calorimetry ap chemistry. calorimetry calorimetry is the measurement of heat flow. calorimetry is...

CalorimetryCalorimetry

AP ChemistryAP Chemistry

CalorimetryCalorimetry Calorimetry is the measurement Calorimetry is the measurement

of heat flow.of heat flow.

It allows us to calculate the It allows us to calculate the amount of energy required to amount of energy required to heat up a substance or to make heat up a substance or to make a substance change states.a substance change states.

Calorimetry TermsCalorimetry Terms

Molar Heat of FusionMolar Heat of Fusion —— The The heat heat absorbedabsorbed by one mole of a by one mole of a substance when changing from a substance when changing from a solidsolid to a to a liquidliquid. .

For water, it = 6.0 kiloJoules/mole For water, it = 6.0 kiloJoules/mole

Heat of solidificationHeat of solidification is opposite is opposite of heat of fusion (heat is of heat of fusion (heat is releasedreleased).).

Molar Heat of VaporizationMolar Heat of Vaporization —— The The heat absorbed by one mole of a heat absorbed by one mole of a substance when changing from a substance when changing from a liquidliquid to a to a gas.gas.

For water, it = 40.7 kiloJoules/mole.For water, it = 40.7 kiloJoules/mole.

Heat of condensationHeat of condensation is the opposite is the opposite of heat of vaporization (heat is of heat of vaporization (heat is releasedreleased))Heat Required For a Phase Change – Or Latent Heat

Process

Heat Absorbed or Released = q

q = (moles) x (Molar Heat Fusion/Vaporization)

Calculating Heat Required To Calculating Heat Required To Change StateChange State

Example #1Example #1: How much heat is : How much heat is needed to melt 56.0 grams of ice needed to melt 56.0 grams of ice into liquid (the molar heat of fusion into liquid (the molar heat of fusion for ice is 6.0 kJ/mol)?for ice is 6.0 kJ/mol)?

56.0 g 1 mole H56.0 g 1 mole H22O 6.0 kJO 6.0 kJ = = 18.0 g 18.0 g 1 mole 1 mole

= 18.7 kJ will be absorbed= 18.7 kJ will be absorbed q = + 18.7kJq = + 18.7kJ

Example #2Example #2 How much heat energy will be How much heat energy will be

released when 200grams steam released when 200grams steam condenses back to a liquid water?condenses back to a liquid water?

Molar Heat Molar Heat condensationcondensation = 40.7kJ/mol = 40.7kJ/mol

q = (moles) x (Molar Heat condensation)

200gram 1 mole 40.7 kJ

18gram 1 mole= 452 kJ released

So, q = - 452kJ

Heating a Substance with Heating a Substance with No Phase ChangeNo Phase Change

Also known as a sensible heat Also known as a sensible heat process.process.

Depends on Specific Heat CapacityDepends on Specific Heat Capacity Specific Heat CapacitySpecific Heat Capacity--The amount --The amount

of energy required to raise one gram of energy required to raise one gram of a substance one degree Celcius.of a substance one degree Celcius.

Every substance and every phase of Every substance and every phase of that substance has its own unique that substance has its own unique specific heat capacity.specific heat capacity.

Water’s Specific Heat (as a liquid) Water’s Specific Heat (as a liquid) = 4.184 Joules/gram = 4.184 Joules/gram ooCC

or unit cans be J/gKor unit cans be J/gK

Specific heat (J/gSpecific heat (J/gooC) = heat capacity C) = heat capacity (J/(J/ooC)C)

You will see both terms used.You will see both terms used. You can also use Molar Heat CapacityYou can also use Molar Heat Capacity This is the heat energy required to This is the heat energy required to

raise 1 mole of a substance 1raise 1 mole of a substance 1ooC or 1K.C or 1K. Molar heat capacity can be determined Molar heat capacity can be determined

from the specific heat by multiplying it from the specific heat by multiplying it by the molar mass of the substance.by the molar mass of the substance.

Energy to Change Energy to Change TemperatureTemperature

q = (mass) ( C ) ( Tq = (mass) ( C ) ( T ) )

HeatMeasured in Joules

Mass In grams

Specific Heat Capacity

Change in TemperatureTfinal – Tinitial

In OCelcius or kelvin

Exothermic vs. EndothermicExothermic vs. Endothermic Exothermic ChangeExothermic Change – Heat Energy is – Heat Energy is releasedreleased

to the surroundings. to the surroundings. q and q and ∆H are (-)∆H are (-)– Molecules Molecules slowslow down, extra energy is down, extra energy is

transferred to surrounding.transferred to surrounding.– CoolingCooling phase changes are exothermic phase changes are exothermic

•Endothermic Change – Heat energy is absorbed by the system. q and ∆H are (+)q and ∆H are (+)

