Chapter 16Reaction Energy

Section 1 – Thermochemistry Section 2 – Driving Force of Reactions

Chapter 16

Define temperature and state the units in which it is measured.

Define heat and state its units.

Perform specific-heat calculations.

Explain enthalpy change, enthalpy of reaction, enthalpy of formation, and enthalpy of combustion.

Solve problems involving enthalpies of reaction, enthalpies of formation, and enthalpies of combustion.

Thermochemistry

Thermochemistry

Thermochemistry: study of the transfer of energy as heat that accompanies chemical reactions and physical changes.

Virtually every physical change & chemical reaction is accompanied by a change in energy.◦ What two ways can energy be involved?

Absorbed (endothermic) or released (exothermic)

Solid Liquid Gas Plasma

What is a phase change?• change from one state of matter to another• during a phase change – temp. remains constant

Heat: the energy transferred between samples of matter because of a difference in their temperatures.

◦ SI unit of heat and energy: Joule (J)

Energy transferred as heat moves spontaneously from matter at a higher temp lower temp

Heat

Specific Heat (C): amount of energy required to raise the temperature of 1 gram by 1°C or by 1 K.

◦ Units: J/(g•°C) or J/(g•K)

◦ The temperature difference as measured in either Celsius degrees or kelvins is the same.

◦ Specific heat allows us to compare various materials to one another

Specific Heat

A higher C value:◦ More energy required to raise/lower temp.; heats /cools relatively

slow◦ Can hold more energy overall

A lower C value:◦ Less energy required to raise/lower temp.; heats/cools relatively

quickly◦ Hold less energy overall (

Meaning of C

Can the specific heat of a substance change with a phase change?

◦ Yep◦ Ice vs. water vs. steam – all have different values

Meaning of C

measure of the average kinetic energy of the particles in a sample of matter.◦ The greater the kinetic energy of the particles in a

sample, the hotter it feels.

Remember: How do we convert Celsius to Kelvin?

K = 0C + 273

Temperature

Q = heat energy lost/gained (J)C = specific heat (J/g oC)m = mass of the sample (g)∆T = difference between the initial and final

temperatures. (oC or K)

Energy transferred depends on:◦ The material itself (all substances have unique C’s)◦ The mass

Heat absorbed/released

Q = m x C x ∆T

During a phase change, temperature remains constant Can’t use previous equation

Heat of Fusion: Hfus

◦ Heat energy needed to melt/freeze 1 mol

Heat of Vaporation: Hvap

◦ Heat energy need to boil/condense 1 mol of a substance

Phase Changes

Q = mol x Hfus / Hvap

Step 1 (_________________) :

Step 2 (_________________):  Step 3 (_________________):

Step 4: (_________________):

Step 5: (_________________):

Calorimeter

Calorimeter: used to measure heat absorbed or released.

Energy released from the reaction is measuredfrom the temp change ofthe water

Calorimeter

A 4.0 g sample of glass was heated from 274 K to 314 K, and was found to have absorbed 32 J of energy as heat.a. What is the specific heat of this type of glass?

b. How much energy will the same glass sample gain when it is heated from 314 K to 344 K?

Example Problem 1

32 J

(4.0 g)(40. K)0.20 J/(g K)p

qc

m T

Given: m = 4.0 g ∆T = 314 – 274 = 40. K q = 32 J

Unknown: a. cp in J/(g•K)

b. q for ∆T of 314 K → 344 KSolution:

a.

Example Problem 1 Solution

0.20 J(4.0 g)(30 K)

(g K)24 Jq

b.

Example Problem 1 Solution

pq c m T

0.20 J(4.0 g)(344 K 314 K)

(g K)q

**See Practice Worksheet for more examples

Assume we have 85 grams of ice at -10 °C . How much energy is needed to convert it to water vapor at 120oC?

