of C 3 H 8 → V of CO 2 All volumes are to be compared at the same temperature and pressure Therefore, volume ratios can be used like mole ratios to find the unknowns.
Practice Problem Assuming all volume measurements are made at the same temperature and pressure, what volume of hydrogen gas is needed to react completely with 4.55 L of oxygen gas to produce water vapor? 9.10 L H 2
What volume of oxygen gas is needed to react completely with 0.626 L of carbon monoxide gas, CO, to form gaseous carbon dioxide? Assume all volume measurements are made at the same temperature and pressure. 0.313 L O 2
Volume-Mass and Mass-Volume Calculations gas volume A → moles A → moles B → mass B or mass A → moles A → moles B → gas volume B You must know the conditions under which both the known and unknown gas volumes have been measured The ideal gas law is useful for calculating values at standard and nonstandard conditions
SAMPLE PROBLEM Calcium carbonate, CaCO 3, also known as limestone, can be heated to produce calcium oxide (lime), an industrial chemical with a wide variety of uses. The balanced equation for the reaction follows. Δ CaCO 3 (s) → CaO(s) + CO 2 (g) How many grams of calcium carbonate must be decomposed to produce 5.00 L of carbon dioxide gas at STP?
1. Analyze Given: balanced chemical equation desired volume of CO 2 produced at STP = 5.00 L Unknown: mass of CaCO 3 in grams
2. Plan The known volume is given at STP This tells us the pressure and temperature The ideal gas law can be used to find the moles of CO 2 The mole ratios from the balanced equation can then be used to calculate the moles of CaCO 3 needed (Note that volume ratios do not apply here because calcium carbonate is a solid)
Practice Problem What mass of sulfur must be used to produce 12.61 L of gaseous sulfur dioxide at STP according to the following equation? S 8 (s) + 8O 2 (g) → 8SO 2 (g) 18.0 g S 8
How many grams of water can be produced from the complete reaction of 3.44 L of oxygen gas, at STP, with hydrogen gas? 5.53 g H 2 O
Tungsten, W, a metal used in light-bulb filaments, is produced industrially by the reaction of tungsten oxide with hydrogen. WO 3 (s) + 3H 2 (g) →W(s) + 3H 2 O(l) How many liters of hydrogen gas at 35°C and 0.980 atm are needed to react completely with 875 g of tungsten oxide? 292 L H 2
What volume of chlorine gas at 38°C and 1.63 atm is needed to react completely with 10.4 g of sodium to form NaCl? 3.54 L Cl 2 How many liters of gaseous carbon monoxide at 27°C and 0.247 atm can be produced from the burning of 65.5 g of carbon according to the following equation? 2C(s) + O 2 (g) → 2CO(g) 544 L CO
Graham’s Law of Effusion Rates of effusion and diffusion depend on relative velocities of gas molecules Velocity varies inversely with mass (lighter molecules move faster)
Average kinetic energy ½ mv 2 For two gases, A and B, at same temp: ½ M A v A 2 = ½ M B v B 2 M A and M B = molar masses of A and B Multiply by 2 M A v A 2 = M B v B 2
Suppose you wanted to compare the velocities of the two gases You would first rearrange the equation above to give the velocities as a ratio
Take square root of each side This equation shows that velocities of two gases are inversely proportional to the square roots of their molar masses
b/c rates of effusion are directly proportional to molecular velocities, can write
Graham’s Law of Effusion The rates of effusion of gases at the same temperature and pressure are inversely proportional to the square roots of their molar masses
Application of Graham’s Law Graham’s experiments dealt with densities of gases Density varies directly with molar mass So…square roots of molar masses from equation can be replaced with square roots of densities
Sample Problem Compare the rates of effusion of hydrogen and oxygen at the same temperature and pressure.
1. Analyze Given: identities of two gases, H 2 and O 2 Unknown: relative rates of effusion
2. Plan molar mass ratio → ratio of rates of effusion The ratio of the rates of effusion of two gases at the same temperature and pressure can be found from Graham’s law
3. Compute Hydrogen effuses 3.98 times faster than oxygen.
Practice Problems A sample of hydrogen effuses through a porous container about 9 times faster than an unknown gas. Estimate the molar mass of the unknown gas. 160 g/mol Compare the rate of effusion of carbon dioxide with that of hydrogen chloride at the same temperature and pressure. CO 2 will effuse about 0.9 times as fast as HCl
If a molecule of neon gas travels at an average of 400 m/s at a given temperature, estimate the average speed of a molecule of butane gas, C 4 H 10, at the same temperature. about 235 m/s