Loading in 5 sec....

Chapter 8: ThermochemistryPowerPoint Presentation

Chapter 8: Thermochemistry

- 141 Views
- Uploaded on
- Presentation posted in: General

Chapter 8: Thermochemistry

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Chapter 8: Thermochemistry

Chapter Outline

8.1 Energy

8.2 Energy Changes and Energy Conservation

8.3 Internal Energy and State Functions

8.4 Expansion Work

8.5 Energy and Enthalpy

8.9 Hess’s Law

8.10 Standard Heat of Formation

8.11 Bond Dissociation Energies

Thermochemistry

- Thermodynamics is the science of the relationship between heat and other forms of energy.

- Thermochemistry is the study of the quantity of heat absorbed or evolved by chemical reactions.

Energy

Types of Energy

1. Kinetic energy

2. Potential energy

3. Chemical energy

4. Heat energy

5. Electric energy

6. Radiant energy

Energy

- There are three broad concepts of energy:

- Kinetic Energy is the energy associated with an object by virtue of its motion.
- Potential Energy is the energy an object has by virtue of its position in a field of force.
- Internal Energy is the sum of the kinetic and potential energies of the particles making up a substance.

Energy

- Kinetic Energy: An object of mass m and speed or velocity has kinetic energy Ek equal to

- This shows that the kinetic energy of an object depends on both its mass and its speed.

A Problem to Consider

- Consider the kinetic energy of a person whose mass is 130 lb (59.0 kg) traveling in a car at 60 mph (26.8 m/s).

- The SI unit of energy, kg.m2/s2, is given the name Joule.

Energy

- Potential Energy: This energy depends on the “position” (such as height) in a “field of force” (such as gravity).

- For example, water of a given mass m at the top of a dam is at a relatively high “position” h in the “gravitational field” g of the earth.

A Problem to Consider

- Consider the potential energy of 1000 lb of water (453.6 kg) at the top of a 300 foot dam (91.44 m).

Energy

- Internal Energy is the energy of the particles making up a substance.

- The total energy of a system is the sum of its kinetic energy, potential energy, and internal energy, U.

Energy

- The Law of Conservation of Energy: Energy may be converted from one form to another, but the total quantities of energy remain constant.

work done

by the

system

energy

change

FIRST LAW OF THERMODYNAMICS∆E = q + w

Energy is conserved!

(endothermic), +q

heat transfer out

(exothermic), -q

w transfer in

(+w)

w transfer out

(-w)

SYSTEM

∆E = q + w

ENTHALPY

Most chemical reactions occur at constant P, so

Heat transferred at constant P = qp

qp = ∆H where H = enthalpy

and so ∆E = ∆H + w (and w is usually small)

∆H = heat transferred at constant P ≈ ∆E

∆H = change in heat content of the system

∆H = Hfinal - Hinitial

ENTHALPY

∆H = Hfinal - Hinitial

If Hfinal > Hinitial then ∆H is positive

Process is ENDOTHERMIC

If Hfinal < Hinitial then ∆H is negative

Process is EXOTHERMIC

Heat of Reaction

- In chemical reactions, heat is often transferred from the “system” to its “surroundings,” or vice versa.

- The substance or mixture of substances under study in which a change occurs is called thethermodynamic system (or simply system.)
- Thesurroundingsare everything in the vicinity of the thermodynamic system.

Heat of Reaction

- Heat is defined as the energy that flows into or out of a system because of a difference in temperature between the system and its surroundings.

- Heat flows from a region of higher temperature to one of lower temperature; once the temperatures become equal, heat flow stops.

Heat of Reaction

- Heat is denoted by the symbol q.
- The sign of q is positive if heat is absorbed by the system.
- The sign of q is negative if heat is evolved by the system.

- Heat of Reaction is the value of q required to return a system to the given temperature at the completion of the reaction.

Heat of Reaction

- An exothermic process is a chemical reaction or physical change in which heat is evolved (q is negative).
- An endothermic process is a chemical reaction or physical change in which heat is absorbed (q is positive).

