CHAPTER 16. REACTION ENERGY
Download
1 / 41

CHAPTER 16. REACTION ENERGY SECTION 1. THERMOCHEMISTRY - PowerPoint PPT Presentation


  • 140 Views
  • Uploaded on

CHAPTER 16. REACTION ENERGY SECTION 1. THERMOCHEMISTRY. Thermochemistry is the study of the transfers of energy as heat that accompany chemical reactions and physical changes. Heat is energy transferred between samples of matter because of a difference in their temperature.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about ' CHAPTER 16. REACTION ENERGY SECTION 1. THERMOCHEMISTRY' - nile


An Image/Link below is provided (as is) to download presentation

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 - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

CHAPTER 16. REACTION ENERGY

SECTION 1. THERMOCHEMISTRY

  • Thermochemistryis the study of the transfers of energy as heat that accompany chemical reactions and physical changes.

  • Heatis energy transferred between samples of matter because of a difference in their temperature.


  • Temperatureis a measure of the average kinetic energy of the particles in a sample of matter.

  • Heat will move from a sample at higher temperature to one at lower temperature.

  • Like other forms of energy, the SI unit of heat is the joule.


Enthalpy of Reaction (commonly referred to as “heat of reaction”)

Enthalpy can be considered as stored energy in a substance. It is not measured directly. Rather, we measure changes in enthalpy during chemical reactions.

Change is indicated using the Greek letter delta = Δ

H is the symbol for enthalpy.


ΔH = enthalpy change = amount of heat energy absorbed by a system during a process at constant pressure.

Enthalpy change is the difference between the enthalpies of products and reactants.

∆H = Hproducts– Hreactants


Exothermic Reactions

If the products have less enthalpy than the reactants, then ΔH < 0, and heat is released. Heat can be indicated as if it were a product of the reaction. This is an exothermic reaction.


  • Ex.:

  • 2H2(g) + O2(g) → 2H2O(g) + 483.6 kJ

  • If 2 mol H2 react with 1 mol O2 to produce 2 mol H2O, 483.6 kJ of heat are produced.

  • If different numbers of moles react, the energy released will be changed in proportion.



Endothermic Reactions

If the products have more enthalpy than the reactants, then ΔH > 0, and heat is absorbed. Heat can be indicated as if it were a reactant. This is an endothermic reaction.

Ex. – cold pack in which NH4NO3 dissolves in water. The reaction absorbs 26 kJ per mol.

NH4NO3(s) + 26 kJ → NH4NO3(aq)



Thermochemical equations are usually written by giving a ∆H value rather than writing the energy as a reactant or product.

For the two examples given previously:

2H2(g) + O2(g) → 2H2O(g)

ΔH = -483.6 kJ

NH4NO3(s) → NH4NO3(aq)

ΔH = +26 kJ

Note that ∆H is negative for an exothermic reaction and positive for an endothermic reaction.


Enthalpy of Formation

  • Themolar enthalpy of formationis the enthalpy change that occurs when one mole of a compound is formed from its elements in their standard state at 25°C and 1 atm.

  • Standard state = state found at atmospheric pressure and 25° C. Ex. – O2 is a gas, Hg is liquid.


To signify standard states, a 0 sign is added to the enthalpy symbol, and the subscript f indicates a standard enthalpy of formation:

Elements in their standard states by definition have = 0



  • Ex.: formation of some compounds.the of carbon dioxide is

  • –393.5 kJ per mol of gas produced.

  • This means that for the reaction

    • C(s) + O2(g) → CO2(g) at 25° C,

    • ΔH = -393.5 kJ/mol

    • Ethyne (acetylene) has

    • = +226.7 kJ/mol. It is very unstable.


Calculating Enthalpies of Reaction formation of some compounds.

  • Hess’s law:the overall enthalpy change in a reaction is equal to the sum of enthalpy changes for the individual steps in the process. It is independent of the path by which reactants go to products.


This lets you calculate formation of some compounds.H for a reaction that is difficult to measure.

It also lets you calculate H from tables of enthalpies of formation (Hf0).


  • In using this law: formation of some compounds.

  • If a reaction is reversed, the sign of H is reversed.

  • If it is necessary to change all the coefficients of a reaction, change H in the same proportion.


Ex. 1 (p. 540): calculate the enthalpy of formation of methane gas, CH4, from its elements, hydrogen gas and solid carbon:

C(s) + 2H2(g) → CH4(g)

The calculation can be done using the enthalpies of combustion of C, H2, and CH4.


These reactions are: methane gas, CH

C(s) + O2(g) → CO2(g)

H2(g) + ½O2(g) → H2O(l)

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)


We can get the desired overall reaction if we double the second reaction (and also double H), reverse the third reaction (and reverse the sign of H), and add them together. To get the overall H, we add the three H values.


C( second reaction (and also double s) + O2(g) → CO2(g) H = -393.5

2H2(g) + O2(g) → 2H2O(l)

H = 2(-285.8) = -571.6

CO2(g) + 2H2O(l) → CH4(g) + 2O2(g)

H = +890.8

C(s) + 2H2(g) → CH4(g)

H = -74.3 kJ


If you know the second reaction (and also double Hf0 for all reactants and products, H can be calculated for any reaction by subtracting the total Hf0s of the reactants from those of the products (all adjusted for the number of moles).


