Thermochemistry

Chapter 16

Thermochemistry

The study of energy involved during chemical reactions.

Energy sources:~ chemical ~ nuclear~ solar~ geothermal~ wind/water

Heat: the energy of motion of molecules

Kinetic molecular theory: substances are composed of particles that are continually moving and colliding with other particles.

Kinetic energy: energy of motion Potential energy: stored energy

Temperature:

transfer of heat to a substance because of faster molecular movement (as long as there is no phase change)

Measured with a thermometer

A temperature change is explained as a change in kinetic energy

Temperature depends on the quantity of heat (q) flowing out or in of the substance.

Heat (q)

q=mc ∆t q=heat m=mass ∆t=change in temperature (tf-ti) c=specific heat capacity (J/(g oC)

Specific heat capacity is the quantity of heat required to raise the temperature of a unit mass of a substance by one degree Celsius.

Table of Specific HeatsTable of Specific Heats

Law of conservation of energy

∆E universe = O The total energy of the universe is

constant, it is not created or destroyed, however it can be transferred from one substance to another.

∆E universe = ∆E system + ∆E surroundings

First Law of thermodynamics

Any change in energy of a system is equivalent by an opposite change in energy of the surroundings.

∆E system = - ∆E surroundings

According to this law, any energy released or absorbed by a system will have a transfer of heat, q.

So, q system = -q surroundings

Sample Problem

15 g of ice was added to 60.0g of water. The Ti of water was 26.5 oC, the final temperature of the mixture was 9.70 oC. How much heat was lost by the water?

q=mc ∆tq=(60.0g) (4.18 J/g oC) (9.70-26.5 oC)q= - 4213.44 J =-4.21 kJ

Calculate the amount of heat needed to increase the temperature of 250g of water from 20oC to 56oC.

q = m x c x (Tf - Ti)q = 250g x 4.18 J/g oC x (56oC - 20oC)q = 37 620 J = 38 kJ

Calculate the specific heat capacity of copper given that 204.75 J of energy raises the temperature of 15g of copper from 25oC to 60oC.

q = m x c x (Tf - Ti)q = 204.75 Jm = 15gTi = 25 oCTf = 60 oC

204.75 J = 15 g x c x (60oC - 25oC)204.75 J= 15 g x c x 35oC204.75 J= 525 g oC x cc = 204.75 J ÷ 525 g oC = 0.39 J/ g oC

Watch flash about heat flow

http://www.mhhe.com/physsci/chemistry/animations/chang_7e_esp/enm1s3_4.swf

Homework:

Handout

Enthalpy (H) Total kinetic and

potential energy (internal energy) of a system under constant pressure.

System: area where reaction takes place

Its particles average motion define its properties.

Surroundings: outside of the system

Open-system

both matter and energy can freely cross from the system to the surroundings and back.

Ex: an open test tube

Closed-System

energy can cross the boundary, but matter cannot.

Ex: a sealed test tube

Isolated-System

neither matter nor energy can cross between the system and the surroundings.

Ex: The universe there are no

surroundings to exchange matter or energy with (as far as we know!)

The internal energy of a reactant or product cannot be measured, but their change in enthalpy (heat of reaction) can.

∆ H = Hproducts – Hreactants

A change in enthalpy occurs during phase changes, chemical reactions and nuclear reactions.

∆ H system = q surroundings

If energy is released out of the system then it is considered to be an exothermic enthalpy change (exit).

If energy is absorbed into the system then it is considered to be an endothermic enthalpy change (enter).

Endothermic ReactionsEndothermic Reactions

Endothermic Reactions

Method 2:2 HgO (s) 2 Hg (l) + O2(g) ΔH=181.67 kJ

Method 3:2 HgO (s) + 181.67 kJ 2 Hg (l) + O2(g)

Exothermic ReactionsExothermic Reactions

Exothermic Reactions

Method 2:4Al(s)+ 3O2(g) 2 Al2O3(g) ΔH=-1675.7 kJ

Method 3:4 Al(s) + 3O2(g) 2 Al2O3(g) +1675.7 kJ

Calorimeter

Instrument used to measure amount of energy involved in a chemical reaction.

It is equivalent to an isolated or closed system. (nothing may enter or exit the system)

The energy change is not measured within the system, but the energy transferred to its surroundings.

A basic calorimeter

Two styrofoam cups nestled within one another (insulation), then filled with a specific quantity of water.

A chemical reaction or phase change takes place inside and a thermometer is placed within to measure any change in temperature that occurs to the system.

Assumptions

1. It is an isolated or closed system and there is no heat transfer between the calorimeter and its surroundings.

2. The amount of heat absorbed or released by the calorimeter itself is too small to influence calculations.

3. Any dilute solutions involved in the reaction are treated as if they are water.

Bomb calorimeter

Cannot use basic design for combustion reactions.

