© 2014 pearson education, inc. thermochemistry: chemical energy

© 2014 Pearson Education, Inc.

Thermochemistry:Thermochemistry:Chemical EnergyChemical Energy


Energy and Its Conservation

Law of Conservation of Energy: Energy cannot be created or destroyed; it can only be converted from one form into another. First Law of Thermodynamics

Energy: The capacity to supply heat or do work.

Kinetic Energy (EK): The energy of motion.

Potential Energy (EP): Stored energy.

∆E = q + w


Energy and Its Conservation

Thermal Energy: The kinetic energy of molecular motion, measured by finding the temperature of an object.

Heat: The amount of thermal energy transferred from one object to another as the result of a temperature difference between the two.


Calorimetry

The energy released or absorbed during a chemical reaction.


Some definitions

Exothermic: energy is given off (temperature increases)

Endothermic: energy is required (temperature decreases)

Enthalpy, H: Internal energy in kJ/mol

- H: exothermic + H: endothermic


Types of CalorimetersCoffee cup Bomb


Coffee Cup Calorimetry

Measure the heat flow at constant pressure (∆H).


Bomb Calorimetry

Measure the heat flow at constant volume (∆E).


Coffee-cup calorimetry

q = -mcT

q: quantity of heat in joules, J

m: mass of liquid, g

c: specific heat capacity, 4.18J/goC

T: change in temperature in oC, Tfinal – Tinitial

q = nΔH or ΔH = q/n


There are two kinds of coffee-cup calorimetry problems.

1st kind: Grams of a chemical are added to a given mass or volume of water.

What to do:

Grams of water will be plugged into m.

T, Last temperature minus first temperature.

Solve for q and change joules to kJ.

Grams of the other chemical convert to moles.

Solve for H by putting kJ over moles and divide.


Example #1

In a coffee-cup calorimeter, 5g of sodium are dropped in 300g of water at 20oC the temperature rises to 35oC. Calculate the enthalpy change for this reaction.


Example #2

A 1.5g sample of calcium is dropped into 275mL of water at 20oC in a coffee-cup calorimeter. The temperature rises to 23oC. The density of water is 1g/mL. What is the energy change for this reaction in kJ/mol?


There are two kinds of coffee-cup calorimetry problems.

2nd kind: Two volumes of a solution are given.

What to do:

Add them and plug into m (the density of most solutions = 1g/mL, so the mL = the grams)

T, Final temperature minus Initial temperature.

Solve for q and change joules to kJ.

Pick one of the chemicals and multiply (L)(M) to find moles.



Example #3

450mL of 2M HCl are mixed with 450mL of 2M NaOH in a coffee-cup calorimeter. The temperature rises from 23oC to 25oC. Calculate the enthalpy change for this reaction.


Example #4

In a coffee-cup calorimeter, 275mL of 0.5M Ca(OH)2 react with 275mL of 0.5M H2SO4. The temperature increases from 25oC to 30oC. Calculate the energy in kJ/mol.


Worked Example Calculating the Amount of Heat Released in a Reaction

According to the balanced equation, 852 kJ of heat is evolved from the reaction of 2 mol of Al. To find out how much heat is evolved from the reaction of 5.00 g of Al, we have to find out how many moles of aluminum are

in 5.00 g.

Solution

The molar mass of Al is 26.98 g/mol, so 5.00 g of Al equals 0.185 mol:

Because 2 mol of Al releases 852 kJ of heat, 0.185 mol of Al releases 78.8 kJ of heat:

✔ Ballpark Check

Since the molar mass of Al is about 27 g, 5 g of aluminum is roughly 0.2 mol, and the heat evolved is about (852 kJ/2 mol)(0.2 mol), or approximately 85 kJ.

Strategy

How much heat in kilojoules is evolved when 5.00 g of aluminum reacts with a stoichiometric amount of Fe 2O3?


Bomb Calorimetry

q = -CT

q: quantity of heat, kJ

C: heat capacity, kJ/oC

T: change in temperature (increases positive, decreases negative)


Two kinds of Bomb also:

1st kind: Enthalpy (kJ/mol) is given

What to do:

Convert grams to moles

Multiply the moles by the enthalpy. This answer plugs in for q

Find C or T (whichever it asks for)


Example #5

A 5g sample of hydrogen is burned in a

bomb calorimeter. The enthalpy of

combustion for hydrogen is -242kJ/mol.

What is the heat capacity of the calorimeter if the temperature rises 5oC?


