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General Chemistry II | Unit II | Part 1 | 1
General Chemistry II, Unit II: Study Guide (part 1) CDS Chapter 21: Reaction Equilibrium in the Gas Phase
• Introduction o Some chemical reactions have a significant amount of reactant left over when equilibrium is reached.
• Observation 1: Reaction equilibrium o N! g + 3H! g → 2NH! g
§ If we start with 1 mole of N2 and 3 moles of H2, the balanced equation predicts 2 moles of NH3 will be produced.
§ Experimentally, only 1.91 moles of NH3 are produced. Remember: there is no limiting reactant involved (since neither reactant is present in less than the stoichiometric ratio established by the equation).
§ Conclusion: this reaction achieves reaction equilibrium in which all three gases are present in the gas mixture.
§ The amount of NH3 produced can be varied by changing: • The temperature of the reaction; • The volume of the reaction container; or • The relative starting amounts of the reactants.
o Some reactions achieve an equilibrium very close to the complete reaction. This means that, at equilibrium, there are virtually no reactants remaining.
o A qualitative model of equilibrium § From phase transitions studies: equilibrium occurs when the rate of the forward process is
matched by the rate of the reverse process § Therefore, in gas reactions, at equilibrium the forward reaction rate is equal to the reverse
reaction rate. § How do we show that the forward and reverse reactions continue to occur at equilibrium?
• Vary the volume. • N2O4 (g) → 2 NO2 (g)
o When the volume is increased, the amount of NO2 at equilibrium is larger and the amount of N2O4 is smaller.
• If the forward and reverse reactions simply stopped when equilibrium was reached, the molecules would have no way of “knowing” of the change in volume, so the amounts of gas would not have changed.
• Change in volume affects both the forward and reverse reaction rates. § If the rate of the forward reaction is faster than the rate of the reverse reaction, there will be
more product than reactant at equilibrium. • Observation 2: Equilibrium constants
o N2O4 (g) → 2 NO2 (g) § In this experiment, we vary the number of moles of N2O4 and measure the equilibrium
pressures of both the reactant and product gases.
General Chemistry II | Unit II | Part 1 | 2 § The amount of NO2 at equilibrium seems to be directly proportional to the starting amount of
N2O4. However, this graph suggests otherwise:
§ The equilibrium pressure of NO2 does not increase proportionally with the initial amount of
N2O4. The slow increase suggests a square root relationship. However, this graph suggests otherwise:
§ The near-‐proportional relationship between the initial amount of N2O4 and the pressure of
N2O4 (from figure 21.1) suggests that we should instead plot the pressure of NO2 against the square root of the pressure of N2O4 (P NO! = 𝑐 P N!O! ).
§ The above equation can be rewritten by solving for c and setting c equal to Kp: K! =! !!! !
! !!!! !.
• The constant Kp (the reaction equilibrium constant) is independent of the initial conditions and the equilibrium partial pressures of both the reactant and product:
General Chemistry II | Unit II | Part 1 | 3 • Something to note: the product (NO2) appears in the numerator and its exponent is its
stoichiometric coefficient in the chemical equation. Similarly, the reactant (N2O4) appears in the denominator and its exponent is its stoichiometric coefficient.
o N! g + 3H! g → 2NH! g
§ Predicted reaction equilibrium constant: K! =!(!!!)!
! !! ∙! !! !
§ Continuing from Observation 1, we will now vary the starting amounts and container volume for this reaction:
§ Kp clearly remains constant for each set of initial conditions. Therefore, the reaction equilibrium
constant can be found using an equation in which the partial pressures of the products (each raised to their corresponding stoichiometric coefficient) are multiplied together in the numerator, and the partial pressure of the reactants (each raised to their corresponding stoichiometric coefficient) are multiplied in the denominator.
• Dynamic Equilibrium and Reaction Rates o The rate law can be predicted from the coefficients in the reaction equation.
# Step Explanation
1 Rate!"#$%#& = k!"#$%#& N!O! Rate!"#"!$" = k!"#"!$" NO! !
Since the forward and reverse reactions are elementary processes, these are their rate laws.
