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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)