-Molecules move faster as they absorb energy

-Phase changes that require energy (or heat) are endothermic

Changes In StateChanges In State

Solid

Liquid Gas

Freezing

Melting

SublimationDeposition

Vaporization

Condensation

Example #3Example #3 How much energy is needed to heat How much energy is needed to heat

80grams of water from 1080grams of water from 10ooC to 55C to 55ooC C at constant pressure?at constant pressure?

q = m C T = m C (Tfinal – Tinitial )

q = (80grams) ( 4.184 J/goC) (55oC – 10oC)

q = + 15062 joules

divide by 1000 to get kilojoules

15062 J 1 kJ =

1000J

15.06 kJ absorbed

Example #4Example #4 -How much energy is needed -How much energy is needed to change 150grams of ice from 0to change 150grams of ice from 0ooC to 50C to 50ooC?C?

This problem requires two steps. Since water is solid ice at 0oC, we need to melt the ice and then heat it up to 50oC.

Step 1 – Calculate heat required to melt 150grams ice

Step 2 - Calculate heat required to heat liquid water from 0oC to 50oC

q = mC T = (150g)(4.184 J/goC)(50oC)

= 31380 J convert to kJ = 31.38kJ

150g 1 mole 6.0 kJ = 50 kJ 18grams 1 mole

*Add both heat values together for your final answer

50 kJ + 31.38kJ = 81.38 kJ heat absorbed.

Multiple Step Calorimetry Multiple Step Calorimetry ProblemsProblems

Use q = mC T

Use

q = Moles x Molar Heat vap/fus

SolidHeats melting

LiquidHeats

Vaporization

Gas Heats

•Add each individual energies together for total

a

b

cd

e

qtotal = a + b + c + d + e

CalorimetryCalorimetry Experimental way of measuring heat Experimental way of measuring heat

generation/consumption by essentially generation/consumption by essentially catching all the heat energy in a water catching all the heat energy in a water bath or water bath + metal apparatusbath or water bath + metal apparatus..

Coffee Cup CalorimetryCoffee Cup Calorimetry Styrofoam cup insulates the contents so Styrofoam cup insulates the contents so

any heat generated or consumed in the any heat generated or consumed in the water can be measured by the water can be measured by the temperature changetemperature change

q = -(Hrxn) = mCq = -(Hrxn) = mCH2OH2O∆T∆T Reactions must take place in water, then Reactions must take place in water, then

you measure the change in temperature.you measure the change in temperature.

Bomb-CalorimeterBomb-Calorimeter Used to measure heats of reaction.Used to measure heats of reaction. Usually do combustion reactions in it.Usually do combustion reactions in it. Metal innards absorb heat. You have Metal innards absorb heat. You have

to keep track of heat absorbed by to keep track of heat absorbed by metal and water.metal and water.

q = mCq = mCH2OH2O∆T + C∆T + Cp p ∆T ∆T qqwaterwater + q + qapparatusapparatus

CCpp is the heat capacity of the apparatus – is the heat capacity of the apparatus – determined experimentallydetermined experimentally. .

Conservation of EnergyConservation of Energy

Under conditions of constant pressureUnder conditions of constant pressure

Heat Lost = Heat gainedHeat Lost = Heat gained

qqp-lostp-lost = q = qp-gainedp-gained

For a Reaction heat gained by water = heat released by reaction.

ExampleExample- When 100mls of 0.1M HCl is - When 100mls of 0.1M HCl is mixed with 100mls of 0.1M NaOH in a mixed with 100mls of 0.1M NaOH in a coffee cup calorimeter, the temperature coffee cup calorimeter, the temperature increases from 25increases from 25ooC to 29.8C to 29.8ooC. Assume C. Assume that the coffee cup is a perfect that the coffee cup is a perfect insulator, pressure is constant and the insulator, pressure is constant and the density of the solution is 1g/ml and density of the solution is 1g/ml and specific heat capacity for the mixture is specific heat capacity for the mixture is that of water, 4.18 J/gK.that of water, 4.18 J/gK.

What is the enthalpy change for the What is the enthalpy change for the reaction?reaction?

∆H = q = mC∆T

Since density is 1g/mL , all volumes are Since density is 1g/mL , all volumes are gram values.gram values.

Total volume must be used = 200mLs = Total volume must be used = 200mLs = 200grams.200grams.

∆∆T = TT = Tfinalfinal – T – Tinitial initial = 29.8= 29.8ooC – 25C – 25ooC = 4.8C = 4.8ooCC

q = mC ∆T = (200g)(4.18J/gK)(4.8K)q = mC ∆T = (200g)(4.18J/gK)(4.8K) q = 4013 J = 4.013 kJq = 4013 J = 4.013 kJ ∆∆H = - 4.013 kilojoulesH = - 4.013 kilojoules Since temperature of water increased, it is Since temperature of water increased, it is

exothermic, which means a - ∆Hexothermic, which means a - ∆H