“into” a system

Dq > 0

Heat of Reaction- Exothermicity
- “out of” a system
Dq < 0

- “out of” a system

Surroundings

Surroundings

Energy

Energy

System

System

Enthalpy and Enthalpy Change

- The heat absorbed or evolved by a reaction depends on the conditions under which it occurs.

- Usually, a reaction takes place in an open vessel, and therefore at the constant pressure of the atmosphere.
- The heat of this type of reaction is denoted qp, the heat at constant pressure.

Enthalpy and Enthalpy Change

- An extensive property is one that depends on the quantity of substance.
- Enthalpy is a state function, a property of a system that depends only on its present state and is independent of any previous history of the system, or independent of path.

- Enthalpy, denoted H, is an extensive property of a substance that can be used to obtain the heat absorbed or evolved in a chemical reaction.

Enthalpy and Enthalpy Change

- The change in enthalpy for a reaction at a given temperature and pressure (called the enthalpy of reaction) is obtained by subtracting the enthalpy of the reactants from the enthalpy of the products.

Enthalpy and Enthalpy Change

- The change in enthalpy is equal to the heat of reaction at constant pressure. This represents the entire change in internal energy (DU) minus any expansion “work” done by the system.

Enthalpy and Enthalpy Change

- The internal energy of a system, U, is precisely defined as the heat at constant pressure plus the work done by the system:

- Enthalpy and Internal Energy

- In chemical systems, work is defined as a change in volume at a given pressure, that is:

Enthalpy and Enthalpy Change

- Since the heat at constant pressure, qp, represents DH, then

- So ΔH is essentially the heat obtained or absorbed by a reaction in an open vessel where the work portion of ΔU is unmeasured.

Thermochemical Equations

- A thermochemical equation is the chemical equation for a reaction (including phase labels) in which the equation is given a molar interpretation, and the enthalpy of reaction for these molar amounts is written directly after the equation.

Thermochemical Equations

- In a thermochemical equation it is important to note phase labels because the enthalpy change, DH, depends on the phase of the substances.

Thermochemical Equations

- The following are two important rules for manipulating thermochemical equations:

- When a thermochemical equation is multiplied by any factor, the value of ΔH for the new equation is obtained by multiplying the ΔH in the original equation by that same factor.
- When a chemical equation is reversed, the value of ΔH is reversed in sign.

Hess’s Law

- Hess’s law of heat summationstates that for a chemical equation that can be written as the sum of two or more steps, the enthalpy change for the overall equation is the sum of the enthalpy changes for the individual steps.

The net enthalpy involved in a reaction is the same whether the reaction takes place in a single step or in a series of steps.

- Hess’s law: if a reaction is carried out in a number of steps, H for the overall reaction is the sum of H for each individual step.
- For example:
- CH4(g) + 2O2(g) CO2(g) + 2H2O(g)H = -802 kJ
- 2H2O(g) 2H2O(l) H= - 88 kJ
- CH4(g) + 2O2(g) CO2(g) + 2H2O(l) H = -890 kJ

Another Example of Hess’s Law

Given:

C(s) + ½ O2(g) CO(g)DH = -110.5 kJ

CO2(g) CO(g) + ½ O2(g)DH = 283.0 kJ

Calculate DH for:C(s) + O2(g) CO2(g)

Hess’s Law & Energy Level Diagrams

Forming H2O can occur in a single step or in a two steps.

∆Htotal is the same no matter which path is followed.

Hess’s Law Energy Level Diagrams

Forming CO2 can occur in a single step or in a two steps.

∆Htotal is the same no matter which path is followed.

- Could you use these data to obtain the enthalpy change for the following reaction?

- For example, suppose you are given the following data:

Hess’s Law the following reaction?

- If we multiply the first equation by 2 and reverse the second equation, they will sum together to become the third.

Standard Enthalpies of Formation the following reaction?

- The term standard state refers to the standard thermodynamic conditions chosen for substances when listing or comparing thermodynamic data: 1 atmosphere pressure and the specified temperature (usually 25oC).

- The enthalpy change for a reaction in which reactants are in their standard states is denoted ΔHo(“delta H zero” or “delta H naught”).

Standard Enthalpies of Formation the following reaction?

- The standard enthalpy of formation of a substance, denotedDHfo, is the enthalpy change for the formation of one mole of a substance in its standard state from its component elements in their standard state.

