Energy Relationships in Chemical Reactions

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Energy is the capacity to do work

• Thermal/kinetic energy is the energy associated with the random motion of atoms and molecules

• Chemical Potential energy is the energy stored within the bonds of chemical substances

Energy is the capacity to do work

• Nuclear energy is the energy stored within the collection of neutrons and protons in the atom

• Electrical energy is the energy associated with the flow of electrons

Ek = ½ mv2

kinetic energy depends on BOTH mass and speed.

Compare two chemical systems.

O2(g) at same speed as H2(g)

More Ek because of greater mass of molecules

Less Ek because mass of molecules is less.

Example 2 : compare

O2(g) with H2(g) at same EK

Average speed of molecules is slower

Average speed of molecules is faster

Ek = ½ mv2

kinetic energy depends on BOTH mass and speed.

equal velocity – unequal mass

Contributions to the kinetic energy:

• The molecule moving through space Ek(translation)

• The molecule rotating Ek(rotation)

• The bound atoms vibrating Ek(vibration)

• The electrons moving within each atom Ek(electron)

Contributions to the potential energy:

• Forces between the bound atoms vibrating,

Ep(vibration)

• Forces between nucleus and electrons and between electrons in each atom, Ep(atom)

• Forces between the protons and neutrons in each nucleus, Ep(nuclei)

• Forces between nuclei and shared electron pair in each bond, Ep(bond)

forces of attraction that exist between molecules. That is: the force between one molecule of water and another molecule of water. These are the forces that must be overcome when a phase change occurs. We must break these forces when we melt a solid or evaporate a liquid.

forces within a molecule that hold it together. That is: covalent bonds. These must be overcome if we take a molecule apart.

Thermochemistry is the study of heat change in chemical reactions.

mass & energyExchange:

closed

energy

isolated

nothing

SYSTEMSURROUNDINGS

Thermochemical Definitions

System: That part of the Universe whose change we are going to measure.

Surroundings: Every thing else that is relevant to the change is defined as the “surroundings”.

Heat (Q): Is the energy transferred between a system and it’s surroundings as result in the differences in their temperatures only!

Heat is the transfer of thermal energy between two bodies that are at different temperatures.

Energy Changes in Chemical Reactions

Temperature is a measure of the thermal energy.

Temperature = Thermal Energy

900C400C

greater thermal energy

Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings.

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

H2O (g) H2O (l) + energy

Thermite reaction: 2 Al + Fe2O3 → Al2O3 + 2 Fe

Endothermic process is any process in which heat has to be supplied to the system from the surroundings.

energy + 2HgO (s) 2Hg (l) + O2 (g)

energy + H2O (s) H2O (l)

Thermodynamics

State functions are properties that are determined by the state of the system, regardless of how that condition was achieved.

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.

energy, pressure, volume, temperature

First Law of Thermodynamics

( Law of Conservation of Energy )

“The Total Energy of the Universe is Constant”

ΔE of Universe = Δ ESystem + Δ ESurroundings = 0

The specific heat capacity (c) of a substance is the amount of heat (Q) required to raise the temperature of one gram of the substance by one degree Celsius.

Q = m c t

Q = quantity of heat (J)

m = mass (g)

c = specific heat capacity (J/g oC or kJ/kg oC or J/g K)

∆t = temperature change (oC or K)

How much heat is given off when an 869 g iron bar cools from 940C to 50C?

Q = m c Δt

= 3.5 x 104 J

Q = (869 g) x (0.449 J/g • 0C) x (890C)

IRON BAR COOLS => LOSES HEAT

If 36.5 kJ of heat is required to raise the temperature of 1.50 kg of a metal from 30.0 oC to 55.0 oC , find the specific heat capacity of the metal.

36.5 kJ=

(1.50 kg)(25.0oC)

= 0.973 kJ/kgoC

∆t =Q

=5.40 kJ

(0.150 kg)(4.19 kJ/kgoC)

= 8.59 oC

tf = 15.0 oC + 8.59 oC

= 23.6 oC

Calculate the final temperature for a 150 mL sample of water at 15.0 0 C if it absorbs 5.40 kJ of heat energy.

∆t =Q

=100.5 J

(50 g)(0.897 J/goC)

= 2.24 oC

100.5 J of energy is added to four 50.0 g samples of the following metals at room temperature: aluminium, gold, silver and sodium. Arrange the metals from lowest to highest final temperature.

