SolutionsSolutions

CE 541CE 541

““A solution is a homogeneous throughout and is A solution is a homogeneous throughout and is composed of two or more pure substances. composed of two or more pure substances. They are weakly bounded mixtures of a solute They are weakly bounded mixtures of a solute and a solvent”and a solvent”

Solute: is usually the component in less quantitySolute: is usually the component in less quantity

Solvent: is usually the component in greater Solvent: is usually the component in greater quantityquantity

The solute dissolves in the solvent and is The solute dissolves in the solvent and is considered soluble in the solventconsidered soluble in the solvent

Aqueous solutions: are solutions where water is Aqueous solutions: are solutions where water is used as the solventused as the solvent

Types of SolutionsTypes of Solutions

Gas in a LiquidGas in a Liquid Liquid in a LiquidLiquid in a Liquid Solid in a LiquidSolid in a Liquid

Solute Solvent Examples

Gas Liquid Carbonated beverages (CO2 in water)

Liquid Liquid Anti-freeze in car radiators (ethyl glycol in watr)

Liquid Solid Dental fillings (mercury in silver)

Solid Liquid Sugar in water

Solid Solid Solder (tin in lead)

Examples of

Solutions

Solute

Gas Liquid Solid

Solvent

GasOxygen and other gases in nitrogen (air)

Water vapor in air (humidity)

Iodine sublimates into air

LiquidCarbon dioxide in water (carbonated water)

Ethanol (common alcohol) in water; various hydrocarbons in each other (petroleum)

Sucrose (table sugar) in water; sodium chloride (table salt) in water; gold in mercury, forming an amalgam

Solid

Hydrogen dissolves rather well in metals; platinum has been studied as a storage medium. This effect was used in the cold fusion experiments.

Hexane in paraffin wax, mercury in gold.

Steel, duralumin, other metal alloys

Factors Affecting Solubility Factors Affecting Solubility and Rate of Solutionand Rate of Solution

1. Factors that affect the actual solubility 1. Factors that affect the actual solubility of a given solute in a solvent:of a given solute in a solvent: Properties of soluteProperties of solute Properties of solventProperties of solvent TemperatureTemperature PressurePressure

2. Factors that affect the rate (how fast) 2. Factors that affect the rate (how fast) at which a given solute dissolves in a at which a given solute dissolves in a given solvent:given solvent: Particle size of soluteParticle size of solute Rate of stirringRate of stirring TemperatureTemperature

Assignment 2: Discuss the Assignment 2: Discuss the two sets of factors.two sets of factors.

ExampleExample

Calculate the solubility in grams per Calculate the solubility in grams per liter of a certain gas in water at a liter of a certain gas in water at a partial pressure of 3.5 atm and 0partial pressure of 3.5 atm and 0 C. C. The solubility is 0.530 g/l at a total The solubility is 0.530 g/l at a total pressure of 1.00 atm and 0pressure of 1.00 atm and 0 C. C.

SolutionSolution

Using Dalton's Law of Partial PressureUsing Dalton's Law of Partial Pressure

PPtotaltotal = P = Pgasgas P Pwatewaterr

PPwaterwater at 0 at 0 C is 0.006 atm C is 0.006 atm

PPgasgas = 1.00 – 0.006 = 0.994 atm = 1.00 – 0.006 = 0.994 atm

SolubilitySolubility11 = 0.53 g/l = 0.53 g/l PP11 = 0.994 = 0.994

SolubilitySolubility22 = ? = ? PP22 = 3.50 atm = 3.50 atm

SolubilitySolubility22 = Solbility= Solbility11 Pressure Factor Pressure Factor

= 0.53 g/l = 0.53 g/l (3.5 atm / 0.994 atm) = (3.5 atm / 0.994 atm) = 1.87 g/l1.87 g/l

Saturated, Unsaturated and Saturated, Unsaturated and Supersaturated SolutionsSupersaturated Solutions

1. Saturated Solutions1. Saturated Solutions

Are solutions which are in dynamic Are solutions which are in dynamic equilibrium (equilibrium () with undissolved solutes) with undissolved solutes

They can be prepared by adding an They can be prepared by adding an excess of solute to a given amount of excess of solute to a given amount of solvent and allowing sufficient time for a solvent and allowing sufficient time for a maximum amount of solute to dissolve maximum amount of solute to dissolve with excess solute presentwith excess solute present

In this case:In this case:

Rate of dissolution (dissolved solute) = Rate of dissolution (dissolved solute) = Rate of crystallization (undissolved Rate of crystallization (undissolved solute)solute)

2. Unsaturated Solutions2. Unsaturated Solutions

Are solutions in which the Are solutions in which the concentration of solute is less than that concentration of solute is less than that of the Saturated (equilibrium) of the Saturated (equilibrium) Solutions, under the same conditions. Solutions, under the same conditions.

