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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 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. - PowerPoint PPT Presentation
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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
• 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...
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...
MOLE vs. MOLARITY
• mole (mol)• 1 mol = 6.02 x 10^23
particles of a particular substance
• molarity (M)• molarity or molar co
ncentration is the number of moles of solute per liter of solutionthe dissolved
substanceA solvent is what a solute dissolves in to make a solution
molarity is defined as...
molarity = moles of solute / liters of solution
So, is molarity an intensive or extensive property?
intensive property
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 0.073 mol of glucose in the total solution
How this works out...
6126
6126
46.1soln 1L
soln 1000
50
073.0
OHMCmolarity
mLx
mL
OHmolCmolarity
Practice, Practice, Practice...
• How many grams of potassium dichromate are required to prepare a 250mL solution whose concentration is 2.16 M?
Strategy...
1. First, figure how many moles are in the solute
then,
2. Convert those moles to grams
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?
moles 0.54
soln of 25.01
K16.2K of moles
soln of L molarity xsolute of mol
rearrange soln of L
sol of mol
722722
LxL
OCrmolOCr
M
2. Convert moles to grams (0.54 moles)
gmol
OCr
OCr
159mol 1
294.2g x 54.0
to...equal is thisK of grams ofamount theFind
294.2g toequal is massmolar
.K of massmolar theFind
722
722
Got to find that MOLAR MASS, baby!!!
ACIDS
• An ACID is a
substance that ionizes
in H20 to produce H+
1.TASTE: Sour
2.COLOR: Changes litmus paper from blue to red – a litmus test determines how acidi
c or basic a substance is
ACIDS3. 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
)()()()(2 22 gHsMgClsMgaqlHC
ACIDS4. Acids react with carbonates and bicarbonates (Na2C
O3, CaCO3, and NaHCO3) to produce CO2 gas
5. Acids are proton donors
6. Aqueous acid solution conduct electricity
)()()()()(2 2223 gCOlOHaqCaClsCaCOaqHCl
ClOHHHCl 32 0
)()()( aqClaqHaqHCl
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
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
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 O2-)
Back to Enthalpy...
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)
Standard Enthalpy of Formation• The elevation of every coast a
nd 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.
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.
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), O2(g), H2(g), C (graphite), Hg (l), Mg(s), N2(g), etc.
• All other states of a substance (i.e. liquid O2) and other compounds made from that substance (i.e. O3 gas) are valued using the most stable form as a reference point.
Standard Enthalpy of Formation
• DEFINITION:– the enthalpy of a reaction carried out at 1 atm and at 25 degr
ees 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:
formation ofenthalpy Standared fH
Once we know these values, we can calculate the standard enthalpy of a reaction, or...
rxnH
Standard Enthalpy of a Reaction rxnH
)reactants()products(H
is...equation dgeneralize The
fHfH nnrxn
• 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
The most stable forms of substances will = 0
EXAMPLEGiven the equation 3 O2(g)→2 O3(g) ΔH = +284.4 kJ,
calculate ΔH for the following reaction. 3/2 O2(g) → O3(g).REMEMBER...
)reactants()products(H
is...equation dgeneralize The
fHfH nnrxn
O2 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) = rxnH = X products = +284.4 kJ
You can check using the table on P. 247
i
P. 247• The ΔH°f for O3(g) is +142.2 • The ΔH°f for O2(g) is 0Given the equation 3 O2(g)→2 O3(g) ΔH = +284.4 kJThere are two O3 in the equation (142.2 kJ x 2 = 284.4 kJ)
– we must remember the coefficients!!!!• Then, the question asks us to calculate ΔH for the
following reaction. 3/2 O2(g) → O3(g). • There is only one O3.
• Since 3/2 O2(g) → O3(g) is ½ of 3 O2(g)→2 O3(g) the enthalpy of the reaction will be ½ as well:
½ (+284.4kJ)...or +142.2kJ
Change of Enthalpy of Reaction • The combustion of thiophene, C4H4S(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 C4H4S(l) is -2523kJ. Use this information and data from Table 6.4 (p.247) to establish ΔH°f for C4H4S(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.
