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Irreversible ReactionsChemical reactions in which proceed to completion where essentially
all of the reactants are converted into products
Example:
2H2(g) + O2(g) → 2H2O(g)
Such a reaction essentially occurs in one direction, called the “forward” direction as shown by the → symbol
2NO2(g) → N2O4(g)dark brown colorless
After adding pure NO2 to a closed vessel (the syringe), the dark brown color gradually fades but does not disappear completely!
Eventually the color stops changing when it reaches a light brown color!
Clearly the reaction has stopped short of completion but instead has reached a state of equilibrium where the concentration of reactants
and products are constant
Reversible ReactionsIn many chemical reactions there is not a complete conversion of
reactants into products because the reaction not only occurs in the “forward” (→) direction, but also in the “reverse” (←) direction where the products come together to reform the reactants. This is shown by
drawing a symbol between the reactants and products.
Example:
2NO2(g) N2O4(g)
The dark brown color does not completely disappear since as soon as the colorless N2O4 starts to form from the forward (→) reaction the
reverse reaction (←) begins to reform the dark brown NO2
Strictly speaking in a closed system ALL chemical reactions will reach equilibrium
However, in some reactions the products are so favored that the reaction essentially proceeds to completion where the amounts of
reactants at equilibrium is negligible
Example:2H2(g) + O2(g) 2H2O(g)
We say the equilibrium position is shifted far to the right
In other reactions the reactants are so favored that the reaction essentially does not occur so that the amounts of products at
equilibrium is negligible
Example:2CaO(s) 2Ca(s) + O2(g)
Here we say the equilibrium is shifted far to the left
Equilibrium is a Dynamic Condition!
To the eye it appears that a system at equilibrium is static and unchanging since the concentrations of reactants and products are
constant
However, at the molecular level the forward and reverse reactions are still occurring but at the same rate
Analogy:
When traffic is crossing a bridge in opposite directions at a uniform speed, the total number of cars on the bridge at any instant will be
approximately the same even though the traffic is still moving
Attaining EquilibriumWhen a reversible reaction begins and only the reactants are present,
only the forward reaction occurs
However, as the reactants are used up and the products accumulate, the forward reaction rate decreases and the reverse reaction rate
increases
As the reaction proceeds the forward rate decreases and the reverse rate increases. Eventually, the forward and reverse reaction rates
become equal and the system reaches equilibrium
Similarly, the reactant concentrations decrease and the products concentration increase until equilibrium is reached and the concentrations of
reactants and products stabilize
The Law of Mass ActionAn empirical law relating the concentrations of reactants and products at
equilibrium
For the general equilibrium reaction:
aA + bB cC + dD
The Law of Mass Action is represented by the following equilibrium expression:
K = [C]c[D]d / [A]a[B]b
where the square brackets correspond to the concentrations at equilibrium, the superscripts are the coefficients in the balanced reactions and K is a
constant called the equilibrium constant
Calculating Equilibrium ConstantsThe value of the equilibrium constant at a given temperature (K
depends on temperature) can be easily calculated in the equilibrium concentrations of the reactants and products at equilibrium are
known
Equilibrium constant values are normally written without units
Equilibrium Constants and Equilibrium PositionsThe equilibrium constant is unique at a given temperature, regardless
of the initial concentrations of gases that are mixed together
While the equilibrium constant, which depends on the ratio of the concentrations will always be the same at a given temperature, the
actual equilibrium concentrations can vary
Each set of possible equilibrium concentrations is called an equilibrium position
There is only one equilibrium constant at a given temperature but there are an infinite number of possible equilibrium positions
The particular equilibrium position attained by a system depends on the initial concentrations but the equilibrium constant does not
Equilibrium Expressions Involving Gases
Equilibria involving gases can also be described using pressures. This comes from the ideal gas equation:
K is used for an equilibrium constant expressed in terms of concentrations while Kp is used for an equilibrium constant expressed
in terms of pressures
Relationship between K and Kp
Here the temperature is in Kelvin and the Universal Gas Constant R = 0.08206 L.atm.mol-1.K-1
Homogeneous EquilibriaReactants and products are all in the same state, most commonly the
gaseous state:
2O3(g) 3O2(g)
Heterogeneous Equilibrium ConstantsThe concentrations of pure solids and liquids in heterogeneous
equilibria are constant so are not included in the equilibrium constant expression
aA(s) + bB(g) cC(l) + dD(g)
K = [D]d / [B]b
This rule does not apply to solutions since their concentrations canchange
Applications of the Equilibrium Constant
Knowing the equilibrium constant for a reaction we can determine:
1. Whether a set of concentrations represents a system in equilibrium
2. The tendency of a reaction to reach equilibrium (but not the rate it will reach equilibrium)
3. The final equilibrium position that will be attained from a given set of initial concentrations
Interpreting Equilibrium ConstantsThe magnitude of the equilibrium constant can be interpreted as a
measure of the inherent tendency of a reaction to occur
Reactions with large K correspond to equilibria which lie far to the right and therefore essentially go to completion
Reactions with small K correspond to equilibria which lie far to the left and therefore essentially to not occur
Reactions with a K close to 1 correspond to equilibria which contain roughly equal amounts of reactants and products at equilibrium
Equilibria with Large K2SO2(g) + O2(g) 2SO3(g)
K = [SO3]2 / [SO2]2[O2] = 3.4 x 102
Reaction essentially goes to completion
Equilibria with Small KCOCl2(g) CO(g) + Cl2(g)
K = [CO][Cl2] / [COCl2] = 2.2 x 10-10
Little reaction takes place
The size of K and the time to reach equilibrium are not related!