3. Supersaturated Solutions3. Supersaturated Solutions

Are solutions in which the Are solutions in which the concentration of solute is greater than concentration of solute is greater than that possible in Saturated (equilibrium) that possible in Saturated (equilibrium) Solutions, under the same conditions.Solutions, under the same conditions.

Concentrations of Concentrations of SolutionsSolutions

1. percent by mass1. percent by mass% by mass = (mass of solute / mass of solution) % by mass = (mass of solute / mass of solution)

1001002. parts per million, ppm2. parts per million, ppm

ppm = (mass of solute / mass of solution) ppm = (mass of solute / mass of solution) 1,000,0001,000,0003. molarity3. molarity

M = molarity = (moles of solute / liter of solution)M = molarity = (moles of solute / liter of solution)4. Molality4. Molality

m = molality = (moles of solute / kilogram of m = molality = (moles of solute / kilogram of solvent)solvent)5. Normality5. Normality

N = normality = (equivalents of solute / liter of N = normality = (equivalents of solute / liter of solution)solution)

Reaction RatesReaction Rates

The law of mass action states that:The law of mass action states that:

""the rate of a chemical reaction is the rate of a chemical reaction is proportional to the active mass of proportional to the active mass of the reactantsthe reactants""

The active mass is related to relative The active mass is related to relative molar concentration of the reactants molar concentration of the reactants in moles per liter for solutionsin moles per liter for solutions

aA + bB aA + bB cC + dD cC + dD

For the General ReactionFor the General Reaction

The overall rate of reaction is proportional The overall rate of reaction is proportional to the concentration of the reactants in to the concentration of the reactants in moles per liter raised to certain powermoles per liter raised to certain power

Rate Rate [A] [A]xx [B] [B]yy

[A] = concentration of A in moles / liter[A] = concentration of A in moles / liter

[B] = concentration of B in moles / liter[B] = concentration of B in moles / liter

x and y = whole number, fractional x and y = whole number, fractional numbers, negative numbers, or zero as numbers, negative numbers, or zero as determined by experimentationdetermined by experimentation

Then:Then:

Rate = k [A]Rate = k [A]xx [B] [B]yy

k = a proportionality constant, called the k = a proportionality constant, called the specific rate constantspecific rate constant

Sometimes x and y are equal to the Sometimes x and y are equal to the coefficients of the balanced equation; coefficients of the balanced equation; that is a and b. The values of x and y that is a and b. The values of x and y have to be determined experimentally. have to be determined experimentally. The value of x and y is the reaction order The value of x and y is the reaction order of each reactant. The sum of x and y is of each reactant. The sum of x and y is the overall reaction order.the overall reaction order.

Example Example Given the following chemical equation and rate Given the following chemical equation and rate equation, determine the reaction order of each reactant equation, determine the reaction order of each reactant and the overall reaction order.and the overall reaction order.

ClCl22 + CHCl + CHCl33 HCl + CCl HCl + CCl44

Rate = [ClRate = [Cl22]]0.50.5[CHCl[CHCl33]]Chlorine + Chloroform Chlorine + Chloroform Hydrochloric Acid + Carbon Hydrochloric Acid + Carbon tetra-chloridetetra-chlorideThe reaction is The reaction is half orderhalf order for chlorine and for chlorine and first orderfirst order for for chloroform, with the overall reaction order being 1.5.chloroform, with the overall reaction order being 1.5.

A + B A + B C CRate = k [A]Rate = k [A]22[B][B]33

The reaction is second order for A and third order for B, The reaction is second order for A and third order for B, with overall reaction order being 5.with overall reaction order being 5.