Ex. 2 (p. 541) second reaction (and also double Calculate H for

NO(g) + 2O2(g) → NO2(g)

Using Tab. A14:

nitrogen monoxide

Hf0 = +90.29 kJ/mol

nitrogen dioxide

Hf0 = +33.2 kJ/mol


This means that: second reaction (and also double

½N2(g) + ½O2(g) → NO(g) H = +90.29

½ N2(g) + O2(g) → NO2(g) H = +33.2

Rewrite these so that the NO is on the left; keep NO2 on the right:

NO(g) → ½N2(g) + ½O2(g) H = -90.29

½ N2(g) + O2(g) → NO2(g) H = +33.2

Net: NO(g) + 2O2(g) → NO2(g)

H = -90.29 + 33.2 = - 57.1


Note that second reaction (and also double H =

Hf0(NO2) -Hf0(NO)

product reactant


  • SECTION 2. DRIVING FORCE OF REACTIONS second reaction (and also double

  • In this section we consider the question of whether a reaction will proceed spontaneously.

  • For a reaction at constant temperature and pressure, this is determined by two factors:

  • enthalpy, which we have discussed

  • entropy


Entropy second reaction (and also double (abbreviation: S) can be considered a measure of randomness of the particles of a system. The higher the entropy, the more disordered the system is, and the more difficult to describe the location of the particles.


Entropy 75329
Entropy – 75329 second reaction (and also double


In general, a gas has the most entropy, followed by a liquid, then a solid. A pure crystalline solid has zero entropy at absolute zero

( = ).

The unit of entropy are joules per kelvin (J/K) or kJ/K. For one mole, it is kJ/(mol•K).


In general, a gas has the most entropy, followed by a liquid, then a solid. A pure crystalline solid has zero entropy at absolute zero

( = 0 K = -273˚C ).

The unit of entropy are joules per kelvin (J/K) or kJ/K. For one mole, it is kJ/(mol•K).


Entropy and Reaction Tendency liquid, then a solid. A pure crystalline solid has zero entropy at absolute zero

  • Some processes that result in an increase in entropy (S is positive):

  • melting of ice

  • forming a solution

  • mixing gases

  • An increase in entropy tends to make a reaction favorable, since systems spontaneously go to more disordered states.


Ex.: The decomposition of ammonium nitrate: liquid, then a solid. A pure crystalline solid has zero entropy at absolute zero

2NH4NO3(s) 2N2(g) + 4H2O(l) + O2(g)

  • On the left side are 2 mol of solid ammonium nitrate.

  • The right-hand side of the equation shows 3 mol of gaseous molecules plus 4 mol of a liquid.

  • The arrangement of particles on the right-hand side of the equation is more random than the arrangement on the left side.


Standard entropy changes for some reactions
Standard Entropy Changes for Some Reactions liquid, then a solid. A pure crystalline solid has zero entropy at absolute zero


A liquid, then a solid. A pure crystalline solid has zero entropy at absolute zerodecrease in enthalpy (H is negative) tends to make a reaction favorable, since it means moving to more stable chemicals.

If the S and H effects are acting in opposite directions, how do we decide if the reaction is favorable?


Free Energy liquid, then a solid. A pure crystalline solid has zero entropy at absolute zero

The combined effects of enthalpy and entropy are expressed in the function free energy (abbreviated G, for Gibbs free energy).

G = H - TS (T = Kelvin temp.)

units: kJ or kJ/mol


J. Willard Gibbs liquid, then a solid. A pure crystalline solid has zero entropy at absolute zero

en.wikipedia.org


If liquid, then a solid. A pure crystalline solid has zero entropy at absolute zeroG < 0, the reaction is favorable (it could spontaneously go to the right).

Here are the possible combinations of H and S:


liquid, then a solid. A pure crystalline solid has zero entropy at absolute zeroG = H - TS

Note that as T increases, the effect of entropy becomes more important.


Relating enthalpy entropy and free energy changes 75328
Relating Enthalpy, Entropy, and Free-Energy Changes liquid, then a solid. A pure crystalline solid has zero entropy at absolute zero75328


Ex. 1: C liquid, then a solid. A pure crystalline solid has zero entropy at absolute zero2H4(g) + H2(g) → C2H6(g)

H = -136.9 kJ/mol (favorable)

S = -0.121 kJ/(mol•K) (unfavorable)

At 298 K: G = H - TS

= -136.9 – 298(-0.121)

= -136.9 + 36.1 = -100.8 kJ/mol

Since this is negative, the reaction is favorable.


Ex. 2: CH liquid, then a solid. A pure crystalline solid has zero entropy at absolute zero4(g) + H2O(g) →

CO(g) + 3H2(g)

H = +206.1 kJ/mol (unfavorable)

S = +0.215 kJ/(mol•K) (favorable)

At 298 K: G = H - TS

= +206.1 – 298(0.215)

= +206.1 – 64.1 = +142.0 kJ/mol

Since this is positive, the reaction is unfavorable.


ad