Example #6

30g of magnesium are burned in a bomb

calorimeter that has a heat capacity of

18kJ/oC. If the enthalpy of combustion for

magnesium is -602kJ/mol, how much will the

temperature rise?


Still Bomb

2nd kind: Enthalpy is asked for (or energy change in kJ/mol)

What to do:

Convert grams to moles.

Plug in C and T and solve for q.



Example #7

If 10g of Ca are burned in a bomb

calorimeter that has a heat capacity of

12.8kJ/oC, the temperature increases by

12.38oC. What is the enthalpy of

combustion of calcium?


Example #8

An 18g sample of sulfur is burned in a bomb

calorimeter. The heat capacity of the

calorimeter is 8.9kJ/oC and the temperature

increases 18.74oC. What is the enthalpy of

combustion for sulfur?


What is “Thermodynamics”?

The study of energy and its interconversions.


Some terms:Enthalpy: Internal energy (energy stored in

the bonds) H kJ/mol (+ endothermic, - exothermic)

Entropy: disorder of a substance S J/mol.K

gases aqueous liquids solids

most disordered to least disordered

Gibbs Free Energy: stored energy less the degree of disorder G kJ/mol


How to calculate H, S & G

Ho = Hfo for products – Hf

o for reactants

So = So of products – So for reactants

Go = Gfo of products – Gf

o for reactants

O superscripted means standard states (1atm, 273K, etc)


Standard Heat of Formation (∆H°f): The enthalpy change for the formation of 1 mol of a substance in its standard state from its constituent elements in their standard states.

∆H°f = -74.8 kJCH4(g)C(s) + 2 H2(g)

Standard states

1 mol of 1 substance

Calculating Enthalpy Changes Using Standard Heats of Formation


c C + d Da A + b B

∆H° = ∆H°f (Products) - ∆H°f (Reactants)

ReactantsProducts

∆H°= [c ∆H°f (C) + d ∆H°f (D)] - [a ∆H°f (A) + b ∆H°f (B)]



Using standard heats of formation, calculate the standard enthalpy of reaction for the photosynthesis of glucose (C6H12O6) and O2 from CO2 and liquid H2O.

C6H12O6(s) + 6 O2(g)6 CO2(g) + 6 H2O(l) ∆H° = ?

H°= [∆H°f (C6H12O6(s))] - [6 ∆H°f (CO2(g)) + 6 ∆H°f (H2O(l))]

= 2802.5 kJ

[(6 mol)(-393.5 kJ/mol) + (6 mol)(-285.8 kJ/mol)]

∆H°= [(1 mol)(-1273.3 kJ/mol)] -



How to calculate H, S & G

So and Go can be calculated the same way as Ho

So = So of products – So for reactants

Go = Gfo of products – Gf

o for reactants

O superscripted means standard states (1atm, 273K, etc)


Example #1

Calculate Ho, So and Go for the reaction below.

C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l)


Predicting Entropy

To decide which substance has more entropy, first look at its state (solid, liquid, etc.). The more disordered state will have more disorder.

If they have the same state, the larger molecule will have more entropy.

Both gaseous, higher temperature, less pressure, more disorder.


Example #2

Which in each pair has more entropy?

A. H2O(g) H2O(l)

B. C6H12O6 (s) C12H22O11(s)

C. HCl(aq) HCl(g)


Predicting S change in a reaction.

Look at both sides of the reaction, identify the state of the chemicals, how many moles of chemicals are present (more particles, more disorder)and how big the particles are. These are in order of importance.

If the right has more disorder, S goes up.If the right has less disorder, S goes down.


Example #3

Predict the sign for S for each reaction.

A. 12CO2(g) + 11H2O(l) C12H22O11(s) + 12O2(g)

B. 2H2O(g) 2H2(g) + O2(g)


Worked Example Predicting the Sign of ΔS for a Reaction

Look at each reaction, and try to decide whether molecular randomness increases or decreases. Reactions that increase the number of gaseous molecules generally have a positive ΔS, while reactions that decrease the number

of gaseous molecules have a negative ΔS.

Solution

a. The amount of molecular randomness in the system decreases when 2 mol of gaseous reactants combine to give 1 mol of liquid product, so the reaction has a negative ΔS° .

b. The amount of molecular randomness in the system increases when 9 mol of gaseous reactants give 10 mol of gaseous products, so the reaction has a positive ΔS° .