The thermal capacity (C, unit → J/oC) of a substance is the amount of heat (Q) required to raise the temperature of a given quantity of the substance by one degree Celsius.

Q = C Δt

A calorimeter is known to absorb 5.48 KJ of energy for every degree increase in temperature. If the temperature of the calorimeter increases from 10.0oC to 78.0oC, how much energy is absorbed?

Q = C Δt

Q = (5.48 KJ/oC)(78oC – 10oC)

Q = 373 KJ

The heat of fusion of a substance is the amount of energy required to change the state of the substance from solid to liquid.

Calculate the amount of energy needed to melt 50.0 g of H2O(s) at 0.00 oC.

ΔH = nΔHfus

50.0 g

18.02 g/molx 6.01 kJ/mol

= 16.7 kJ

The heat of vaporization of a substance is the amount of energy required to change the state of the substance from liquid to gas.

ΔH = n ΔHvap

Calculate the amount of energy needed to evaporate 200 g of H2O(l) at 100.0 oC.

Answer: 451 kJ

Energy in J

liquid

melting point

boiling point

H2O(s) H2O(l) H2O(g)

Q = (msolid)(csolid)(T)

Q = (mliquid)(cliquid)(T)

Q = (mgas)(cgas)(T)

PHASE TRANSITIONS

ΔH = (nsolid)(ΔHfusion)

ΔH = (nliquid)(ΔHvap)

Energy in J

PHASE TRANSITIONS

Energy is increasing

Energy in J

PHASE TRANSITIONS

Energy is decreasing

A. The flat lines on a heating curve represent

1) a temperature change

2) a constant temperature

3) a change of state

B. The sloped lines on a heating curve represent

1) a temperature change

2) a constant temperature

3) a change of state

A. Water condenses at a temperature of

1) 0°C 2) 50°C 3) 100°C

B. At a temperature of 0°C, water

1) freezes 2) melts 3) changes to a gas

C. When a gas condenses, heat is

1) released 2) absorbed

D. Freezing is

1) endothermic 2) exothermic

Is energy absorbed (1) or released (2) in each of the following:

____A. Ice to liquid water

____B. Water vapor to rain

____C. Water to ice

When it rains, the air becomes

1) warmer 2) cooler 3) does not change

Complete using the terms gains or loses

In the cooling coils of a refrigerator, liquid Freon

___________ heat from the food and changes to a

Food ___________heat and becomes colder

In the back of the refrigerator, Freon _________

heat and condenses back to a liquid

losesloses

To reduce a fever, an infant is packed in 1250 g of ice. If the ice at 0.00 °C melts and warms to body temperature (37.0°C) how much heat in joules is absorbed?

Step 1: Diagram the change of state

Step 2: Calculate the heat to melt ice (fusion)

ΔH = n Δ Hfus

= (1250 g /18.02 g/mol) x 6.01 KJ/mol

= 416.897 KJ

Step 3: Calculate the heat from 0°C to 37°C

Q = m c ΔT

= 1250 g x 37.0°C x 4.19 J / g °C = 193 325 J

Qtotal = 416 897 J + 193,325 J = 6.10 x 105 J

The measurement of heat in a reaction.

m c ΔT = m c ΔT

Qlost = Qgained

For processes occurring at constant pressure the change in energy equals the heat gained or lost.

A 25.64g sample of a solid was heated to 100.000C and added to a calorimeter containing 50.00g of water. The water temperature increased from 25.100C to 28.490C. What is the specific heat capacity of the solid?

Mass(g) c(J/g*ºC) Tinitial Tfinal T

25.64 ? 100.00 28.49 71.51

50.00 4.184 25.10 28.49 3.39

csolid x25.64g x 71.510C = 4.19J/g oC x50.00g x 3.390C

csolid = 4.19J/g ºC x50.00g x 3.39ºC

25.64g x 71.51ºC= 0.387J/g oC

m c ΔT = m c ΔT

Qlost = Qgained

How much cold water at 15.0oC is needed to cool a 300 kg bath from 70.0 oC to a final temperature of 45.0oC?

