Chemistry 1AChapter 6

Some Chemical Changes Release Energy

Combustion of Methane

CH4(g) + 2O2(g)→ CO2(g) + 2H2O(l) +

Some Chemical Changes Absorb Energy

Energy Terms• Energy = the capacity to do work• Work, in this context, may be defined as what is

done to move an object against some sort of resistance.

Two Types of Energy

• Kinetic Energy = the energy of motion= 1/2 mμ2

• Potential Energy = energy by virtue of position or state

Law of Conservation of Energy

Endergonic Change

more stable + energy → less stable systemlesser capacity + energy → greater capacity

to do work to do worklower PE + energy → higher PE

coin in hand + energy → coin in air above hand

Coin and Potential Energy

Bond Breaking and Potential Energy

Exergonic Change

less stable system → more stable + energy

greater capacity → lesser capacity + energyto do work to do work

higher PE → lower PE + energy

coin in air above hand → coin on ground + energy

Bond Making and Potential Energy

Which higher energy? Is it kinetic or potential?

• 428 m/s Ar atoms or 456 m/s Ar atoms?• 428 m/s Ar atoms or 428 m/s Kr atoms?• Na+ close to Cl− or Na+ and Cl− far

apart?• ROOR or 2 RO• H(g) and O2(g) or HO2(g)• Solid CO2 or gaseous CO2

Units of Energy

• Joule (J) =

• 4.184 J = 1 cal• 4.184 kJ = 1 kcal• 4184 J = 1 Cal (dietary calorie)• 4.184 kJ = 1 Cal

2

2

kg ms

Approximate Energy of Various Events

More Terms

• External Kinetic Energy = Kinetic energy associated with the overall movement of a body

• Internal Kinetic Energy = Kinetic energy associated with the random motion of the particles within a body

External and Internal Kinetic Energy

Heat

• Heat = Energy transfer from a region of higher temperature to a region of lower temperature due to collisions of particles.

Heat Transfer

Radiant Energy• Radiant Energy is electromagnetic

energy that behaves like a stream of particles.

• It has a dual Nature– Particle

• photons = tiny packets of radiant energy• 1017 photons/second from a flashlight bulb

– Wave• oscillating electric and magnetic fields• describes effect on space, not true nature

of radiant energy

A Light Wave’s Electric and Magnetic Fields

Radiant Energy Spectrum

Endergonic Change

more stable + energy → less stable systemlesser capacity + energy → greater capacity

to do work to do worklower PE + energy → higher PE

Exergonic Change

less stable system → more stable + energy

greater capacity → lesser capacity + energyto do work to do work

higher PE → lower PE + energy

Bond Breaking and Potential Energy

Bond Making and Potential Energy

Exergonic (Exothermic) Reaction

weaker bonds → stronger bonds + energyless stable → more stable + energy

higher PE → lower PE + energy

Exothermic Reaction

Endothermic Reaction

stronger bonds + energy → weaker bonds

more stable + energy → less stablelower PE + energy → higher PE

NH4NO3(s) + energy → NH4+(aq) + NO3

−(aq)

Energy and Chemical Reactions

Factors that Affect Heats of Chemical Reactions

• Nature of the reaction, including the states of the reactants and products

• Amount of reactant

• Conditions for the reaction– Constant pressure or constant volume– Temperature and pressure

2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l)ΔH° = −3.08 x 103 kJ

Conversion Factors from ΔH°

2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l)ΔH° = −3.08 x 103 kJ

Relationship Between ΔH and the Chemical Equation

• Whatever you do to the equation, do the same to the ΔH. – Reverse the equation…change the

sign of the ΔH. – Multiply the coefficients by some

number…multiply the ΔH by the same number.

Gas

Gas Model

• Gases are composed of tiny, widely-spaced particles. – For a typical gas, the average

distance between particles is about ten times their diameter.

Gas Model (cont.)• Because of the large distance between the

particles, the volume occupied by the particles themselves is negligible (approximately zero).

– For a typical gas at room temperature and pressure, the gas particles themselves occupy about 0.1% of the total volume. The other 99.9% of the total volume is empty space. This is very different than for a liquid for which about 70% of the volume is occupied by particles.