Strategy

Predict whether ΔS° is likely to be positive or negative for each of the following reactions:a. b.


Spontaneity

A spontaneous reaction occurs without the need of outside energy.

Do not confuse spontaneous with instantaneous.

Spontaneous happens on its own, instantaneous happens right away.


How can you tell?

S + tends to be spontaneousH – tends to be spontaneous

The only way to know for sure is if G is -, then it is guaranteed to be spontaneous.

Go = Ho - T So

G = H - T S


Example #4

Predicting the signs of ΔH, ΔS and ΔG

A. When solid ammonium chloride is dissolved in water, it forms an aqueous solution and the temperature drops.

B. When hydrogen gas is mixed with oxygen gas and a spark ignited, a fire ball to the ceiling occurs. 2H2(g) + O2 (g) 2H2O(g)


Worked Conceptual Example Predicting the Signs of ΔH, ΔS, and ΔG for a Reaction

First, decide what kind of process is represented in the drawing. Then decide whether the process increases or decreases the entropy (molecular randomness) of the system and whether it is exothermic or endothermic.

Solution

The drawing shows ordered particles in a solid subliming to give a gas. Formation of a gas from a solid increases molecular randomness, so ΔS is positive. Furthermore, because we’re told that the process is nonspontaneous, ΔG is also positive. Because the process is favored by ΔS (positive) yet still nonspontaneous, ΔH must be unfavorable

(positive). This makes sense, because conversion of a solid to a liquid or gas requires energy and is always endothermic.

Strategy

What are the signs of ΔH, ΔS, and ΔG for the following nonspontaneous transformation?


Calculating G using H & S

Be sure to change S into kJ from J.

Change temperature to Kelvin by adding 273 to celsius.

Plug in and solve for the unknown.


Example #5

For the reaction:

H2O(g) H2O(l)

Ho = -44kJ/mol So = 119J/mol.K

Calculate Go at 10oC, 0oC and -10oC.


More info about G

If G is negative the reaction is spontaneous (runs forwards)

If G is positive it is not spontaneous forwards (can be spontaneous backwards)

If G = 0 it is at equilibrium (no net forward or backward movement)


Temperature and spontaneity

You can change the temperature of reactions and they will become spontaneous at their new temperature.

Phase changes:

The boiling point and freezing (melting) points are temperatures where G =0.


Free Energy

Reversible


Example #6

For the reaction:

H2O(g) H2O(l)

Ho = -44kJ/mol So = 119J/mol.K

At what temperature does this reaction become spontaneous?


Example #7

For the phase change:

Br2(l) Br2(g)

Ho = 31kJ/mol and So = 93J/mol.K

What is the normal boiling point of bromine?


Worked Example Using the Free-Energy Equation to Calculate Equilibrium Temperature

The spontaneity of the reaction at a given temperature can be found by determining whether ΔG is positive or negative at that temperature. The changeover point between spontaneous and nonspontaneous can be found by

setting ΔG = 0 and solving for T.

Solution

At 25 °C (298 K), we have

Because ΔG is positive at this temperature, the reaction is nonspontaneous. The changeover point between spontaneous and nonspontaneous is approximately

The reaction becomes spontaneous above approximately 1120 K (847 ° °°C).

Strategy

Lime (CaO) is produced by heating limestone (CaCO3) to drive off CO2 gas, a reaction used to make Portland cement. Is the reaction spontaneous under standard conditions at 25 ° °°C? Calculate the temperature at which the reaction becomes

spontaneous.


Haber Process:

Multiple-Step Process

N2H4(g)2 H2(g) + N2(g)

Hess’s Law: The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps in the reaction.

2 NH3(g)3 H2(g) + N2(g) ∆H°= -92.2 kJ

∆H°1 = ?

2 NH3(g)N2H4(g) + H2(g)

∆H°reaction = -92.2 kJ2 NH3(g)3 H2(g) + N2(g)

∆H°2 = -187.6 kJ

Calculating Enthalpy Changes Using Hess’s Law


∆H°1 = ∆H°reaction - ∆H°2

= -92.2 kJ - (-187.6 kJ) = +95.4 kJ

∆H°1 + ∆H°2 = ∆H°reaction

Calculating Enthalpy Changes Using Hess’s Law


It often takes some trial and error, but the idea is to combine the individual reactions so that their sum is the desired reaction. The important points are that:

• All the reactants [CH4(g) and O2(g)] must appear on the left.