mc∆t = mcΔt (300 kg)(4.19 kJ/kgoC)(25.0oC) = (m)(4.19 kJ/kgoC)(30.0oC)

m = 250 kg

Qlost = Qgained

A calorimeter system is known to absorb 15.2 kJ for each degree rise in temperature. 250.0 g of a metal is heated to 450.0 oC and placed in the calorimeter. It causes the temperature to rise from 22.0 oC to 28.0 oC. Calculate the specific heat capacity of the metal. ( C = 15.2kJ )

mc∆t = C∆t

(.250 kg)(c)(422 oC) = (15.2 kJ/oC)(6.0 oC)

c = 0.864 kJ/kg oC

Qlost = Qgained

Thermal Capacity ( C ) of a calorimeter is defined as the Calorimeter value for each change in degree celcius whichIncludes the water and the calorimeter ( absorbing heat energy)

90.0 g of a substance at 70.0 oC is placed into 150 mL of water at 20.0oC in an 80.0 g aluminum calorimeter. If the final temperature is 32.0oC, calculate the specific heat capacity of this substance.

mc∆t =

mc∆t + mc∆t

H2O(l) + CALORIMETERSUBSTANCE ??

Qlost = Qgained

(90.0 g)( c )(38oC) = (150g)(4.19 J/g oC)(12 oC)

+ (80.0g)(0.897 J/goC)(12oC)

c = 2.46 J/g oC

hot cooltf

90.0 oC tf 20.0 oC

Δt = (90 – tf) Δt = (tf – 20)

(copper) (H2O + cal.)

mcΔt = mcΔt + mcΔt

(.06kg)(.385kJ/kgoC)(90-tf) = (.12kg)(4.19kJ/kgoC)(tf-20) + (.075kg)(.897kJ/kgoC)(tf-20)(.0231)(90-tf) = (.5028)(tf-20) + (.067275)(tf-20)

60.0 g of copper at 90.0 oC is placed into 120.0 g of H2O(l) at 20.0 oC in a 75.0 g aluminum calorimeter. Calculate the final temperature.

Qlost = Qgained

2.079 .0231tf = .5028tf –10.056 +.067275tf –1.3455

.593175tf = 13.4805

tf = 22.7oC That was tough!But not as tough as

80.0g of steam at 120 oC is bubbled through 800.0g of H2O(l) at 40.0 oC. Calculate the final temperature of the water.

tfTemp.

time (s)

(steam) (water)

mc∆t + nΔHvap + mcΔt = mc∆t

Qlost = Qgained

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. Cannot measure enthalpy.

H = change in enthalpy; heat given off or absorbed during a reaction at

constant pressure

If the heat transfer involves a chemical reaction then Q is called the heat of reaction. (ΔH)

It is the energy required to return a system to the given temperature at the completion of the reaction.

Q = ΔH UNITS: kJ

• Melting (fusion) ΔHfus

• Evaporating Δ Hvap

• Combustion Δ Hcom

• Formation Δ Hf

• Decomposition Δ Hdec

• Solvation Δ Hsol

• Neutralization Δ HN

• Reaction Δ Hrxn

• Condensation Δ Hcon

• Freezing Δ Hfr

A molar heat can be defined as the amount of energy released or absorbed when one mole of a substance is put through a defined process.

UNITS: kJ/mol

If the heat transfer involves a specified amount of substance burning in oxygen then q is called the heat of combustion. (ΔHcom)

Q = ΔH = n Δ Hcom

Molar heat of combustion

• How much energy is released by the burning of one mole of methane?

• How much energy is released by the burning of two moles of methane?

• How much energy is released by the burning of 0.12 mol of methane?

-890.5 KJ/mol

2 x -890.5 KJ/mol

0.12 x -890.5 KJ/mol

-1781 KJ

-106.86 KJ

• How many moles of octane must there have been if 6987 KJ of energy was absorbed by a calorimeter?

• One mole of octane releases 5470.1 KJ.

• 6987 KJ x 1 mol =

5470.1 KJ 1.277 mol

In an experiment 4.98 g of methane is combusted in a bomb calorimeter. The temperature of 3000 g of water increases from 15.2 oC to 74.2 oC. What is the molar heat of combustion of methane?

