CHAPTER 11

THERMOCHEMISTRY

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11.1 – The Flow of Energy - HEAT

What is thermochemistry?It is the study of the heat changes involved in physical changes (melting, vaporizing, etc.) and chemical changes (combustion, decomp., synthesis, etc.)

 

Chemical Potential Energy

Chemical potential energy is the energy stored within the structural units of chemical substances.– Gasoline and other fuels provide power

in the form of heat energy that can be converted to other forms of energy.

What is heat?

Heat is a form of energy (energy is the capacity to do work).

It can be measured in the Standard International units of Joules (J),

Heat is represented by the variable q.

More about heat!

Heat itself cannot be detected by the senses or by instruments. Only changes of heat can be detected.

Conversions:

1 calorie (cal) = 4.186 Joules (J) 1000 cal = 1 Calorie (Cal) = 1 kilocalorie (kcal) Remember that variable C (specific heat) is not

the same as Cal C=specific heat, Cal=kcal, C°=Celsius degree Do not get the “C’s” mixed up! Heat is not the same thing as temperature

What is a calorie?

One calorie is the amount of heat energy required to raise the temperature of 1 gram of water by 1ºC.

What happens to the heat during a change?

This is the LAW OF CONSERVATION OF ENERGY! Means…?

Heat can be transferred from one object to another, but energy can never be created or destroyed (it is just converted from one type of energy to another type)

What is heat capacity?

It is the energy required to change the temperature of a substance/object

Some substances heat up quickly and are good conductors (copper), while other objects are resistant to heating up and are good insulators (glass).

The higher the heat capacity, the better insulator it is (and poorer conductor of heat) –beach vs ocean

SPECIFIC heat capacity: the amount of energy required to raise 1 g of a specific substance by 1ºC.

What is the SYSTEM?

The system is whatever is being analyzed…for example…freezing water to ice…the system is the water itself.

What are the SURROUNDINGS?

The surroundings includes everything else in the universe EXCEPT the system…in the example…the surroundings are the beaker that the cup is in, the room, objects in the room, etc.

Exothermic

A process where heat is … LOST is called an EXOTHERMIC PROCESS Heat “exits” the system = EXO-thermic

More Exothermic

NaOH + HCl NaCl + HOH + heat In this example, the reaction (system) between

sodium hydroxide and hydrochloric acid releases heat to the beaker (surroundings) and it feels hot

Here, heat is shown as a product to indicate that heat is produced during the reaction

Endothermic Process

A process where heat is gained (absorbed) is called an ENDOTHERMIC PROCESS

Heat goes “into” the system = ENDO-thermic Example of an endothermic chemical change…

decomposition of sodium bicarbonate: 2NaHCO3 + heat Na2CO3 + CO2 + HOH

In this example, the reaction (system) takes heat from the beaker (surroundings) and it feels cold

Here, heat is shown as a reactant to indicate that heat is GAINED during the process

How can these heat energy changes be calculated or quantified?

H = enthalpy (heat) change +H = endothermic = gain heat -H = exothermic = loss heat Look up values for the change, in a reference table

(usually expressed in kJ/mol or kcal/mol) To find total amount of heat energy (q): q = n H Here n = moles of the substance

According to graph below, at phase change temperature cannot change

Calculating Specific Heat

q = CmΔT Where: m = mass (g)

C = specific heat (J/g·C°)T = change in temperature

(C°) C = q / mΔT

11.2- Calorimetry

Calorimetry is a method to measure heat released or absorbed during chemical reactions or phase changes.

Calorimetry is the accurate and precise measurement of heat change for chemical and physical processes.

How does calorimetry work?

In chemical and physical processes, the heat absorbed or released by the system is equal to the heat absorbed or released by the surroundings.

An insulated container, called a calorimeter, is used to measure temperature changes during physical or chemical processes.

Calorimeter diagram

Heat/Enthalphy Terms & Equations

Heat is represented by q Enthalpy is represented by ΔH (read delta H) In our text all heat changes are carried out at

constant pressure – thus

q = ΔH

The Heat Calculation

q = H = m · C · TWhere: m = mass (g)

C = specific heat (J/g·C°)T = change in temperature (C°)

Another common unit for specific heat is cal/g·C°Kelvin degrees may be used interchangeably with

Celsius degrees since T for Kelvin and Celsius will be the same.