Factors Affecting the Rate Factors Affecting the Rate of a Chemical Reactionof a Chemical Reaction

Nature of ReactantsNature of Reactants Concentration of ReactantsConcentration of Reactants TemperatureTemperature CatalystsCatalysts

Assignment 3: Discuss the Assignment 3: Discuss the Four FactorsFour Factors

Chemical EquilibriaChemical Equilibria(Reversible and Irreversible (Reversible and Irreversible

Reactions)Reactions)Some reactions are irreversible in practice, Some reactions are irreversible in practice, meaning that the chemical equilibrium is not meaning that the chemical equilibrium is not established and that the reaction is complete. established and that the reaction is complete. When can this happen?When can this happen?

Products are removedProducts are removed Rate of reverse reaction is very slow (negligible)Rate of reverse reaction is very slow (negligible)

What products act as a driving force for a What products act as a driving force for a reaction to go irreversibly?reaction to go irreversibly?

GasGas PrecipitatePrecipitate Non-ionized or partially ionized substance, such Non-ionized or partially ionized substance, such

as wateras water

GASGAS

Gas removed as soon as it formsGas removed as soon as it forms

MgCOMgCO33 + 2HCl + 2HCl MgCl + H MgCl + H22O + COO + CO22

If gas remains in contact with the If gas remains in contact with the reactants, as in a closed container, then a reactants, as in a closed container, then a reversible reaction occurs and an reversible reaction occurs and an equilibrium is establishedequilibrium is established

PRECIPITATEPRECIPITATE

The precipitation of a substance acts to The precipitation of a substance acts to remove it from the reactionremove it from the reaction

AgNOAgNO33 + HCl + HCl AgCl AgCl + HNO + HNO33

The reaction is reversible as long as the The reaction is reversible as long as the precipitate is in contact with the solution precipitate is in contact with the solution but equilibrium favors the productsbut equilibrium favors the products

WATERWATER

NaOH + HBr NaOH + HBr NaBr + H NaBr + H22OO

The equilibrium is established but the The equilibrium is established but the reaction strongly favors the productsreaction strongly favors the products

Reversible ReactionsReversible ReactionsA + B A + B C + D C + D

The system is in chemical equilibrium when:The system is in chemical equilibrium when:

Rate at which C and D molecules react to Rate at which C and D molecules react to form A and B molecules = Rate at which A form A and B molecules = Rate at which A and B molecules react to form C and D and B molecules react to form C and D moleculesmolecules

For any equilibrium reaction, a constant known For any equilibrium reaction, a constant known as the equilibrium constant (K) can be obtained as the equilibrium constant (K) can be obtained experimentally if all quantities in the expression experimentally if all quantities in the expression can be determined.can be determined.

For the Law of Mass ActionFor the Law of Mass Action

Rate Forward Rate Forward [A][B] = k [A][B] = kff [A][B] [A][B]

Similarly, Rate Reverse = kSimilarly, Rate Reverse = krr [C][D] [C][D]

kkff and k and krr are the specific rate constants are the specific rate constants for the forward and reverse reactions, for the forward and reverse reactions, respectively.respectively.

At equilibrium:At equilibrium:

Rate of forward reaction = Rate of Rate of forward reaction = Rate of reverse reactionreverse reaction

thenthen

since ksince kff and k and krr are constants, then (k are constants, then (kff / k / krr) is also ) is also constant. constant.

K is the equilibrium constant which has a K is the equilibrium constant which has a certain value at a given temperature for a given certain value at a given temperature for a given reaction.reaction.Generally,Generally,IfIf

then,then,

]][[]][[ DCkBAk rf

]][[

]][[

BA

DC

k

k

r

f

]][[

]][[

BA

DCK

dDcCbBaA

ba

dc

BA

DCK

][][

][][

ExampleExample

Write the expression of K for the Write the expression of K for the following reactions:following reactions:

)3.....(

)2.....(2

)1.....(

23

22

22

COCaOCaCO

HIIH

COClClCO

SolutionSolution

in equation (3), since CaCOin equation (3), since CaCO33 and CaO are and CaO are solids, they are not considered in the solids, they are not considered in the equilibrium expression because their equilibrium expression because their concentrations are constant at a given concentrations are constant at a given temperature and hence they are included temperature and hence they are included in the value for the constant K.in the value for the constant K.