Qlost = Qgained

n ΔHc = m c ΔT

(4.98g/16.05g/mol) Hc = (3000g)(4.19 J/goC)(59.0oC)

Δ Hc = (3000g) (4.19 J/goC) (59.0oC)

(0.31028mol)

Δ Hcom = 2.39 x 103 KJ/mol

(burning of fuel) (H2O + cal.)

n Δ HC = mc∆t + mc∆t

5.00g26.04 g/mol = (1.5 kg)(4.19kJ/kgoC)(37.9oC)

+ (.40kg)(.897kJ/kgoC)(37.9oC)

Δ Hcom

Hc =1.31 x 103 kJ/mol

5.00 g of C2H2 (g) is burned in a 400.0 g aluminum calorimeter which contains 1500 g of H2O(l) at 20.0 oC. If the final temperature is 57.9 oC, calculate the molar heat of combustion for acetylene?

Qlost = Qgained

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) + 890.5 KJ

CH4(g) + 2 O2(g)→ CO2(g) + 2 H2O(l) ΔH=-890.5 KJ

Endothermic or Exothermic ?

N2(g) + O2(g) + 182.6 kJ → 2NO(g)

N2(g) + O2(g) → 2NO(g) ΔH= +182.6 kJ

Endothermic or Exothermic ?

Endothermic reactions

Reaction coordinate

reactants

products500

Exothermic reactions

Reaction coordinate

reactants

products

(ΔH)600

ΔH = ENTHALPY OF PRODUCTS – ENTHALPY OF REACTANTS

For the reactions above:

Endothermic

ΔH = 500 kJ – 400 kJ

= + 100 kJ

Exothermic

ΔH = 400 kJ – 600 kJ

= - 200 kJ

• Energy is absorbed

• Energy is on reactant side

• ΔH is positive (+)

• Product molecules have greater total enthalpy

• Reactant bonds are stronger

• Environment cools down

• Energy is released

• Energy is on product side

• ΔH is negative (-)

• Reactant molecules have greater total enthalpy

• Product bonds are stronger

• Environment warms up

H2O (s) H2O (l) H = +6.01 kJ

• The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

• If you reverse a reaction, the sign of H changes

H2O (l) H2O (s) H = -6.01 kJ

• If you multiply both sides of the equation by a factor n, then H must change by the same factor n.

2H2O (s) 2H2O (l) H = 2 x +6.01 = +12.02 kJ

H2O (s) H2O (l) H = 6.01 kJ

• The physical states of all reactants and products must be specified in thermochemical equations.

Thermochemical Equations

H2O (l) H2O (g) H = 44.0 kJ

How much heat is evolved when 266 g of white phosphorus (P4) burn in air?

P4 (s) + 5O2 (g) P4O10 (s) H = -3013 kJ

266 g P4

1 mol P4

123.9 g P4

x3013 kJ1 mol P4

x = - 6470 kJ

the heat change that results when one mole of a compound is formed from its elements (in most stable form) at a pressure of 1 atm.

Standard enthalpy of formation

ΔHf0 standard

Sample Problem Writing Formation Equations

PROBLEM: Write balanced equations for the formation of 1 mol of the following compounds from their elements in their standard states and include H0

SOLUTION:

(a) Silver chloride, a solid at standard conditions.

(b) Calcium carbonate, a solid at standard conditions.

Use the table of heats of formation for values.

H0f = -127.0kJ(a) Ag(s) + 1/2Cl2(g) AgCl(s)

H0f = -1207.6kJ(b) Ca(s) + C(s) + 3/2O2(g) CaCO3(s)

Write formation reactions for each of the following:

1. HClO4(l)

2. N2O4(g)

3. HCOOH(l)

4. KClO3 (s)

5. NO2(g)

6. AgBr(s)

½ H2(g) + ½ Cl2(g) + 2 O2(g) → HClO4(l) + 40.6 kJ

1 N2(g) + 2 O2(g) + 11.1 kJ → N2O4(g)

1 H2(g) +1 C(s) + 1 O2(g) → HCOOH(l) + 425.0 kJ

1 K(s) + ½ Cl2(g) + 3/2 O2(g) → 1 KClO3(s) + 397.7 kJ

½ N2(g) + 1 O2(g) + 33.2 kJ → 1 NO2(g)

1 Ag(s) + ½ Br2(l) → 1 AgBr (s) + 100.4 kJ

Hess’s Law of Heat Summation

The enthalpy change of an overall process is the sum of the enthalpy changes of its individual steps.

Example: Problem: Calculate the energy involved in the oxidation of elemental sulfur to sulfur trioxide from reactions:

1) S (s) + O2 (g) SO2 (g) Δ H1 = -296.0 kJ

2) 2 SO2 (g) + O2 (g) 2 SO3 (g) Δ H2 = -198.2 kJ

3) S (s) + 3/2 O2 (g) SO3 (g) Δ H3 =

1/2 -99.1 KJ

-395.1 KJ

If you add two or more equations to get a new equation, you must add the ΔH’s to get the ΔH for the new equation.