The Heat Calculation

Reactions are generally carried out in water in a Styrofoam cup. The volume of water is equal to the mass of water since the density of water is 1.0 g/mL. The difference between the initial and final temperature give T, and the specific heat of water is known.

Thermochemical Equations

When calcium oxide (CaO) mixes with water, heat is given off. This is the reaction that occurs during cement mixing!

CaO(s) + H2O(l) = Ca(OH)2 + 65.2 kJ A chemical equation that contains a heat term

is a thermochemical equation.

Thermochemical Equations

When you look up H in a table

-H = exothermic = product

+H = endothermic = reactant

Thermochemical Equation Calculations

2NaHCO3(s) + 129 kJ = Na2CO3(s) + H2O(g) + CO2(g)

Calculate the kJ required to decompose 2.24 mol NaHCO3

Knowns: 2.24 mol NaHCO3, H = 129 kJ

129 kJ

2 mol NaHCO3

Thermochemical Equation Calculations

and

2.24 mol NaHCO3 129 kJ = 144 kJ

2 mol NaHCO3

Thermochemical Equations

For exothermic reactions, the potential energy of the reactant(s) is higher than the potential energy of the products(s)

The physical states of the reactants and products must be considered! This is very important! Make sure to note the physical state before looking up enthalpy/heat values.

Thermochemical Equations

Heats of combustion are the heats for the complete burning of one mole of substance. (p. 305)

The combustion of natural gas, mostly methane, CH4, is used to heat many homes.

CH4(g) + 2O2(g) = CO2(g) + 2H2O(g) + 890 kJ OR

CH4(g) + 2O2(g) = CO2(g) + 2H2O(g) H = -890 kJ

11.3 – Heat in Changes of State

If you add heat to an ice/water mixture, the temperature will remain at 0C until all the ice is melted This is a phase change plateau!

The molar heat of fusion, Hfus, is the heat absorbed by 1 mol of substance during melting.

The molar heat of solidification , Hsolid, is the heat lost when 1 mol of a substance solidifies.

11.3 – Heat in Changes of State

Hfus = -Hsolid

The heat of fusion and heat of solidification are equal but with opposite signs.

It takes 6.01 kJ of heat being absorbed to melt 1 mol ice. Hfus = +6.01 kJ/mol

It takes 6.01 kJ of heat being released to solidify 1 mol ice. -Hsolid _= -6.01kJ/mol

Heat of Vaporization

When water at 100°C is heated, the temperature remains at 100°C until all of the water is vaporized. This is a phase change plateau!

The molar heat of vaporization, Hvap, is the amount of heat needed to vaporize 1 mol liquid.

The Hvap of water is 40.7 kJ/mol. It takes that much heat to vaporize 1 mol water.

Heat of Condensation

The molar heat of condensation, Hcond, is the amount of heat to condense 1 mol of vapor.

Hvap = -Hcond These values are equal but opposite in sign.

Heat of Solution

Heat changes can occur when a solute (the substance dissolving) dissolves in a solvent (the substance doing the dissolving – often water).

The heat change caused by dissolution of one mol of substance is the molar heat of solution, Hsoln.

Heat of Solution

A hot pack is a practical application of an exothermic reaction.

CaCl2(s) = Ca2+(aq) + 2Cl1-(aq)

Hsoln = -82.8 kJ/mol An endothermic reaction:

NH4NO3(s) = NH41+(aq) + NO3

1-(aq)

Hsoln = 25.7 kJ/mol

H2O(l)

H2O(l)

11.4 Calculating Heat ChangesHess’s Law

Most reactions take place as a series of steps. Diamond and graphite

C(diamond) = C(graphite) C(graphite) = C(diamond)

This doesn’t happen! If it did, diamonds wouldn’t be worth any more than graphite!

The enthalpy change cannot be measured directly because the reaction is too slow!

Hess’s Law

Hess’s Law states that if you add two or more thermochemical equations together to give a final equation, then you can also add the heats of reaction to give the final heat of reaction.

This is Hess’s Law of Heat Summation!

Hess’s Law

Sometimes it is virtually impossible to determine the heats of reaction in the laboratory. CO (carbon monoxide) is such a reaction.

CO heat of reaction can be determined mathematically by combining two reactions that can be performed experimentally -

More Hess’s Law!

Sometimes the given equations contain fractions. That is common because we are usually looking at a substance other than oxygen or hydrogen.