__ C4H4S(l) + ___ O2(g) → __ CO2(g) + ___ S02(g) + ___ H20(l)1 6 4 1 2
Change of Enthalpy of Reaction
• The combustion of thiophene, C4H4S(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 C4H4S(l) is -2523kJ. Use this information and data from Table 6.4 (p.247) to establish ΔH°f for C4H4S(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
)reactants()products(H fHfH nnrxn
Change of Enthalpy of Reaction • The combustion of thiophene, C4H4S(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 C4H4S(l) is -2523kJ. Use this information and data from Table 6.4 (p.247) to establish ΔH°f for C4H4S(l).
STEP II:
i
PRODUCTS
4 (________kJ/mol CO2(g))2 (________kJ/mol H2O(l))1 (________kJ/mol SO2(g))
REACTANTS
X kJ/mol C4H4S(l)
6 (________kJ/mol O2(g))
-393.5
-285.8-296.8 0.00
Change of Enthalpy of Reaction • The combustion of thiophene, C4H4S(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 C4H4S(l) is -2523kJ. Use this information and data from Table 6.4 (p.247) to establish ΔH°f for C4H4S(l).
STEP II:
i
[4(-393.5) + 1(-296.8) + 2(-285.8)] – [1(ΔH) + 6(0)] = -2523
PRODUCTS -
REACTANTS = ΔHRXN
Change of Enthalpy of Reaction • The combustion of thiophene, C4H4S(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 C4H4S(l) is -2523kJ. Use this information and data from Table 6.4 (p.247) to establish ΔH°f for C4H4S(l).
STEP III:
SOLVE for "X"
i
-2442.4 - ΔH(reactants) = -2523
ΔHC4H4S(l) = 81.3kJ/mol
REVIEW...
• You will solve... – specific heat questions– questions about reactions in (constant-pressu
re) bomb calorimeters– and, questions regarding Change of Enthalpy
of a reaction
STUDY GUIDE - TEST I
A 3.30g sample of the sugar glucose, C6H12O6(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 GLUCOSE2. SET UP THE EQUATION FOR ΔEsys
3. FIND THE ΔE FOR ALL THE INDIVIDUAL COMPONENTS OF THAT MAKE UP THE SYSTEM
4. SUBTRACT TO FIND ΔE OF GLUCOSE
STEP IFIND THE MOLAR MASS OF GLUCOSE...
ELEMENT # of ATOMS atomic mass per atom
TOTAL MASS OF ELEMENT
C 6 12.01 72.06
H 12 1.008 12.096
O 6 16.00 96
180.156g/mol
STEP II - SET UP THE EQUATION FOR ΔEsys
• Δqrxn + ΔqH20 = Δqsys (in this case, we have a calorimeter acting as an isolated system)
• Δqsys = Δqcal
• Δqrxn + ΔqH20 = Δqcal
(since it is a closed system, we can set it to zero)
• -Δqcal + Δqrxn + ΔqH20 = 0
• NOW, solve for the change of heat for the individual components
STEP III - Individual Components
• Let's start with the ΔE of the Calorimeter
• heat capacity of "cal" = 847 J/°C
• Δt = 11.7°C
• Δqcal = C(Δt)
• Δqcal = (847 J/°C)(11.7°C)
• Δqcal = 9909.9 J
• (-)9909.9 J + Δqrxn + ΔqH20 = 0
STEP III - Individual Components
• Let's move on to the water surrounding the rxn• Specific Heat of water = 4.184 J/g°C• Mass of water = 850.0 g• Same Δt = 11.7°C• ΔqH2O = ms(Δt)• ΔqH2O = (850.0g)(4.184J/g°C)(11.7°C)• ΔqH2O = 41,609.88J• 9909.9 J + Δqrxn + 41609.88 J = 0• Δqrxn = - 51,519.78 J
STEP IV - find the ΔE per mol of glucose
• Bring back the molar mass of glucose...
• Convert Δqrxn from J to kJ
• ΔE for the combustion of 1 mol of glucose
-51.5 kJ/0.018 mol = ? kJ/1 mol
ΔE = -2861.1 kJ
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...
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
Question 5 - the trick is in the wording...
• If they heat like waterand...• If their masses are like waterThen,• We calculate the enthalpy of the reaction li
ke waterMEANING: Treat this problem as if these tw
o solutions are two portions of water reacting together...
Question 5
• ΔHrxn = qHCl + qNaOH
• (50g)(4.184J/g°C)(29.8°C-23°C)
• ΔHrxn = 2845 J = 2.85 kJ
] x 2[