Sample Problem Using Hess’s Law to Calculate an Unknown H

SOLUTION:

PROBLEM: Two gaseous pollutants that form auto exhaust are CO and NO. An environmental chemist is studying ways to convert them to less harmful gases through the following equation:

CO(g) + NO(g) CO2(g) + 1/2N2(g) H = ?

Given the following information, calculate the unknown H:

Equation A: CO(g) + 1/2O2(g) CO2(g) H = -283.0kJ

Equation B: N2(g) + O2(g) 2NO(g) H = +180.6kJ

Equations A and B have to be manipulated by reversal and/or multiplication by factors in order to sum to the first, or target, equation.

Divide Equation B by 2 and reverse it.

H = -90.3kJ

CO(g) + 1/2O2(g) CO2(g) H = -283.0kJ

NO(g) 1/2N2(g) + 1/2O2(g)

H = -373.6kJCO(g) + NO(g) CO2(g) + 1/2N2(g)

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

C(s) + O2(g) → CO2(g) H = -393.5 kJ

S(s) + O2(g) → SO2(g) H = -296.1 kJ

CS2(l) + 3 O2(g) → CO2(g) + 2 SO2(g) H = -1072.0 kJ

1. Write the enthalpy of formation reaction for CS2

C(s) + 2 S(s) → CS2 (l)

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

C(s) + O2(g) → CO2(g) H0 = -393.5 kJ

2 S(s) + 2 O2(g) → 2 SO2(g) H0 = -296.1x2 kJ

CO2(g) + 2SO2(g) → CS2(l) + 3 O2(g) H0 = +1072.0 kJ+

C(s) + 2 S(s) → CS2(l) ΔH0 = +86.3 kJ

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atm.

aA + bB cC + dD

Ho dHo (D)fcHo (C)f= [ + ] - bHo (B)faHo (A)f[ + ]

Ho nHo Pf= nHo Rf-

Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

●Establish an arbitrary scale with the standard enthalpy of formation (Hf

o) of a element as a reference point for all enthalpy expressions.

●The standard enthalpy of formation of any element in its most stable form is zero.

Ho (O2) = 0f

Ho (O3) = 142 kJ/molf

Ho (C, graphite) = 0f

Ho (C, diamond) = 1.90 kJ/molf

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol.

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0 nH0 (products)f= nH0 (reactants)f-

H0 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0 = [ (12 x –393.5) + (6 x –285.8) ] – [ 2x49.04 ] = -6534.88 kJ

-6535kJ2 mol

= - 3267.0 kJ/mol C6H6

A sample of benzene burns and releases -6542 kJ of energy. What is the heat of formation of Benzene (C6H6). The reaction produces carbon dioxide and liquid water.

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0 nH0 (products)f= nH0 (reactants)f-

-6542 = 6H0 (H2O)f12H0 (CO2)f[ + ] - 2H0 (C6H6)f[ ]

-6542 = [ (12 x –393.5) + (6 x –285.8) ] – [ 2ΔH0f ]

-6542 + 6436.8 = -2ΔHof

-105.2 = -2 ΔHof

ΔHof = +52.6 KJ/mol

In any chemical reaction

REACTANTS → PRODUCTS

Energy must be absorbed to break the chemical bonds of the reactant molecules.

Energy must be released when the chemical bonds are formed in the building of the product molecules.

C3H8 (g) + O2 (g) → CO2 (g) + H2O (g)1 3 45

Represent the molecules with structural formulae

O = C = O

O = C = O +

O = C = O

H – C – H

H – C – H +

H – C – H

H – O – H

Bond Type Energy (kJ/mol)

C - C 348

C - H 415

O = O 499

C = O 742

O - H 465

• Stability is determined by how much energy is needed to take the compound apart.– Look up the heats of decomposition, for one

– the higher the heat of decomposition the more stable the compound

• calcium sulphate ΔHd = 1434.5 KJ

• copper (II) sulphide ΔHd = 53.1 KJ

• benzene ΔHd = -49.1 KJ

• nitrogen monoxide ΔHd = -91.3 KJ

Which compound is the most stable?

Energy Relationships in Chemical Reactions

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