• All the products [CO2(g) and H2O(l)] must appear on the right.

• All intermediate products [CH2O(g) and H2O(g)] must occur on both the left and the right so that they cancel.

• A reaction written in the reverse of the direction given [ ] must have the sign of its ΔH° reversed (Section 8.7).

• If a reaction is multiplied by a coefficient [ is multiplied by 2], then ΔH° for the reaction must be multiplied by that same coefficient.

Worked Example Using Hess’s Law to Calculate ΔH °

Strategy

Methane, the main constituent of natural gas, burns in oxygen to yield carbon dioxide and water:

Use the following information to calculate ΔH° in kilojoules for the combustion of methane:



Solution

Continued


As in Worked Example 8.6, the idea is to find a combination of the individual reactions whose sum is the desired reaction. In this instance, it’s necessary to reverse the second and third steps and to multiply both by 1/2 to make the overall equation balance. In so doing, the signs of the enthalpy changes for those steps must be changed and

multiplied by 1/2. Note that CO2(g) and O2(g) cancel because they appear on both the right and left sides of equations.


Strategy

Water gas is the name for the mixture of CO and H2 prepared by reaction of steam with carbon at 1000 °°C:

The hydrogen is then purified and used as a starting material for preparing ammonia. Use the following information to calculate ΔH° in kilojoules for the water-gas reaction:


Solution

The water-gas reaction is endothermic by 131.3 kJ.

Worked Example Using Hess’s Law to Calculate ΔH °Continued


Example #8

Using the reactions:

C(s) + O2(g) CO2(g) H = -393.5kJ/mol

CO(g) + ½ O2(g) CO2(g)H = -283kJ/mol

Calculate the enthalpy for:

C(s) + ½ O2(g) CO(g)


Example #9

Calculate H for the reaction:

2C(s) + H2(g) C2H2(g)

Given the following:

2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O (l)

H = -2599.2kJ/mol

C(s) + O2(g) CO2(g) H = -393.5kJ/mol

2H2(g) + O2(g) 2H2O (l) H = -571.6kJ/mol


Example #10

Given the following data:

H2(g) + ½ O2(g) H2O(l) H = -286kJ

N2O5(g) + H2O (l) 2HNO3(l) H = -77kJ

½ N2(g) + 3/2 O2(g) + ½ H2(g) HNO3 (l) H = -174kJ

Calculate the H for:

2 N2(g) + 5 O2(g) 2 N2O5(g)


Hess’s Law and Gibb’s Free Energy

Hess’s Law works the same way for Gibb’s Free Energy, you just use G values instead of H values.


Example #11

Calculate G for the reaction

2CO(g) + O2(g) 2CO2(g)

From the following:

2CH4(g) + 3O2(g) 2CO(g) + 4H2O(g)

G = -1088kJ/mol

CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

G = -801kJ/mol


Example #12Calculate G for the reaction

C2H4 (g) + H2O(l) C2H5OH(l)

From the following:

4C(s) + 6H2(g) + O2(g) 2C2H5OH(l)

G = -700kJ/mol

2H2(g) + O2(g) 2H2O(l) G = -474kJ/mol

2C(s) + 2H2(g) C2H4(g) G = 136kJ/mol


More Ways to find G!

Go = - R T lnKG = Go + RT lnQ

K: equilibrium constants (Kc, Kp, Ka, Kb, Kw, Ksp….)

Q: reaction quotientR: 0.008314kJ/mol.KT: temperature in Kelvin


Example #13

At 127oC the synthesis of ammonia has the following equilibrium concentrations:

[NH3] = 0.031M

[N2] = 0.85M

[H2] = 0.0031M

N2(g) + 3H2(g) 2NH3(g)

Calculate Go for this reaction using the equilibrium concentrations.


Example #14

Calculate G for the synthesis of methanol at 25oC where carbon monoxide is present at 5atm and hydrogen gas is present at 3atm.

CO(g) + 2H2(g) CH3OH(l)

Is this reaction spontaneous with these pressures at this temperature?


Example #15

Determine K for the following reaction:

6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g)


Example #16

For a reaction Ho = -600kJ/mol and So = 150J/mol.K and 25oC.

Calculate K for this reaction.


Using standard heats of formation, calculate the standard enthalpy of reaction for the photosynthesis of glucose (C6H12O6) and O2 from CO2 and liquid H2O.

C6H12O6(s) + 6 O2(g)6 CO2(g) + 6 H2O(l) ∆H° = ?