)3].....([

)2.....(]][[

][

)1.....(]][[

][

2

22

2

2

COK

IH

HIK

ClCO

COClK

Le Chatelier’s PrincipleLe Chatelier’s Principle

““If an equilibrium system is subjected If an equilibrium system is subjected to a change in conditions of to a change in conditions of ConcentrationConcentration, , TemperatureTemperature, or , or PressurePressure, the system will change to a , the system will change to a new equilibrium position, where new equilibrium position, where possible, in a direction that will tend possible, in a direction that will tend to restore the original conditions.”to restore the original conditions.”

ConcentrationConcentration When the concentration of one of the When the concentration of one of the

substance in a system at equilibrium is substance in a system at equilibrium is increased, the principle predicts that the increased, the principle predicts that the equilibrium will shift so as to use up equilibrium will shift so as to use up partially the added substance.partially the added substance.

Decreasing the concentration of one Decreasing the concentration of one substance in a system at equilibrium will substance in a system at equilibrium will cause the equilibrium to shift so as to cause the equilibrium to shift so as to replenish partially the substance removed.replenish partially the substance removed.

In all cases, the equilibrium constant, K, will In all cases, the equilibrium constant, K, will remain constant with the concentration of remain constant with the concentration of the reactants or products varying.the reactants or products varying.

If we have:If we have:

1.1. increase in concentration of either A or B will increase in concentration of either A or B will shift the equilibrium to the products sideshift the equilibrium to the products side

2.2. increase in concentration of either C or D will increase in concentration of either C or D will shift the equilibrium to the reactants sideshift the equilibrium to the reactants side

3.3. decrease in concentration of either A or B decrease in concentration of either A or B will shift the equilibrium to the reactants sidewill shift the equilibrium to the reactants side

4.4. decrease in concentration of either C or D decrease in concentration of either C or D will shift the equilibrium to the products sidewill shift the equilibrium to the products side

DCBA

TemperatureTemperature ““If the temperature of a system at If the temperature of a system at

equilibrium is changed, the equilibrium equilibrium is changed, the equilibrium will shift so as to change the will shift so as to change the temperature towards its original value.”temperature towards its original value.”

A. Exothermic ReactionsA. Exothermic Reactions

1.1. increase in temperature will shift the increase in temperature will shift the equilibrium to reactants sideequilibrium to reactants side

2.2. decrease in temperature will shift decrease in temperature will shift equilibrium to products side equilibrium to products side

energyheatDCBA .

B. Endothermic ReactionsB. Endothermic Reactions

1.1. increase in temperature will shift the increase in temperature will shift the equilibrium to the products sideequilibrium to the products side

2.2. decrease in temperature will shift decrease in temperature will shift equilibrium to the reactants sideequilibrium to the reactants side

The equilibrium constant, K, will The equilibrium constant, K, will change when temperature is change when temperature is changed.changed.

energyheatDCBA .

PressurePressure

Increasing the pressure on a system Increasing the pressure on a system at equilibrium will shift the at equilibrium will shift the equilibrium in the direction which will equilibrium in the direction which will decrease the volume. Decreasing the decrease the volume. Decreasing the pressure will have the opposite effect.pressure will have the opposite effect.

If no change in volume in going from If no change in volume in going from reactants to products, pressure will reactants to products, pressure will have no effect on the equilibrium. have no effect on the equilibrium. The equilibrium constant, K, does not The equilibrium constant, K, does not change with change in pressure.change with change in pressure.

ExamplesExamples

ConcentrationConcentration

Changing the concentration of an Changing the concentration of an ingredient will shift the equilibrium to ingredient will shift the equilibrium to the side that would reduce that the side that would reduce that change in concentration.change in concentration.

This can be illustrated by the This can be illustrated by the equilibrium of carbon monoxide and equilibrium of carbon monoxide and hydrogen gas, reacting to form hydrogen gas, reacting to form methanol.methanol.

CO + 2 HCO + 2 H22 ⇌ CH⇌ CH33OHOH

Suppose we were to increase the Suppose we were to increase the concentration of CO in the system. Using concentration of CO in the system. Using Le Chatelier's principle we can predict Le Chatelier's principle we can predict that the amount of methanol will increase, that the amount of methanol will increase, decreasing the total change in CO. If we decreasing the total change in CO. If we are to add a species to the overall are to add a species to the overall reaction, the reaction will favor the side reaction, the reaction will favor the side opposing the addition of the species. opposing the addition of the species. Likewise, the subtraction of a species Likewise, the subtraction of a species would cause the reaction to fill the “gap” would cause the reaction to fill the “gap” and favor the side where the species was and favor the side where the species was reduced. reduced.

