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MOLARITY. INTENSIVE and EXTENSIVE PROPERTIES Extensive Properties - are properties that depend on how much matter is being considered –for example: volume.

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Presentation on theme: "MOLARITY. INTENSIVE and EXTENSIVE PROPERTIES Extensive Properties - are properties that depend on how much matter is being considered –for example: volume."— Presentation transcript:

1 MOLARITY

2 INTENSIVE and EXTENSIVE PROPERTIES Extensive Properties - are properties that depend on how much matter is being considered –for example: volume the property of space a substance takes up is a value dependent on how much of the substance there is. –Can you think of others...

3 INTENSIVE and EXTENSIVE PROPERTIES Intensive Properties - properties that do NOT depend on how much matter is being considered –for example: boiling point the boiling point of liquid water into water vapor is the same regardless of how much water there is. –can you think of others...

4 MOLE vs. MOLARITY mole (mol) 1 mol = 6.02 x 10^23 particles of a particular substance molarity (M) molarity or molar concentration is the number of moles of solute per liter of solution the dissolved substance A solvent is what a solute dissolves in to make a solution

5 molarity is defined as... molarity = moles of solute / liters of solution So, is molarity an intensive or extensive property? intensive property

6 Let's say......you have a 1.46 molar glucose solution. (this means you have 1.46 moles of the solute, which is glucose, dissolved in a solvent to make 1 L of the final solution) *Of course, we don't always work with solutions that have a volume of 1L. So, in this situation, if we had only 50 mL of the solution, there would be mol of glucose in the total solution

7 How this works out...

8 Practice, Practice, Practice... How many grams of potassium dichromate are required to prepare a 250mL solution whose concentration is 2.16 M? Strategy First, figure how many moles are in the solute then, 2. Convert those moles to grams

9 1. How many moles in make up the solute? How many grams of potassium dichromate are required to prepare a 250mL solution whose concentration is 2.16 M?

10 2. Convert moles to grams (0.54 moles) Got to find that MOLAR MASS, baby!!!

11 ACIDS An ACID is a substance that ionizes in H 2 0 to produce H+ 1.TASTE: Sour 2.COLOR: Changes litmus paper from blue to red –a litmus test determines how acidic or basic a substance is

12 ACIDS 3. Reactive with certain metals: –Acids react with zinc, iron, and magnesium to produce hydrogen –magnesium is used as a common treatment for stomach aches neutralizes the HCl in your stomach

13 ACIDS 4. Acids react with carbonates and bicarbonates (Na 2 CO 3, CaCO 3, and NaHCO 3 ) to produce CO 2 gas 5. Acids are proton donors 6. Aqueous acid solution conduct electricity

14 BASES Bases - substances that ionize in water to produce OH- ions 1. TASTE: bitter 2. TOUCH: slippery 3. Changes litmus paper from red to blue 4. aqueous base solutions ALSO conduct electricity 5. Proton Acceptors

15 Shared characteristics between A's and B's If both acids and bases are strong electrolytes, they will become neutralized. MEANING of electrolyte: –becoming ionized in a solution (charged) –when in a solution, electrolytes conduct electricity *If there is a strong acid and a weak base or a weak acid and a strong base, the reaction will not neutralize

16 ACID-BASE NEUTRALIZATION ACID + BASE → Salt + Water –HCl(aq) + NaOH(aq) → NaCl(aq) + H20(l) –BTW... What do the letters in parantheses mean? (aq) aqueous solutions are compounds dissolved in another substance (l) liquid is a state of matter DEFINITION of Neutralization: A reaction between an acid and a base generaly producing water and salt (the salt being an ionic compound made up of cations other than H+ and an anion other than OH- or O 2 -)

17 Back to Enthalpy...

18 Reactions may occur in a series of steps, but... HESS'S LAW: –When reactants are converted into products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. ANALOGY: You take an elevator from the first floor to the sixth floor of a building. The gain in your gravitational potential energy (which corresponds to the enthalpy change for the overall process) is the same whether you go directly to the sixth floor or stop at each floor on your way up (breaking the trip into a series of steps)

19 Standard Enthalpy of Formation The elevation of every coast and land surface of this planet is related to an arbitrary value we call the "sea level." It is defined as zero. Sea levels change and are changing, so the value that represents the height of a land surface like, Mt. Everest, has no correlation to its distance from the center of the Earth.

20 Standard Enthalpy of Formation If the actual Atlantic Ocean rises high enough to flood the streets of NY, the distance from the sea level to the height of Mt. Everest would no longer really be 8,848 m even though its distance from the center of the Earth did not change. The point is, we take a commonly agreed upon starting point (sea level) and determine the elevation of land surfaces based on their distance from that point.