2 k! N!O! = k! NO! ! In dynamic equilibrium, the rates are equal to each other.
3 k!k!=
NO! !
N!O!
A simple rearrangement of the previous expression. This looks similar to the Kp expression, except it uses gas concentrations instead of partial pressures.
4 NO! =n!!!V
=P!!!RT
N!O! =n!!!!V
=P!!!!RT
Apply the Ideal Gas Law to make the connection between gas concentrations and pressures.
5 k!k!=
NO! !
N!O!=P!!!!
P!!!!=
1RT
Substitution of the Ideal Gas Law relationships above into equation 3.
6 K! = RT
k!k!=P!!!!
P!!!!
Multiplying by RT yields the same expression as Kp.
• Observation 3: Temperature Dependence of the Reaction Equilibrium o H2 (g) + I2 (g) -‐> 2 HI (g)
§ Here, we measure the equilibrium partial pressures at a variety of temperatures.
General Chemistry II | Unit II | Part 1 | 4 § Since the equilibrium constant increases with temperature, the pressure must also increase
dramatically with temperature. § The temperature and pressure measurements show no simple relationship between
temperature and the equilibrium constant Kp. From thermodynamics, the equilibrium constant
varies with temperature according to lnK! =!∆!°!"
= ∆!°!.
§ Since ΔH° and ΔS° don’t strongly depend on temperature, a linear relationship appears between lnKp and 1/T:
• The negative slope means that Kp increases with temperature. This is because -‐ ΔH°/R,
the slope, must be negative. § For exothermic reactions, the slope should be positive and the equilibrium constant should
decrease with increasing T. • Observation 4: Changes in Equilibrium and Le Chatelier’s Principle
o Ideally, our goal is to be able to control the equilibrium of a gas reaction – that would allow us to force a reaction to produce as much or as little reaction as we want.
o Discoveries so far: § The equilibrium pressure of the product of a reaction increases with increasing quantity of
reactant. § The equilibrium pressure of the product of a reaction varies with the volume of the reaction
container. However, this variation is not easily predictable or explainable. § In order to explain this, we can rewrite the equilibrium constant to show the volume of the
container (for N! g + 3H! g → 2NH! g ):
K! =n NH! ! RT
V!
n N!RTV ∙ n H! ! RT
V! =
n NH! !
n N! ∙ n H! ! RTV
!
K!RTV
!=
n NH! !
n N! ∙ n H! !
• When the volume increases, the left side of the equation decreases – therefore, the right side must also decrease, so the denominator (N2 and H2) must increase while the numerator (NH3) decreases. This shifts the equilibrium from products to reactants.
o Changes in temperature § Kp increases with T for endothermic reactions
• The products are increasingly favored. § Kp decreases with T for exothermic reactions.
·····
= ��
= ��
� ��
General Chemistry II | Unit II | Part 1 | 5 • The reactants are increasingly favored. • In exothermic reactions, the reverse reaction is endothermic.
§ The equilibrium shifts in the direction of the endothermic reaction when the temperature increases.
o Le Chatelier’s Principle – when a reaction at equilibrium is stressed by a change in conditions, the equilibrium will be reestablished in such a way as to counter the stress.
o Changes in volume § Increasing the volume reduces the partial pressures of all the gases present and thus reduces
the total pressure. § The reaction responds to the stress of a volume increase by offsetting the pressure decrease
with an increase in the number of moles of gas at equilibrium.
McMurry & Fay 13.1–13.11
Equilibrium constant Kc (13.2) The number obtained by multiplying the equilibrium concentrations of all the products and dividing by the product of the equilibrium concentrations of all the reactants, with the concentration of each substance raised to the power of its coefficient in the balanced chemical equation. For the reaction 𝑎A + 𝑏B ⇌ 𝑐C + 𝑑D:
𝐾! =C ! D !
A ! B !