Gas Model (cont.)

• The particles have rapid and continuous motion.– For example, the average velocity of a

helium atom, He, at room temperature is over 1000 m/s (or over 2000 mi/hr). The average velocity of the more massive nitrogen molecules, N2, at room temperature is about 500 m/s.

– Increased temperature means increased average velocity of the particles.

• The particles are constantly colliding with the walls of the container and with each other. – Because of these collisions, the gas

particles are constantly changing their direction of motion and their velocity. In a typical situation, a gas particle moves a very short distance between collisions. Oxygen, O2, molecules at normal temperatures and pressures move an average of 10−7 m between collisions.

Gas Model (cont.)

• There is no net loss of energy in the collisions. A collision between two particles may lead to each particle changing its velocity and thus its energy, but the increase in energy by one particle is balanced by an equal decrease in energy by the other particle.

Gas Model (cont.)

• The particles are assumed to be point-masses, that is, particles that have a mass but occupy no volume.

• There are no attractive or repulsive forces at all between the particles.

Ideal Gas

Gas Properties and their Units

• Pressure (P) = Force/Area– units

• 1 atm = 101.325 kPa = 760 mmHg = 760 torr• 1 bar = 100 kPa = 0.9869 atm = 750.1 mmHg

• Volume (V)– unit usually liters (L)

• Temperature (T)– ? K = --- °C + 273.15

• Number of gas particles expressed in moles (n)

Decreased Volume Leads to Increased Pressure

P α 1/V if n and T are constant

Relationship between P and V

Boyle’s Law• The pressure of an ideal gas is inversely

proportional to the volume it occupies if the moles of gas and the temperature are constant.

Increased Temperature Leads to Increased Pressure

P α T if n and V are constant

Relationship between P and T

Gay-Lussac’s Law

• The pressure of an ideal gas is directly proportional to the Kelvin temperature of the gas if the volume and moles of gas are constant.

Increased Moles of Gas Leads to Increased Pressure

P α n if T and V are constant

Relationship between n and P

Relationship Between Moles of Gas and Pressure

• If the temperature and the volume of an ideal gas are held constant, the moles of gas in a container and the gas pressure are directly proportional.

Ideal Gas Equation

Constant Pressure or Volume

• Heat at Constant Pressure = ΔH = change in enthalpy

• Heat at Constant Volume = ΔE= change in the total energy

E = KE + PE

Standard Heat Changes

• ΔH at 298.15 K and 1 atm = ΔH°

• ΔE at 298.15 K and 1 atm = ΔE°

Standard Changes in Enthalpy (ΔH°) and Total Internal Energy

(ΔE°)

2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l)ΔH° = −3.08 x 103 kJΔE° = −3.09 x 103 kJ

ΔH° = Heat at constant pressure at 25 °C and 1 atm

ΔE° = Heat at constant volume at 25 °C and 1 atm

Heat at Constant Pressure and Volume(Example 1)

Sign Conventions• Heat evolved is negative.• Heat absorbed is positive.• Work done by the system

leads to heat lost by the system, so it is negative.

• Work done on the systemadds heat to the system, so it is positive.

Heat at Constant Pressure and Volume(Example 2)

Heat at Constant Pressure and Volume(Example 3)

Goal: to Convert from ΔE° to ΔH°

• ΔE° is easier to determine accurately in the laboratory.

• Because most chemical reactions are run at constant pressure, ΔH° values are usually used to describe heats of reaction.

Conversion from ΔE to ΔH

• We saw for the KClO3 and NH3reactions, that

qV = qrxn

qP = qrxn + qwork

• SoqP = qV + qwork

ΔH = ΔE + w

• For chemical changes at constant pressure and temperature, the only work is work done by a gas as it expands (negative) or on a gas as it is compressed (positive).

Relationship between Heat at Constant and Constant Pressure

ΔH = ΔE + wΔH = ΔE + FΔd

ΔH = ΔE + P•ΔV

Relationship between Heat at Constant and Constant Pressure (2)

ΔH = ΔE + PΔVPV = nRT (ideal gas equation)– If pressure and temperature are kept

constant, the only way to increase the volume is to increase moles of gas.