21 Standard Enthalpy of Formation The same is true for states of enthalpy. Just like at sea level where we say the elevation is zero, a substance's standard enthalpy of formation is also zero. You could say that a substances standard enthalpy of formation is the sea level for that substance's change in heat occurring during a reaction. The value is zero for the most stable forms of a substance: Ca(s), O 2 (g), H 2 (g), C (graphite), Hg (l), Mg(s), N 2 (g), etc. All other states of a substance (i.e. liquid O 2 ) and other compounds made from that substance (i.e. O 3 gas) are valued using the most stable form as a reference point.

22 Standard Enthalpy of Formation DEFINITION: –the enthalpy of a reaction carried out at 1 atm and at 25 degrees C –the enthalpy value relates to the heat absorbed or released during the formation of 1 mole of a specific compound The symbol for this value is represented as: Once we know these values, we can calculate the standard enthalpy of a reaction, or...

23 Standard Enthalpy of a Reaction n = the coefficient of the products m = the coefficient of the reactants Σ = "the sum of" f = formation ° = standard state conditions (1 atm and 25 degrees C) m

24 The most stable forms of substances will = 0 Given the equation 3 O 2 (g)→2 O 3 (g) ΔH = kJ, calculate ΔH for the following reaction. 3/2 O 2 (g) → O 3 (g). REMEMBER... O 2 is the most stable form of oxygen. So, it is equal to 0. Therefore, the the change in enthalpy is all about the products. (x products) - (0 reactants) = = X products = kJ You can check using the table on P. 247 i

25 P. 247 The ΔH°f for O 3 (g) is The ΔH°f for O 2 (g) is 0 Given the equation 3 O 2 (g)→2 O 3 (g) ΔH = kJ There are two O 3 in the equation (142.2 kJ x 2 = kJ) –we must remember the coefficients!!!! Then, the question asks us to calculate ΔH for the following reaction. 3/2 O 2 (g) → O 3 (g). There is only one O 3. Since 3/2 O 2 (g) → O 3 (g) is ½ of 3 O 2 (g)→2 O 3 (g) the enthalpy of the reaction will be ½ as well: ½ (+284.4kJ)...or kJ

26 Change of Enthalpy of Reaction The combustion of thiophene, C 4 H 4 S(l), a compound used in the manufacture of pharmaceuticals, produces carbon dioxide and sulfur dioxide gases and liquid water. The enthalpy change in the combustion of one mole of C 4 H 4 S(l) is -2523kJ. Use this information and data from Table 6.4 (p.247) to establish ΔH°f for C 4 H 4 S(l). STEP I: WRITE THE BALANCED EQUATION *To carry any combustion reaction you need oxygen as a reactant. Liquid water will always be included as one of the products. __ C 4 H 4 S(l) + ___ O 2 (g) → __ CO 2 (g) + ___ S0 2 (g) + ___ H 2 0(l)

27 Change of Enthalpy of Reaction The combustion of thiophene, C 4 H 4 S(l), a compound used in the manufacture of pharmaceuticals, produces carbon dioxide and sulfur dioxide gases and liquid water. The enthalpy change in the combustion of one mole of C 4 H 4 S(l) is -2523kJ. Use this information and data from Table 6.4 (p.247) to establish ΔH°f for C 4 H 4 S(l). STEP II: USE THE COEFFICIENTS FROM YOUR BALANCED EQUATION TO DETERMINE THE TOTAL CHANGE IN ENTHALPY IN THE FORMATION OF EVERY SUBSTANCE USED IN THE REACTION. USE THIS INFORMATION TO WRITE YOUR ΔH° RXN EQUATION. i

28 Change of Enthalpy of Reaction The combustion of thiophene, C 4 H 4 S(l), a compound used in the manufacture of pharmaceuticals, produces carbon dioxide and sulfur dioxide gases and liquid water. The enthalpy change in the combustion of one mole of C 4 H 4 S(l) is -2523kJ. Use this information and data from Table 6.4 (p.247) to establish ΔH°f for C 4 H 4 S(l). STEP II: i PRODUCTS 4 (________kJ/mol CO 2(g) ) 2 (________kJ/mol H 2 O (l) ) 1 (________kJ/mol SO 2(g) ) REACTANTS X kJ/mol C 4 H 4 S(l) 6 (________kJ/mol O 2(g) )