- Note that the standard enthalpy of formation for a pure element in its standard state is zero.

Standard Enthalpies of Formation the following reaction?

- The law of summation of heats of formation states that the enthalpy of a reaction is equal to the total formation energy of the products minus that of the reactants.

- Sis the mathematical symbol meaning “the sum of”, and m and n are the coefficients of the substances in the chemical equation.

A Problem to Consider the following reaction?

- Large quantities of ammonia are used to prepare nitric acid according to the following equation:

- What is the standard enthalpy change for this reaction? Use Table 6.2 for data.

A Problem to Consider the following reaction?

- You record the values of DHfo under the formulas in the equation, multiplying them by the coefficients in the equation.

Table 6.2

Table 6.2

Table 6.2

Table 6.2

- You can calculateDHoby subtracting the values for the reactants from the values for the products.

A Problem to Consider the following reaction?

- Using the summation law:

- Be careful of arithmetic signs as they are a likely source of mistakes.

Calculation of the following reaction?H

We can use Hess’s law in this way:

H = nHf(products) - mHf(reactants)

where n and m are the stoichiometric coefficients.

Calculating the following reaction?ΔHrxn

Three different ways:

- Using Heat of Formation (ΔHf) values
- Hess’s Law
- Using Bond Dissociation Energies

Heat of Formation ( the following reaction?ΔHf)

- change in enthalpy that accompanies formation of one mole of a compound from its elements with all substances in standard states at 25C.
- value for a free element in standard state is ZERO.
- look up values in Appendix Four of AP book.

C the following reaction?(graphite) + O2(g) CO2(g)DH0 = -393.5 kJ

rxn

S(rhombic) + O2(g) SO2(g)DH0 = - 296.1 kJ

r xn

CS2(l) + 3O2(g) CO2(g) + 2SO2(g)DH0 = -1072 kJ

rxn

2S(rhombic) + 2O2(g) 2SO2(g)DH0 = -296.1x2 kJ

C(graphite) + 2S(rhombic) CS2 (l)

C(graphite) + 2S(rhombic) CS2 (l)

rxn

rxn

C(graphite) + O2(g) CO2(g)DH0 = -393.5 kJ

+

CO2(g) + 2SO2(g) CS2(l) + 3O2(g)DH0 = +1072 kJ

rxn

DH0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ

rxn

Calculate the standard enthalpy of formation of CS2 (l) given that:

1. Write the enthalpy of formation reaction for CS2

2. Add the given rxns so that the result is the desired rxn.

Δ the following reaction?Hrxn from ΔHf

∆Hrxn = ∑∆H f(products) - ∑∆H f(reactants)

Δ the following reaction?Hrxn from ΔHf

Calculate the enthalpy change for the oxidation of FeO to Fe2O3.

4FeO(s) + O2(g) → 2Fe2O3(s)

Δ the following reaction?Hrxn from ΔHf

Calculate the heat of reaction for the following process:

2SO2 + O2→ 2SO3

Hess’s Law the following reaction?

- the heat of reaction is the same whether the reaction takes place in one step or in a series of steps
- this true because enthalpy is a state function

Hess’s Law the following reaction?

Determine ∆H for the following reaction:

CuO(s) + H2(g) → Cu(s) + H2O(g)

Use the following thermochemical equations:

CuO(s) → Cu(s) + ½ O2(g) ∆H=155kJ

H2O(g) → H2(g) + ½ O2(g) ∆H=242kJ

MgO(s) + 2HCl(g) the following reaction?→ MgCl2(s) + H2O(l)

Mg(s) + 2HCl(g) → MgCl2(s) +H2(g) ∆H=-456.9

Mg(s) + O2(g) → MgO(s)∆H=-601.6

H2O(l) → H2(g) + O2(g) ∆H=+285.8

Bond Dissociation Energy the following reaction?

- Energy required to break a chemical bond
- Energy is USED to break a bond
Energy is RELEASED when a bond forms

∆H the following reaction?rxn using BDE values

∆Hrxn = ∑BDEreactants - ∑BDEproducts

Notice that “reactants” and “products” are switched compared to when using ∆Hf values.