Unit 4 Section A.17-18 In which you will learn about: Molar volume Avogadro’s law Stoichiometry with gases.

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Presentation transcript:

Unit 4 Section A In which you will learn about: Molar volume Avogadro’s law Stoichiometry with gases

A.17 Ideal Gases and Molar Volume Thus far, you have considered gas behavior under differing conditions of volume, pressure, and temperature. – One variable that you have not fully considered is the actual amount of the gas sample; that is, the number of gas molecules contained in a particular sample. – If you have the same volumes of oxygen gas, nitrogen gas, and carbon dioxide gas in three different balloons at the same temperature and pressure, how do the numbers of molecules compare?

Avogadro’s Law That same question was investigated in the early 1800s by the Italian lawyer and mathematical physics professor Amedeo Avogadro – By making careful observations of gas samples like those you have worked with, he proposed: Equal volumes of all gases at the same temperature and pressure contain the same number of molecules = Avogadro’s Law

Molar Volume One useful consequence of Avogadro’s law is that all gases have equal molar volumes if they are measured at the same temperature and pressure. – Molar volume = volume occupied by one mole of a substance STP, molar volume = 22.4 L/mol – There is no similar simple relationship between moles of various solids or liquids and their corresponding volumes

Gas Stoichiometry The realization that all gases have the same molar volume under the same conditions greatly simplifies our thinking about chemical rxns involving gases – For example, consider the following chemical equations that involve gaseous reactions: N 2 (g) + O 2 (g)  2 NO (g) 2 H 2 (g) + O 2 (g)  2 H 2 O (g) – You have learned that coefficients in chemical equations indicate the relative numbers of molecules or moles of reactants and products – Based on Avogadro’s law, similar calculations can also involve volumes of gaseous reactants and products.

Sample Problem 1 What volume (in liters) of NO (g) is produced as 3 L N 2 (g) and 3 L O 2 (g) react? See equation on previous slide. Because the amounts of both the reactants have tripled, the amount of product will also be tripled: 6 L NO (g) will be produced. Although it works in this problem, note that, unlike mass, volumes are not necessarily conserved in a chemical reaction: – 100 mL mL COULD yield 200 mL, not 300 mL

Sample Problem 2 In the following equation, what volume of C 2 H 6 (g) is needed to produce 12 L CO 2 (g)? All measurements are at the same temperature. 2 C 2 H 6 (g) + 7 O 2 (g)  4 CO 2 (g) + 6 H 2 O (g) From the chemical equation, we know that 2 L C 2 H 6 will produce 4 L CO 2. If we want to produce 12 L of CO 2, which is three times more than is represented in the equation, the we must start with three times as much C 2 H 6 = 6 L C 2 H 6.

Sample Problem 3 Using the same scenario as sample problem 2, how many liters of O 2 would be needed to produce 12 L CO 2 ? 12 L CO 2 x (7 L O 2 / 4 L CO 2 ) = 21 L O 2 Mole Ratio!

A.18 Molar Volume and Reactions of Gases (HOMEWORK) 1)What volume would be occupied by 3 mol CO 2 (g) at 0 °C and 1 atm? 2)In a certain gaseous reaction, 2 mol NO react with 2 mol O 2 : 2 NO (g) + O 2 (g)  2 NO 2 (g) a)Given the same conditions of temperature and pressure, what volume of O 2 (g) would react with 4 L NO gas? b)How might chemists use their knowledge of molar volumes to monitor the progress of this reaction?

Homework Cont’d 3) Toxic carbon monoxide (CO) gas is produced when fossil fuels, such as gasoline, burn without sufficient oxygen gas. The CO can eventually be converted to CO 2 in the atmosphere. Automobile catalytic converters are designed to speed up this conversion: Carbon monoxide gas + oxygen gas  carbon dioxide gas a)Write the balanced equation for this conversion. b)How many moles of oxygen gas would be needed to convert 50.0 mol carbon monoxide to carbon dioxide? c)What volume of oxygen gas would be needed to react with 968 L CO? (Assume both gases are at the same temperature and pressure).