Used in research for H for fuels, oils, food, explosives…

Larger and more sophisticated The reaction container is strong

enough to with stand an explosion, hence the name “bomb”

Have fixed components, like volume of water, thermometer…

Heavily insulated or vacuum insulated so no convection or conduction can occur affecting the enthalpy of the system.

Use the equation: q=CΔt

A Bomb CalorimeterA Bomb Calorimeter

Problems

Pg 638 # 5-7 Handout Questions 1-10

Standard Molar Enthalpy of Formation

Quantity of energy released (-) or absorbed (+) when one mole of a compound is formed directly from its elements at standard temperature and pressure.

We use a table to find them. Unit for ΔHf: kJ/mol Watch your states!

Calculating enthalpy changes

Amount of a substance reacting matters, so can use q= nΔH.

Remember n=amount of moles. If you are given a mass (g) and

molar mass (g/mol), then you can solve for n by dividing mass by molar mass. (review from chem 11 stoichiometry section)

Sample Problem:

Show the formation reaction of methanol.

If I had 10.5 g of CH3OH(l), how much energy would be released?

C(s) + 2H2(g) + 1/2O2(g) CH3OH(l) ΔHf = -239 kJ/mol

mm= (12.01) +(4 x 1.01) + (16.00) = 32.05 g/mol

q = nΔH = (10.5 g / 32.05 g/mol) (-239 kJ/mol) = -78.3 KJ

On Your Own:

I have 125.6 g of NaOH(s) formed at standard temperature and pressure, how much energy will be released?

Answer: - 1336.4 kJ

Standard Molar Enthalpy of Combustion

Energy changes involved with combustion reactions of one mole of a substance.

Remember that these reactions are only measured once cooled to 25oC

Combustion is a reaction with oxygen as a reactant

Practice Problem:

Write the combustion reaction for methane (CH4) ΔHcomb for CH4 is -965.1 kJ/mol

How much energy is released when 12.8 g of CH4 is combusted?

CH4(g) + 2O2 (g) 2H2O(l) + CO2 (g) ΔHcomb = -965.1 kJ/mol

q=nΔH = (12.8 g/ 16.05 g/mol) (-965.1 kJ/mol) = - 769.7 kJ

Combustion reaction:

Always form water and carbon dioxide.

Ex: CH4 + 2O2 CO2 + 2H2O

Remember your alkanes: CnH2n+2

Meth: C=1, Eth: C=2, Prop: C=3, But: C=4, Pent: C=5, Hex: C=6…

Practice

Pg 643 # 15-17 Pg 645 # 19,21,23

Enthalpy and changes of state

A change in matter without any change in chemical composition of the system

*** Always involves energy changes, but not temperature changes. There is constant temperature during a phase change. Why?

Because…

No change in kinetic energy, however we know that energy is entering the system because the bonds that are holding the molecules together are being broken or altered. This increases Ep of the molecules

Just think!

We know that there are energy changes during phase changes, because when we sweat our body is trying to cool itself down using evaporation!

Phase changes: (surroundings)

(s) to (l) fusion endo T

(l) to (g) vaporization endo T

(g) to (l) condensation exo T

(l) to (s) solidifying exo T

(g) to (s) sublimation exo T

(s) to (g) sublimation endo T

Latent Heat of Phase ChangeLatent Heat of Phase ChangeΔΔH H meltmelt = - = - ΔΔH H freezfreez

Molar Heat of Fusion or melting Molar Heat of Fusion or melting

The energy that must be The energy that must be absorbedabsorbed in in order to convert order to convert one moleone mole of of solidsolid to to liquidliquid at its at its melting pointmelting point..

Molar Heat of Freezing or solidifyingMolar Heat of Freezing or solidifyingThe energy that must be removed in The energy that must be removed in order to convert order to convert one moleone mole of of liquidliquid to to solidsolid at its at its freezing pointfreezing point..

Molar Heat of Vaporization Molar Heat of Vaporization

The energy that must be The energy that must be absorbedabsorbed in in order to convert order to convert one moleone mole of of liquidliquid to to gasgas at its at its boiling pointboiling point..

Molar Heat of CondensationMolar Heat of CondensationThe energy that must be The energy that must be removedremoved in in order to convert order to convert one moleone mole of of gasgas to to liquidliquid at its at its condensation pointcondensation point..

ΔH vap = - ΔH cond

Molar Enthalpy for changes of state

You can find tables of these values listed

Unit: kJ/mol

Latent HeatLatent Heat – Sample Problem – Sample ProblemProblemProblem: The molar heat of fusion of water is : The molar heat of fusion of water is

6.009 kJ/mol. How much energy is needed to convert 60g of ice 6.009 kJ/mol. How much energy is needed to convert 60g of ice at 0at 0C to liquid water at 0C to liquid water at 0C? C?