Heterogeneous equilibria (13.4)
Homogeneous equilibria have all reactants in products in the same phase, usually gaseous or solution. Conversely, heterogeneous equilibria are those in which the reactants and products are present in more than one phase. The concentrations of pure solids and liquids are not included when writing an equilibrium equation. The equilibrium equation for the reaction CaCO! s ⇌ CaO s +CO! g is simply 𝐾! = CO! since CaCO3 and CaO are both solids.
Judging the Extent of Reaction (13.5)
The numerical value of the equilibrium constant for a reaction indicates the extent to which reactants are converted to products.
Predicting the Direction of Reaction (Reaction Quotient) (13.5)
The reaction quotient is the same as the equilibrium constant Kc, except that the concentration values it uses are not necessarily equilibrium values.
• If Qc < Kc, net reaction goes from left to right (reactants to products). • If Qc > Kc, net reaction goes from right to left (products to reactants). • If Qc = Kc, no net reaction occurs.
Altering an Equilibrium Mixture: Changes in Concentration (13.7)
• The concentration stress of an added reactant or product is relieved by net reaction in the direction that consumes the added substance.
• The concentration stress of a removed reactant or product is relieved by net reaction in the direction that replenishes the removed substance.
General Chemistry II | Unit II | Part 1 | 6
Altering an Equilibrium Mixture: Changes in Pressure and Volume (13.8)
• An increase in pressure by reducing the volume will bring about net reaction in the direction that decreases the number of moles of gas.
• A decrease in pressure by expanding the volume will bring about net reaction in the direction that increases the number of moles of gas.
These observations are due to the Ideal Gas Law: the pressure of an ideal gas is inversely proportional to the volume at constant temperature and quantity.
Catalysts and equilibrium (13.10)
• If a reaction mixture is at equilibrium in the absence of a catalyst, it will still be at equilibrium after a catalyst is added because the forward and reverse rates, though faster, remain equal.
• If a reaction mixture is not at equilibrium, a catalyst accelerates the rate at which equilibrium is reached, but it does not affect the composition of the equilibrium mixture.
• Catalysts do not appear in the equilibrium constant expression.
Link between chemical equilibrium and chemical kinetics (13.11)
Step Explanation
A + B ⇌ C + D Consider this reaction.
Rate forward = 𝑘! A B Rate reverse = 𝑘! C D
The forward and reverse reactions are elementary reactions with these rate laws.
𝑘! A B = 𝑘! C D at equilibrium The forward and reverse rates are equal at equilibrium.
𝑘!𝑘!=
C DA B
Rearrange.
𝐾! =C DA B
The right side of the equation is the same as the equilibrium constant expression for the forward reaction, which equals the equilibrium constant Kc since the reaction mixture is at equilibrium.
𝐾! =𝑘!𝑘!
The equilibrium constant is simply the ratio of the rate constant for the forward and reverse reactions.
CDS Chapter 16: Phase Transitions and Phase Equilibrium • Introduction
o Liquids and solids don’t follow the Kinetic Molecular Theory of Gases § The densities of these condensed phases are thousands of times greater than the typical density
of a gas. § Unlike the Kinetic Molecular Theory, molecules in liquid and solid phases must interact with
each other. • Observation 1: Gas-‐Liquid Phase Transitions
o Recall Charles’ Law: the volume of a fixed sample of gas is proportional to the absolute temperature of the gas, provided that the pressure is constant.
General Chemistry II | Unit II | Part 1 | 7 § The typical Charles’ Law experiment traps gas in a cylinder (with a piston to maintain a constant
pressure) and varies the temperature. With each new temperature, the piston moves to establish a new volume.
o If we keep lowering the temperature of a sample of butane gas, the linear relationship between volume and temperature abruptly stops at 272.6 K – the volume drops to only 0.097 L:
§ This change in physical properties at one temperature is a phase transition. At 272.6 K, butane
converted from gas to liquid. The temperature of this transition is the boiling point. § The boiling point does not depend on how much liquid or gas there is in a sample.
o If we increase the applied pressure, the phase transition occurs at a higher temperature (for butane, 20 K higher). The temperature of the phase transition depends on the applied pressure.
o The boiling point temperature also depends on substance identity. Different substances have vastly different boiling points.