Sometimes the given equation(s) need to be multiplied by an appropriate factor to obtain the correct results. Multiply each coefficient by the factor & multiply the r H by the same factor.

Hess’s Law of Heat Summation

Ho = H 0 products - H 0

reactants

Add up the enthalpies for all the products. Take the coefficient(s) multiplied by the rH and add them all together. Look in Tables for rH values!

Do the same for all the reactants. Subtract the sum of reactants from sum of

products.

Problem : Calculate the Enthapy for the following reaction:

N2(g) + 2O2(g) ---> 2NO2(g) ΔH° = ??? kJ

Using the following two equations:

N2(g) + O2(g) ---> 2NO(g) ΔH° = +180 kJ

2NO2(g) ---> 2NO(g) + O2(g) ΔH° = +112 kJ

Solution:

In order to solve this, we must reverse at least one equation and it turns out

that the second one will require reversal. Here are both with the reversal to

the second:

N2(g) + O2(g) ---> 2NO(g) ΔH° = +180 kJ

2NO(g) + O2(g) ---> 2NO2(g) ΔH° = -112 kJ

Notice that I have also changed the sign on the enthalpy from positive to

negative.

Next, we add the two equations together and eliminate identical items. We

also add the two enthalpies together.

N2(g) + 2O2(g) ---> 2NO2(g) ΔH° = +68 kJ

The answer has been obtained: +68 kJ/mol.

Find the ΔH of the following reaction:

2NOCl(g) ---> 2NO(g) + Cl2(g)

2NO(g) ---> N2(g) + O2(g) ΔH = -180.6 kJ

N2(g) + O2(g) + Cl2(g) ---> 2NOCl(g) ΔH = +103.4 kJ

1) Flip first reaction, flip second reaction:

N2(g) + O2(g) ---> 2NO(g) ΔH = +180.6 kJ

2NOCl(g) ---> N2(g) + O2(g) + Cl2(g) ΔH = -103.4 kJ

2) Add the equations and the ΔH values:

+180.6 + (-103.4) = +77.2

2NOCl(g) ---> 2NO(g) + Cl2(g) ΔH = +77.2 kJ

PROBLEM: During discharge of a lead-acid storage battery, the following

chemical reaction takes place:

Pb + PbO2 + 2H2SO4 ---> 2PbSO4 + 2H2O

Using the following two reactions:

(1) Pb + PbO2 + 2SO3 ---> 2PbSO4 ΔH° = -775 kJ

(2) SO3 + H2O ---> H2SO4 ΔH° = -113 kJ

Determine the enthalpy of reaction for the discharge reaction above.

Solution:

1) Multiply chemical equation (2) by 2:

2SO3 + 2H2O ---> 2H2SO4 ΔH = -226 kJ

2) Switch the reactants and products in chemical reaction (2). Because of that, the

sign of the change in enthalpy becomes positive. Let's number the following

chemical equation as (3):

(3) 2H2SO4 ---> 2SO3 + 2H2O ΔH° = 226 kJ

3) Add chemical equations (1) and (3):

Pb + PbO2 + 2SO3 + 2H2SO4 ---> 2PbSO4 + 2SO3 + 2H2O

4) Then, add the enthalpy changes of equations (1) and (3):

Pb + PbO2 + 2H2SO4 ---> 2PbSO4 + 2H2O ΔH° = -549 kJ

NO2(g) + (7/2) H2(g) ---> 2H2O(l) + NH3(g) ΔH° = ??? kJ

2NH3(g) ---> N2(g) + 3H2(g) ΔH° = +92 kJ

(1/2) N2(g) + 2H2O(l) ---> NO2(g) + 2H2(g) ΔH° = +170 kJ

Determine the enthalpy (ΔH°) of reaction for:

Using the following two equations:

Solution:

1) Notice how there is only one NH3 in the target equation and it's on the right-

hand side. That means we have to flip our first equation and divide it by two. Like

this:

(1/2)N2(g) + (3/2)H2(g) ---> NH3(g) ΔH° = -46 kJ

Notice that the sign changed on the ΔH and its numerical value was cut in half.

2) The target equation has one NO2 and it's on the left-hand side, so we need to flip

the second equation. Like this:

NO2(g) + 2H2(g) ---> (1/2) N2(g) + 2H2O(l)ΔH° = -170 kJ

Notice that the sign changed on the ΔH.

3) The ΔH for the target equation is:

-46 + -170 = -216 kJ