29 Change of Enthalpy of Reaction The combustion of thiophene, C 4 H 4 S(l), a compound used in the manufacture of pharmaceuticals, produces carbon dioxide and sulfur dioxide gases and liquid water. The enthalpy change in the combustion of one mole of C 4 H 4 S(l) is -2523kJ. Use this information and data from Table 6.4 (p.247) to establish ΔH°f for C 4 H 4 S(l). STEP II: i [4(-393.5) + 1(-296.8) + 2(-285.8)] – [1(ΔH) + 6(0)] = PRODUCTS - REACTANTS = ΔH RXN

30 Change of Enthalpy of Reaction The combustion of thiophene, C 4 H 4 S(l), a compound used in the manufacture of pharmaceuticals, produces carbon dioxide and sulfur dioxide gases and liquid water. The enthalpy change in the combustion of one mole of C 4 H 4 S(l) is -2523kJ. Use this information and data from Table 6.4 (p.247) to establish ΔH°f for C 4 H 4 S(l). STEP III: SOLVE for "X" i ΔH(reactants) = ΔH C 4 H 4 S(l) = 81.3kJ/mol

31 REVIEW... You will solve... –specific heat questions –questions about reactions in (constant- pressure) bomb calorimeters –and, questions regarding Change of Enthalpy of a reaction

32 STUDY GUIDE - TEST I A 3.30g sample of the sugar glucose, C 6 H 12 O 6 (s), was placed in a bomb calorimeter, ignited, and burned to form carbon dioxide and water. The temperature of the water changed from 22.4°C to 34.1°C. If the calorimeter contained 850.g water and had a heat capacity of 847J/°C, what is ΔE for the combustion of 1mol glucose? (The heat capacity of the bomb is the energy transfer required to raise the bomb’s temperature by 1oC. STEPS... 1.FIND THE MOLAR MASS OF GLUCOSE 2.SET UP THE EQUATION FOR ΔE sys 3.FIND THE ΔE FOR ALL THE INDIVIDUAL COMPONENTS OF THAT MAKE UP THE SYSTEM 4.SUBTRACT TO FIND ΔE OF GLUCOSE

33 STEP I FIND THE MOLAR MASS OF GLUCOSE... ELEMENT# of ATOMS atomic mass per atom TOTAL MASS OF ELEMENT C H O g/mol

34 STEP II - SET UP THE EQUATION FOR ΔE sys Δq rxn + Δq H 2 0 = Δq sys (in this case, we have a calorimeter acting as an isolated system) Δq sys = Δq cal Δq rxn + Δq H 2 0 = Δq cal (since it is a closed system, we can set it to zero) -Δq cal + Δq rxn + Δq H 2 0 = 0 NOW, solve for the change of heat for the individual components

35 STEP III - Individual Components Let's start with the ΔE of the Calorimeter heat capacity of "cal" = 847 J/°C Δt = 11.7°C Δq cal = C(Δt) Δq cal = (847 J/°C)(11.7°C) Δq cal = J (-) J + Δq rxn + Δq H 2 0 = 0

36 STEP III - Individual Components Let's move on to the water surrounding the rxn Specific Heat of water = J/g°C Mass of water = g Same Δt = 11.7°C Δq H 2 O = ms(Δt) Δq H 2 O = (850.0g)(4.184J/g°C)(11.7°C) Δq H 2 O = 41,609.88J J + Δq rxn J = 0 Δq rxn = - 51, J

37 STEP IV - find the ΔE per mol of glucose Bring back the molar mass of glucose... Convert Δq rxn from J to kJ ΔE for the combustion of 1 mol of glucose kJ/0.018 mol = ? kJ/1 mol ΔE = kJ

38 Question 5 When 50.0mL of 1.00M HCl is mixed with 50.0mL of 1.00M NaOH (both at 23.0°C), the resulting solution increases in temperature to 29.8°C. Assuming that the solution has the density and specific heat of pure water, calculate the enthalpy of the reaction. This question is really easy actually...

39 Question 5 - the trick is in the wording... It states: –"Assuming that the solution has the density and specific heat of pure water..." –For our purposes, are these solutions any different than water? Both have the same specific heat as water Both have the same density as water –meaning: 1 mL is equal to 1 gram

40 Question 5 - the trick is in the wording... If they heat like water and... If their masses are like water Then, We calculate the enthalpy of the reaction like water MEANING: Treat this problem as if these two solutions are two portions of water reacting together...

41 Question 5 ΔH rxn = qHCl + qNaOH (50g)(4.184J/g°C)(29.8°C-23°C) ΔH rxn = 2845 J = 2.85 kJ ] x 2 [


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