• Observation 2: Vapor pressure of a liquid o Obviously, liquids left in an open container will eventually evaporate even if the temperature of the
liquid is well below its boiling point. o The tendency of a liquid to evaporate is known as its volatility – a more volatile liquid evaporates more
readily. § A quantitative measure of volatility can be found by modifying the previous cylinder by adding a
gauge to measure the pressure. The cylinder is filled with water only; the piston is then pulled back, creating an empty space above the liquid water. We would assume that the pressure of this space is zero.
§ Instead, the pressure rises to a constant 23.8 torr. Therefore, there must be gaseous water in the container. Since there was no gas initially, this gas must have come from evaporation of the liquid water.
§ Since not all of the water is evaporated at equilibrium, both the liquid and gas phases are present at the same time. They are in phase equilibrium
o The pressure of the vapor above the liquid rises to 23.8 torr regardless of the volume or initial amount of liquid.
§ However, the amount of water that evaporates must differ when the volume or initial amount is varied. This is because the volume available for the vapor to occupy changes with these modifications – in order to maintain the same pressure (23.8 torr), the quantity of water vapor must change.
General Chemistry II | Unit II | Part 1 | 8 § This reveals that the pressure of the vapor is the most important property in establishing
phase equilibrium. § The vapor pressure of a liquid is the single specific pressure at which it will be in phase
equilibrium with its vapor. o Vapor pressure varies with temperature – and the relationship is not proportional.
§ The vapor pressure also depends on the identity of the substance.
• Observation 3: Dynamic Equilibrium between Liquid and Gas Phases o Why do we always get the same pressure at equilibrium for the same temperature, regardless of other
conditions? o Since the vapor exerts the same pressure in a larger volume (for constant T), the Ideal Gas Law states
that there are more molecules in the vapor after the volume is increased and equilibrium is reestablished.
§ More liquid must have evaporated for this to happen. § The only way that the liquid could have “known” to evaporate when the volume increased is if
the molecules in the liquid were always evaporating – even at equilibrium. There must be a constant movement of molecules from liquid to gas.
§ Since the pressure of the vapor remains constant when volume is fixed, condensation must also always be occurring. This constant pressure at a fixed volume shows that the number of molecules going from liquid to gas must exactly offset the number of molecules going from gas to liquid.
§ This constant movement of molecules between the phases is known as dynamic equilibrium. o What factors are important in dynamic equilibrium?
§ When the volume is increased, the vapor pressure remains constant and the quantity of gas increases.
§ For this to be possible, evaporation initially occurs more rapidly than condensation until equilibrium is achieved.
§ Mechanics of evaporation and condensation • Molecules leave the gas phase and enter the liquid phase by striking the surface of the
liquid. Therefore, the rate of condensation depends on the frequency of molecules striking the liquid surface.
• This strike frequency decreases when the volume is increased because the density of gas molecules decreases (think KMT).
General Chemistry II | Unit II | Part 1 | 9 • The rate of condensation becomes lower than the rate of evaporation, so there is a net
flow of molecules from liquid to gas. This continues until equilibrium is reestablished, at which point the rates of evaporation and condensation are once again equal.
§ The vapor pressure doesn’t depend on the surface area of the liquid since the rate of evaporation and the rate of condensation increase with increasing surface area.
o What determines the rate of evaporation? The increase in vapor pressure with increasing temperature is not solely due to the relationship between P and T in the Ideal Gas Law.
§ While the Ideal Gas Law relationship between P and T is proportional, the relationship between vapor pressure and temperature is much larger than proportional.
§ Since the vapor pressure rises with temperature, there must be a greater quantity of gas in the vapor phase at higher temperature. Therefore, the rate of condensation must be higher.
§ However, at equilibrium the rates of condensation and evaporation must be equal. Therefore, the rate of evaporation is also higher at higher temperatures.