TemperatureTemperature Let us take for example the reaction of Let us take for example the reaction of

nitrogen gas with hydrogen gas. This is a nitrogen gas with hydrogen gas. This is a reversible reaction, in which the two gases reversible reaction, in which the two gases react to form ammonia:react to form ammonia:

NN22 + 3 H + 3 H22 ⇌ 2 NH⇌ 2 NH33 ΔH = -92kJ ΔH = -92kJ

This is an exothermic reaction when producing This is an exothermic reaction when producing ammonia. If we were to lower the ammonia. If we were to lower the temperature, the equilibrium would shift in temperature, the equilibrium would shift in such a way as to produce heat. Since this such a way as to produce heat. Since this reaction is exothermic to the right, it would reaction is exothermic to the right, it would favor the production of more ammonia. favor the production of more ammonia.

Total PressureTotal Pressure Once again, let us refer to the reaction of Once again, let us refer to the reaction of

nitrogen gas with hydrogen gas to form nitrogen gas with hydrogen gas to form ammonia:ammonia:

NN22 + 3 H + 3 H22 ⇌ 2 NH⇌ 2 NH33 ΔH = -92kJ ΔH = -92kJ

Note the number of moles of gas on the left Note the number of moles of gas on the left hand side, and the number of moles of gas on hand side, and the number of moles of gas on the right hand side. We know that gases at the the right hand side. We know that gases at the same temperature and pressure will occupy same temperature and pressure will occupy the same volume. We can use this fact to the same volume. We can use this fact to predict the change in equilibrium that will predict the change in equilibrium that will occur if we were to change the total pressure.occur if we were to change the total pressure.

Suppose we increase total Suppose we increase total pressure on the system: now, by Le pressure on the system: now, by Le Chatelier's principle the Chatelier's principle the equilibrium would move to equilibrium would move to decrease the pressure. Noting that decrease the pressure. Noting that 4 moles of gas occupy more 4 moles of gas occupy more volume than 2 moles of gas, we volume than 2 moles of gas, we can deduce that the reaction will can deduce that the reaction will move towards the products if we move towards the products if we were to increase the pressure.were to increase the pressure.

a. Effect of Adding an a. Effect of Adding an Inert GasInert Gas

An inert gas (or noble gas) such as helium is An inert gas (or noble gas) such as helium is one which does not react with other one which does not react with other elements or compounds. To add an inert gas elements or compounds. To add an inert gas into a closed system at equilibrium may or into a closed system at equilibrium may or may not result in a shift. For example, may not result in a shift. For example, consider adding helium to a container with consider adding helium to a container with the following reaction:the following reaction:

NN22 + 3H + 3H22 ⇌ 2NH⇌ 2NH33

The main effect of adding an inert gas to a The main effect of adding an inert gas to a closed system is that it will increase the closed system is that it will increase the total pressure or volume. An inert gas total pressure or volume. An inert gas would not be directly involved in the would not be directly involved in the reaction, but could result in a shift.reaction, but could result in a shift.

b. Volume Held b. Volume Held ConstantConstant

If volume is held constant, the If volume is held constant, the individual concentrations of the above individual concentrations of the above gases do not change. The gases do not change. The partial partial pressurespressures also do not change, even also do not change, even though we have increased the total though we have increased the total pressure by adding helium. This pressure by adding helium. This means the reaction quotient does not means the reaction quotient does not change, so the system is still at change, so the system is still at equilibrium and no shift occurs.equilibrium and no shift occurs.

c. Volume Allowed to c. Volume Allowed to IncreaseIncrease

If the volume is allowed to increase, the If the volume is allowed to increase, the concentrations, as well as the partial concentrations, as well as the partial pressures, all decrease. Because there pressures, all decrease. Because there are more stoichiometric moles on the are more stoichiometric moles on the lefthand side of the equation, the lefthand side of the equation, the decrease in concentration affects the decrease in concentration affects the lefthand side more than the righthand lefthand side more than the righthand side. Therefore, the reaction would shift side. Therefore, the reaction would shift to the left until the system is at to the left until the system is at equilibrium again.equilibrium again.