H°= [∆H°f (C6H12O6(s))] - [6 ∆H°f (CO2(g)) + 6 ∆H°f (H2O(l))]

= 2802.5 kJ

[(6 mol)(-393.5 kJ/mol) + (6 mol)(-285.8 kJ/mol)]

∆H°= [(1 mol)(-1273.3 kJ/mol)] -



1. Reverse the “reaction” and reverse the sign on the standard enthalpy change.

C(s) + O2(g)CO2(g)

Why does the calculation “work”?

∆H° = 393.5 kJ

Becomes

CO2(g)C(s) + O2(g) ∆H° = -393.5 kJ

C6H12O6(s) + 6 O2(g)6 CO2(g) + 6 H2O(l) ∆H° = 2802.5 kJ(3)(1) (2)



1. Multiply the coefficients by a factor and multiply the standard enthalpy change by the same factor.

6 C(s) + 6 O2(g)6 CO2(g)


∆H°= 6(393.5 kJ) = 2361.0 kJ

Becomes

C(s) + O2(g)CO2(g) ∆H°= 393.5 kJ

C6H12O6(s) + 6 O2(g)6 CO2(g) + 6 H2O(l) ∆H° = 2802.5 kJ(3)(1) (2)



2. Reverse the “reaction” and reverse the sign on the standard enthalpy change.

C6H12O6(s) + 6 O2(g)6 CO2(g) + 6 H2O(l) ∆H° = 2802.5 kJ


∆H°= 285.8 kJ

Becomes

∆H°= -285.8 kJ

(3)(1) (2)

H2(g) + O2(g)H2O(l)2

1

H2O(l)H2(g) + O2(g)2

1



2. Multiply the coefficients by a factor and multiply the standard enthalpy change by the same factor.

6 H2(g) + 3 O2(g)6 H2O(l)


∆H°= 6(285.8 kJ) = 1714.8 kJ

Becomes

C6H12O6(s) + 6 O2(g)6 CO2(g) + 6 H2O(l) ∆H°= 2802.5 kJ(3)(1) (2)

∆H° = 285.8 kJH2(g) + O2(g)H2O(l)2

1



C6H12O6(s) + 6 O2(g)6 CO2(g) + 6 H2O(l)

∆H°= 1714.8 kJ


6 H2(g) + 3 O2(g)6 H2O(l)

C6H12O6(s)6 C(s) + 6 H2(g) + 3 O2(g)

6 C(s) + 6 O2(g)6 CO2(g)

∆H°= 2802.5 kJ

∆H°= -1273.3 kJ

∆H°= 2361.0 kJ

C6H12O6(s) + 6 O2(g)6 CO2(g) + 6 H2O(l) ∆H°= 2802.5 kJ(3)(1) (2)



• The amount of energy that must be supplied to break a chemical bond in an isolated molecule in the gaseous state and is thus the amount of energy released when the bond forms.

or

• Standard enthalpy changes for the corresponding bond-breaking reactions.

Bond Dissociation Energy:

Calculating Enthalpy Changes Using Bond Dissociation Energies


2 HCl(g)H2(g) + Cl2(g)

(2mol)(432 kJ/mol)

∆H°= D(Reactant bonds) - D(Product bonds)

∆H°= (DH-H + DCl-Cl) - (2 DH-Cl)

∆H°= [(1 mol)(436 kJ/mol) + (1mol)(243 kJ/mol)] -

= -185 kJ



Worked Example Using Bond Dissociation Energies to Calculate ΔH °

Identify all the bonds in the reactants and products, and look up the appropriate bond dissociation energies in

Table 8.3. Then subtract the sum of the bond dissociation energies in the products from the sum of the bond

dissociation energies in the reactants to find the enthalpy change for the reaction.

Solution

The reactants have four C–H bonds and three Cl–Cl bonds; the products have one C–H bond, three C–Cl bonds, and three H–Cl bonds. The approximate bond

dissociation energies from Table 8.3 are:

Strategy

Use the data in Table 8.3 to find an approximate ΔH° in kilojoules for the industrial synthesis of chloroform by reaction of methane with Cl2.


Worked Example Using Bond Dissociation Energies to Calculate ΔH °

Subtracting the product bond dissociation energies from the reactant bond dissociation energies gives the approximate enthalpy change for the reaction:

The reaction is exothermic by approximately 330 kJ.

Continued

© 2014 pearson education, inc. thermochemistry: chemical energy

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