§ Since the rate of evaporation depends on the identity of the liquid is, vapor pressures differ for each liquid.
o A model for the relationship between temperature and evaporation § Temperature is a measure of the kinetic energy of molecules in both gases and liquids. This is
shown by the fact that a greater quantity of molecules is able to escape the liquid at higher temperatures (they are able to overcome the liquid’s intermolecular forces).
§ What causes different substances to have different rates of evaporation? • Substances with lower vapor pressures have lower rates of evaporation and
condensation – this means the substance has fewer molecules with high enough KE to escape the liquid at a given temperature.
§ Dynamic equilibrium • At a given temperature, only a fraction of the liquid molecules have enough KE to
evaporate – this fixes the rate of evaporation • The rate of condensation must match the rate of evaporation at equilibrium – this is
only possible at one specific pressure. • Therefore, at a given temperature, only a single pressure will result in phase
equilibrium for that substance.
McMurry & Fay 10.1–10.4
Polar covalent bonds and dipole moments (10.1)
Polar covalent bonds form between atoms of different electronegativity. These bonds are characterized by an unequal sharing of the bonding electrons between the two atoms. These bonds have positive and negative ends, known as dipoles. Molecules as a whole can also be polar because of the net sum of the individual bond polarities and lone-‐pair contributions.
Intermolecular forces (10.2) Water (H2O) has clear intramolecular forces between each of the two hydrogen atoms and the oxygen atom. However, H2O exists as a gas, liquid or solid depending on its temperature – therefore, there are also intermolecular forces that act between molecules to hold them together in these phases.
Ion-‐dipole forces (10.2) An ion-‐dipole force is the result of electrical interactions between an ion and the partial charges on a polar molecule.
General Chemistry II | Unit II | Part 1 | 10
Dipole-‐dipole forces (10.2) Dipole-‐dipole forces occur between neutral but polar molecules as the result of electrical interactions among dipoles on neighboring molecules. These forces are generally weak (3–4 kJ/mol). The strength of a dipole-‐dipole interaction depends on the sizes of the dipole moments involved.
London Dispersion Forces (10.2) Intermolecular forces occur between nonpolar molecules and among the individual atoms of a noble gas. All atoms and molecules, regardless of structure, experience London Dispersion Forces which result from the motion of electrons. While the average distribution of electrons throughout a symmetrical molecule is symmetrical, at any given moment there may be more electrons at one end of the molecule than the other. This instantaneous dipole can affect the electron distributions in neighboring molecules and induce temporary dipoles. LDFs are generally small (1–10 kJ/mol).
Hydrogen bonds (10.2) A hydrogen bond is an attractive interaction between a hydrogen atom bonded to a very electronegative atom (O, N or F) and an unshared electron pair on another electronegative atom. Hydrogen bonds are especially strong because O-‐H, N-‐H and F-‐H bonds are highly polar and hydrogen atoms lack core electrons to shield their nuclei. Hydrogen bonds are strong (40 kJ/mol).
General Chemistry II | Unit II | Part 1 | 11
A Comparison of Intermolecular Forces (10.2)
Viscosity (10.3) Viscosity is the measure of a liquid’s resistance to flow. It is realted to the ease with which individual molecules move around in the liquid (which is, in turn, related to the IMFs). Substances with small, nonpolar molecules expeirence weak IMFs and have low viscosities. Larger, more polar substances experience stronger IMFs and have higher viscosities.
Surface tension (10.3) Surface tension is the resistance of a liquid to spread out and increase its surface area. It is caused by the difference in IMFs experienced by molecules at the surface of a liquid and those experienced by molecules in the interior. Molecules at the surface feel attractive forces on only one side and are thus pulled in toward the liquid – molecules in the interior are surrounded and pulled equally in all direcitons. Surface tension is higher in liquids that have stronger IMFs. Surface tension and viscosity are also temperature-‐dependent – molecules at higher temperatures have more KE to counteract the attractive forces holding them together.
Phase changes (10.4)
Heating curve:
General Chemistry II | Unit II | Part 1 | 12
The heat of fusion (ΔHfusion) describes the amount of energy required to convert a solid into a liquid. The heat of vaporization (ΔHvap) describes the amount of energy required to convert a liquid into a gas.