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

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

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

Calorimeter

Calorimeter: used to measure heat absorbed or released.

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

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?

Temperature

Joule: SI unit of heat and energy

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

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

Heat

Energy transferred depends on:◦ The material◦ The mass

Specific Heat: specific heat of a substance is the amount of energy required to raise the temperature of one gram by one Celsius degree (1°C) or one kelvin (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

Molar Heat Capacities of Elements and Compounds

p

qc

m T

pq c m T

Heat released Specific heat: or absorbed:

cp = specific heat at a given pressure

q = energy lost or gainedm = mass of the sample∆T = difference between the initial and final

temperatures.

Specific Heat Calculation

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

             

Please note how each part of the curve is numbered. These numbers correspond to the steps in our problem-solving.

  Typical Heat Curve

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 – We are below 0°C◦ What is the phase?◦ What is happening to the particles as heat is added?◦ General formula: q=(m)(cp)(∆T) ◦ Answer = 1751 J or 1.8kJ

Step 2 - We are at 0 °C◦ What is happening to the phase?◦ 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 – We are above 0 °C but below 100 °C ◦ What is the phase?◦ What is happening to the particles as we heat?◦ General formula: q=(m)(cp)(∆T) ◦ Answer = 36 kJ

Step 4 - We are at 100 °C◦ What is happening to the phase?◦ Notice how there is no increase in temp.◦ Where is this energy going?◦ General formula: q=(∆H vap)(mol)◦ Answer = 190 kJ

Step 5 – We are above 100 °C ◦ What is the phase?◦ What is happening to the particles as we heat?◦ General formula = (m)(cp)(∆T) ◦ Answer = 3.2kJ

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 up and is in a

single phase, the formula used is : q=(m)(cp)(∆T) When a phase change is occurring all the energy

is involved with the intermolecular forces and because of this we do not see a temperature change. To calculate we must multiply the moles by molar heat; ◦ q=(∆H fusion)(mol) OR◦ q=(∆H vap)(mol)

Typical Heat Curve – Problem Solving; cont’d

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

Enthalpy of Reaction

The difference between the enthalpies of products and reactants.

∆H = Hproducts – Hreactants

Enthalpy change

Thermochemical Equation Equation that includes the quantity of energy

released or absorbed as heat

2H2(g) + O2(g) → 2H2O(g) + 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.

If 2x as many reactants were provide, 2x as many moles of water would be produced and 2x as many kJ of energy would be released.

chemical reaction that releases energy the energy of the products is less than the energy of the reactants◦ example:

2H2(g) + O2(g) → 2H2O(g) + 483.6 kJ

Exothermic

Endothermic chemical reaction that absorbs energy

the products have a larger enthalpy than the reactants

2H2O(g) + 483.6 kJ → 2H2(g) + O2(g)

Enthalpy of a Reaction Enthalpy is often not included in the

reaction expression, instead as a ∆H value◦ Example:

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

◦ Negative ∆H is exothermic – releases heat◦ Positive ∆H is endothermic – absorbs heat

Energy released, ∆H is negative.

Exothermic Reaction

Energy is absorbed, ∆H is positive.

Endothermic

fH0

molar enthalpy of formation: 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.◦ Enthalpies of formation are given for a standard

temperature and pressure so that comparisons between compounds are meaningful.

Enthalpy of Formation

standard states

heat of FORMATION

change in enthalpy

Where do I find ΔHf ?

In your textbook!

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

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

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

Stability

∆Hc : enthalpy change that occurs during the complete combustion of one 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

Combustion/Bomb Calorimeterused to find enthalpy of combustion

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.

This means that the energy difference between reactants and products is independent of the route taken to get from one to the other.

Enthalpy of Reaction

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!!

Enthalpy of Formation

∆H0 = Σ[(ΔH0f of products) × (mol of

products)]

– Σ[(ΔH0f of reactants) × (mol of reactants)]

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

Majority of spontaneous reactions are exothermic◦ Reactions proceed in a direction that leads to a

lower energy state

Some spontaneous reactions are endothermic◦ Example: melting

(particles have more energy in the liquid state)

Entropy: ◦ Increases when a system goes from one state to

another without an enthalpy change

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

The arrangement of particles on the right-hand side of the equation is more random than the arrangement on the left side and hence is less ordered.

Entropy

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

There is a tendency of nature to go from order to disorder◦ Think about your bedroom, does it always stay neat

and organized? Or does it get messier and messier?

◦ Increase in disorder Increase in entropy

◦ The entropy of the universe is ALWAYS increasing

Entropy

Standard Entropy Changes for Some Reactions

As you go from a solid liquid gasWhat happens to the entropy?

As you go from a gas liquid solidWhat happens to the entropy?

Entropy in States of Matter

Entropy

Unit: kJ/(molK) Change in Entropy ΔS

◦ Difference between the entropy of the products and the reactants

Increase in Entropy Positive Value Decrease in Entropy Negative Value

Change in Entropy

Spontaneous Reactions:◦ Least enthalpy◦ Greatest entropy

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

Gibb’s Free Energy

Only the change in free energy can be measured.

At a constant pressure and temperature, the free-energy change, ∆G, of a system is defined as the difference between the change in enthalpy, ∆H, and the product of the Kelvin temperature and the entropy change, which is defined as T∆S:

∆G0 = ∆H0 – T∆S0

Change in Free Energy

Relating Enthalpy and Entropy to Spontaneity

∆G0 = ∆H0 – T∆S0

The expression for free energy change is for substances in their standard states. Unit: kJ/mol

If ∆G < 0 the reaction is spontaneous.

Gibb’s Free Energy

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