 Step 1: -10°C to O°C (solid)◦ What is happening to the particles as heat is added?◦ General formula: q=(m)(cp)(∆T)

◦ Answer = 1751 J 1.8kJ

Step 2: melting◦ Notice how there is no increase in temp.◦ Where is this energy going?◦ General formula: q=(∆H fusion)(mol)

◦ Answer = 28 kJ

Typical Heat Curve – Problem Solving

Step 3: – 0 °C to 100 °C (liquid)◦ What is happening to the particles as we heat?◦ General formula: q=(m)(cp)(∆T)

◦Answer = 36,000 36 kJ Step 4: Boiling

◦ Notice how there is no increase in temp.◦ Where is this energy going?◦ General formula: q=(∆H vap)(mol)

◦Answer = 190 kJ Step 5: 100 °C to 120 °C (gas)

◦ What is happening to the particles as we heat?◦ General formula = (m)(cp)(∆T)

◦Answer = 3,200 J 3.2 kJ

Typical Heat Curve – Problem Solving; cont’d

Now, add up all the steps for total heat.◦Answer = 259 kJ

Summary: When the substance is heating

◦ q=(m)(cp)(∆T) During phase change

◦ q=(mol)(∆H fus) OR◦ q=(mol)(∆H vap)

Typical Heat Curve – Problem Solving; cont’d

Enthalpy (H): Heat◦ H is never directly determined, rather ∆H

Enthalpy(∆H): “change in enthalpy”energy absorbed as heat during a chemical reaction at constant pressure

Enthalpy of reaction (∆Hrxn): heat energy absorbed or released during a chemical reaction

Enthalpy (Heat) of Reaction

The difference between the enthalpies of products and reactants.

∆Hrxn = Hproducts – Hreactants

◦ -∆Hrxn : exothermic, reaction releases heat

◦ +∆Hrxn : endothermic, reaction absorbs heat

Enthalpy change

Thermochemical Equation Equation that includes the quantity of

energy released or absorbed as heat

2H2(g) + O2(g) → 2H2O(g) ∆H = -483.6 kJ

◦ States of matter are ALWAYS shown because it can influence the energy of released or absorbed.

◦ Coefficients are always viewed as number of moles in the reaction.

Energy released-∆H

Exothermic Reaction

Energy is absorbed +∆H

Endothermic

Compounds with a large negative enthalpy of formation are very stable.

Compounds with positive values of enthalpies of formation are typically unstable.

Stability

∆Hcomb : enthalpy change that occurs during the complete combustion of 1 mole of a substance

**Enthalpy of combustion is defined in terms of one mole of reactant

**Enthalpy of formation is defined in terms of one mole of product

Enthalpy of Combustion

fH0

molar enthalpy of formation: enthalpy change that occurs when 1mole of a compound is formed from its elements in their standard state: 25°C and 1 atm.◦ Enthalpies of formation are given for a standard

temperature and pressure so that comparisons between compounds are meaningful.

Enthalpy of Formation

“StandardState”

Heat (enthalpy) of FORMATION

Where do I find ΔHf ?

In your textbook! (or handout)

The values are given as the enthalpy of formation for one mole of the compound from its elements in their standard states.

◦ **because ΔHf is reported for the formation of 1 mol of a substance, it may be necessary to write on ½ mol for one of the reactants

◦ Ex: 1/2N2 (g) + 1/2O2 (g) NO (g)

Determined by using 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.

Determining Enthalpy of Reaction

1. The sign ∆Hrxn of depends on the direction of the reaction

(If you flip the reaction around, the a positive enthalpy would become negative)

2. The magnitude ∆Hrxn of is proportional to the amount of the substance in a reaction.

(If you change of the coefficients then multiply the

∆H, too)

Some Laws we should know*from enthalpy packet

If you know the reaction enthalpies of individual steps in an overall reaction, you can calculate the overall enthalpy without having to measure it experimentally.

What is the heat of formation of methane?C(s) + 2H2(g) → CH4(g)

Example 1

0 393.5 kJcH

0 285.8 kJcH

0 890.8 kJcH

H2(g) + ½O2(g) → H2O(l)

C(s) + O2(g) → CO2(g)

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

The component reactions in this case are the combustion reactions of carbon, hydrogen, and methane:

Calculating using Hess’s Law

This reaction must be reversed! The ΔH is changed to positive!

CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) ∆H0 = +890.8 kJ

cH0 2( 285.8 kJ) 2 [ H2(g) + ½O2(g) → H2O(l) ]

This reaction must x2!The ΔH is x2!

Goal Reaction: C(s) + 2H2(g) → CH4(g)

0 393.5 kJcH

0 2( 285.8 kJ)cH

0 890.8 kJH

0 74.3 kJfH

2H2(g) + O2(g) → 2H2O(l)

C(s) + O2(g) → CO2(g)

C(s) + 2H2(g) → CH4(g)

CO2(g) + 2H2O(l) → CH4(g) + 2O2(g)

ADD IT UP!!

