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[