PΔV = (Δn)RT

ΔH = ΔE + (Δn)RTΔn = ∑ moles (g) products − ∑ moles (g) reactants

Δn = ∑ coef. (g) products − ∑ coef. (g) reactants

Heat at Constant Pressure Conventions

• ΔH° describes kJ per mole of primary reactant (or kJ per the number of moles equal to the coefficient in the balanced equation for each reactant and product).

• qP describes kJ per amount of substance in a change.

Heat at Constant Volume Conventions

• ΔE° describes kJ per mole of primary reactant (or kJ per the number of moles equal to the coefficient in the balanced equation for each reactant and product).

• qV describes kJ per amount of substance in a change.

Ways to get ΔH°• From Tables (ΔH° for

combustion or ΔH° of formation)

• From calorimeter data (bomb or open)

• From other ΔH°s– General Law of Hess– Heat of formation problem

Bomb Calorimeter

Exercise

Derivation of Calorimeter Equation

qlost = −qgained

qrxn = −[qproducts + qcalorimeter + qwater + qsurr]qrxn ≅ −[qcalorimeter + qwater]qrxn = −[qcal + qw]

Heat Capacity and Specific Heat (Capacity)

• Heat capacity, C = the heat energy (kJ) necessary to raise the temperature of an object (such as a calorimeter) by 1 ºC (or 1K).– Units of kJ/ºC or kJ/K…it’s the same number

for each unit• Specific heat (capacity), c = the heat

energy (kJ) necessary to raise the temperature of 1 g of a substance (such as water) by 1 ºC (or 1K).– Units of kJ/g•ºC or kJ/g•K…it’s the same

number for each unit

Bomb Calorimeter Problems – Calculating ΔH° - Part 1• Given – mass of reactant, mass of H2O,

Tinitial, Tfinal, and Ccal

• Steps– Assign variables– Calculate qV

ΔT = T2 – T1 Watch signs.

Bomb Calorimeter Problems – Calculating ΔH° - Part 2

– Calculate ΔE°

– Calculate ΔH°

Bomb Calorimeter Problems –Calculating Ccal - Part 1

• Given – mass of reactant, mass of H2O, Tinitial, Tfinal, and ΔH°

• Steps– Assign variables– Calculate ΔE°

Calorimeter Problems –Calculating CCal - Part 2– Calculate qV

– Calculate Ccal

Law of Hess

• If a reaction can be viewed as a sum of two or more equations, the ΔH° for the net reaction is equal to the sum of the ΔH°’s for the intermediate reactions.

Law of Hess Example

General Law of Hess Problems

• ΔH°net = Σ ΔH°intermediate

• Write intermediate equations and their ΔH°’s.

• Rearrange intermediate equations so they add to yield the desired net equation.

• Whatever you did to the intermediate equations, so the same to their ΔH°’s.

• Add the new intermediate ΔH°’s.

Heat of Formation

• ΔH°f = Heat at constant pressure for the formation of one mole of substance from its elements in their standard states (at 298.15 K and 1 atm).

Standard States of Elements

• Metals – Hg(l) or Symbol(s) – e.g. Zn(s)

• Noble gases – Symbol(g) – e.g. Ne(g)

• Diatomic elements – H2(g), N2 (g), O2(g), F2(g), Cl2(g), Br2(l), I2(s)

• Carbon – C(graphite)• Other elements – S8(s), Se8(s),

P4(s), As4(s), Sb4(s)

Heat of Formation Equation

• Chemical reactions could take place in two steps.– Conversion from reactants to elements in their standard states.

ΔH°1 = - ΣΔH°f (reactants)– Conversion of elements in their standard states to products.

ΔH°2 = ΣΔH°f (products)

• The overall ΔH°rxn would be equal to the sum of the ΔH°s for the two steps.

ΔH°rxn = ΔH°1 + ΔH°2 = ΔH°2 + ΔH°1

ΔH°rxn = ΣΔH°f (products) + (−ΣΔH°f (reactants))

ΔH°rxn = ΣΔH°f (products) − ΣΔH°f (reactants)