Calculating Enthalpy from Heat of Formation

∆H0 = (ΣΔH0f products) – (ΣΔH0

f reactants)

*Need to use reference sheet for Hf values or table in back of textbook

**Keep in mind: elements in their standard state have an enthalpy of formation of zero◦ Ex: Ag(s) = 0 ; N2 (g) = 0

See Practice WKST (Calculating Enthalpy from Heats of Formation)

Explain the relationship between enthalpy change and the tendency of a reaction to occur.

Explain the relationship between entropy change and the tendency of a reaction to occur.

Discuss the concept of free energy, and explain how the value of this quantity is calculated and interpreted.

Describe the use of free energy change to determine the tendency of a reaction to occur.

16.2 Driving Force of Reactions

Spontaneous Reaction Most chemical reactions are exothermic,

liberating heat◦ So products have less energy than the reactants◦ Nature’s tendency is toward a lower energy

state

Spontaneous Reaction

But, some endothermic processes are also spontaneous◦ Example: melting

(particles have more energy in the liquid state)

What conclusion can we come to then?◦ there is another factor that drives a reaction◦ 2 Factors: Enthalpy, but also Entropy

Entropy (S) : the degree of randomness or disorder of the particles in a system

Think about your bedroom, does it always stay neat and organized? Or does it get messier and messier?

Increase in disorder Increase in entropy

Entropy

Nature’s tendency is toward more disorder or greater entropy.

◦ Also means, the entropy of the universe is ALWAYS increasing.

Entropy

When a solid dissolves in a liquid, the entropy in the system increases.

As you go from a solid liquid gas, the entropy in the system ______________.

When you dilute a substance, the entropy in the system _______________.

When you increase the temperature of a system, the entropy ________________ .

Examples of Changes in Entropy

Change in Entropy ΔS◦ Difference between the entropy of the products

and the reactants

Increase in Entropy +ΔS (more disorder)

Decrease in Entropy -ΔS (less disorder)

Unit: kJ/mol . K or J/mol . K

Change in Entropy

Standard Entropy Changes for Some Reactions

Calculate entropy like calculating enthalpy from heats of formation table, but use S values instead!

∆S = Sproducts - Sreactants

See practice WKST, use thermodynamic data sheet for Sf values

Change in Entropy Calculations

Gibb’s Free Energy (G):◦ Combined enthalpy-entropy function that

assesses how these two factors will affect the spontaneity of a reaction

Spontaneous Reactions: the ideal situation◦ Large negative enthalpy – very stable◦ Large positive entropy – greater disorder

Gibb’s Free Energy

Only the change in free energy can be measured (KJ/mol) At a constant pressure and temperature…

∆G0 = ∆H0 – T∆S0

*T must be in Kelvin since ∆S will be in units of KJ/mol k

*∆S should also be in KJ

Change in Free Energy

Relating Enthalpy and Entropy to Spontaneity

If -∆G the reaction is spontaneous +∆G the reaction is not spontaneous

∆H ∆S ∆G

- + - /spontaneous

+ - + /not spontaneous

- - - / Low temps spontaneous

+ + - / High temps spontaneous

See “Free Energy” worksheet to practice calculationSee Practice Worksheet – “Pg. 78 Gibbs Free Energy” to review concepts

Ice - Enthalpy vs Entropy

For the reaction NH4Cl(s) → NH3(g) + HCl(g)

∆H0 = 176 kJ/mol and ∆S0 = 0.285 kJ/(mol•K) at 298.15 K. Calculate ∆G0, and tell whether this reaction is spontaneous in the forward direction at 298.15 K.

Example Problem

Given: ∆H0 = 176 kJ/mol at 298.15 K ∆S0 = 0.285 kJ/(mol•K) at 298.15

KUnknown: ∆G0 at 298.15 KSolution: The value of ∆G0 can be calculated

according to the following equation:∆G0 = ∆H0 – T∆S0

∆G0 = 176 kJ/mol – 298 K [0.285kJ/(mol•K)]∆G0 = 176 kJ/mol – 84.9 kJ/mol∆G0 = 91 kJ/mol

Example Problem Solution