2 2

2

60 1 6.00920.00

18.02 1

g H O mol H O kJkiloJoules

g H O mol

Mass of ice

Molar Mass of

water

Heat of

fusion

Chemical reactions:

If a reaction is exothermic, energy is released to the surroundings (temp increases) and chemical potential energy of the system decreases ∆ H is a negative number

If a reaction is endothermic, energy is absorbed from the surroundings (temp. decreases) and the chemical potential energy of the system increases Δ H is a positive number

It requires 10-100 times more energy for a chemical reaction to take place than a phase change because a chemical reaction requires stronger ionic and covalent bonds to be broken, but a phase change only requires intermolecular bonds to be broken.

Manipulating the Enthalpy Change Term

The value of Δ H given as kJ/ mol refers to kJ per 1 mole of reactant or product as written in the equation.

N2(g) + 3H2(g) 2NH3(g)       Δ H = - 92.4 kJ

92.4 kJ of energy is released for every 1 mole of N2(g), 3 moles of H2(g), 2 moles of NH3(g) produced.

How much energy is released if only 1 mole of ammonia (NH3) gas is produced?

92.4 kJ of energy is released in the production of 2 moles of ammonia gas

Half as much energy will be released if only half the amount of ammonia gas is produced

½ x 92.4 = 46.2 kJ of energy will be released in the production of 1 mole of ammonia.

How much energy is released if 10 moles of nitrogen (N2) gas and 30 moles of hydrogen (H2) gas is used in the reaction?

92.4 kJ of energy is released for every 1 mole of N2(g)

10 times as much energy will be released if 10 times the amount of reactants are used

10 x 92.4 = 924kJ of energy will be released

Problems

Pg 648-649# 26,28-29 Handout #16-18

Heat Curve: Temperature vs. Time

Total energy change of a system

When a given amount of energy is available to heat a substance, sometimes there is more then one change of state involved as the substance goes from an initial to a final temperature.

This requires energy calculations for changing the temperature in a certain state and then calculations for a change of state.

1. To calculate the energy involved in a change of temperature of a substance without a change of state, we use q=mc ∆t

2. To calculate the energy involved in a change of state we would use:

q = nΔHrxn

Sample Problem #1

Calculate the total energy needed or released during the cooling of 1000 g of liquid iron (c= 0.82 J/g oC) to solid iron (c=0.52 J/g oC), where the initial temperature is 1700 oC to 80 oC. The enthalpy of solidification is -15

KJ/mol. The melting point of iron is 1535 oC.

Heating Curve

Temp

(oC)

Time (min)

1700

1535

80

q1

q2

q3

ΔE = cooling liquid + phase change from l to s + solid cooling

= q1 + q2 + q3= q (cooling (l)) + n Δ Hsolid+q(cooling (s))

= mc Δt + n Δ Hsolid + mc Δt = [(1000g)(0.82 J/g C)(1535-1700)] +

[(1000g/55.85 g/mol)(-15 KJ/mol) ] + [ (1000g)( 0.52 J/g C) (80-1535)]

=-135300 J + (-268.6 KJ) + (-756600 J) = -1160 KJ

Sample Problem #2

How much energy is required to raise the temperature of 50.0 g of water (ice) at -15.0°C to a temperature of 175°C?

c H2O(l) = 4.18 J/g C c H2O(g)and (s) = 2.01 J/g C ∆Hmelt = 6.02 kJ/mol ∆Hvap = 40.7 kJ/mol Melting point = 0 oC Boiling point = 100 oC First, start by drawing your own heat

curve.

the energy required to change the temperature of the ice from -15.0°C to 0°C, q1; the energy involved in the change of state as the ice melts at 0°C, q2; the energy involved to heat the liquid water from 0°C to 100°C, q3; the energy involved in the change of state as the water boils, q4; and finally the energy required to heat water vapour at 100°C to 175°C, q5.

Heat Curve

100 oC

0 oC

- 15 oC

175 oC

Temp (oC)

Time (min)

q1 = mcΔt = 50.0 g x (2.06 J/g•°C) x 15.0°C = 1540 J or 1.54 kJ

q2 = nΔHfus = 50.0 g / (18.01 g/mol) x 6.03 kJ/mol = 16.7 kJ

q3 = mcΔt = 50.0 g x (4.18 J/g•°C) x 100.0°C = 20 900 J or 20.9 kJ

q4 = nΔHvap = 50.0 g/(18.01 g/mol) x 40.8 kJ/mol = 113 kJ

q5 = mcΔt = 50.0 g x (2.02 J/g•°C) x 75.0°C = 7 580 J or 7.58 kJ

qtotal = q1 + q2 + q3 + q4 + q5 = 1.54 kJ + 16.7 kJ + 20.9 kJ + 113 kJ + 7.58 kJ = 160 kJ

Practice:

Pg 655 # 30-34 Handout #1 (finish it, #19&20)