Example:
2H2(g) + O2(g) 2H2O(g)
Hydrogen and oxygen have a strong tendency to form water since water has a lower energy. However, the large activation energy
prevents the reaction from occurring at room temperature
The value of K therefore depends on ∆E while the rate of reaction depends on Ea
Reaction Quotient, Q
It is possible to determine whether a mixture of reactants and products are at equilibrium and if not the direction the equilibrium
will shift to attain it
To do this we calculate the reaction quotient, Q from the Law of Mass Action using the initial concentrations rather than the
equilibrium concentrations:
Calculating Equilibrium Pressures and Concentrations
Equilibrium problems typically involve finding the concentrations at equilibrium given the equilibrium constant and the initial
concentrations
Problems vary in difficultly depending on the information given
Procedure for Solving Equilibrium Problems
1. Write the balanced equation for the reaction2. Write the equilibrium expression from the law of mass action
3. List the initial concentrations4. Calculate Q and determine the direction of the shift to equilibrium
(not necessary if the initial concentration of one or more of the reactants and products is zero)
5. Construct an ICE table containing the Initial concentrations, the Change needed to reach equilibrium and then the Equilibrium
concentrations6. Substitute the equilibrium concentrations into the equilibrium
expression and solve for the unknown (try to simplify if K is small)
7. Check your calculated equilibrium concentrations by making sure they give the correct value of K
Quadratic Equations
Many equilibrium and acid-base problems require solving quadratic equations so make sure you review this!
If:
then: 02 cbxax
aacbbx
242
Suppose for a synthesis of hydrogen monofluoride from hydrogen and fluorine, 3.000 mol hydrogen and 6.000 mol of fluorine are
mixed in a 3.000 L flask. Given that the equilibrium constant for the reaction at this temperature is 1.15 x 102, calculate the equilibrium
concentration of each component
Systems with Small Equilibrium Constants
Systems with small equilibrium constants are easier to deal with because we can make approximations which simplify the calculation
Modifying Equilibrium ConditionsIt is possible to shift the position of an equilibrium by modifying the
reaction conditions. Such changes are called “stresses”
Stresses can be applied by changing the concentrations of one or more of the reactants or products as well as changing the
temperature and pressure of the system
Le Châtelier’s PrincipleWhen a stress is placed on a reaction at equilibrium, the position of
the equilibrium shifts in the direction that relieves the stress
Concentration ChangesThe stress of adding a substance to an equilibrium system is relieved by shifting the position of the equilibrium away from that substance
The stress of removing a substance from an equilibrium system is relieved by shifting the position of the equilibrium towards that
substance
These shifts have to occur in order to keep the equilibrium constant Kthe same (assuming T and P or T and V are constant)
N2(g) + 3H2(g) 2NH3(g)
If we add N2, the equilibrium position shifts to the right forming more NH3. The equilibrium constant, however, remains unchanged
Pressure ChangesThere are three ways to change the pressure of an equilibrium
system containing gaseous components
1. Add or remove a gaseous reactant or product
2. Add an inert gas which is not involved in the reaction
3. Change the volume of the system
#1 has already been considered
#2 increases the total pressure but has no effect on the partial pressures of the reactants or products so there is no shift
Pressure Changes through Changes in VolumeWhen the volume of an equilibrium system is reduced, the system responds by reducing its own volume. This is achieved by shiftingthe equilibrium position towards the side containing the fewer
number of molecules thereby decreasing the total number of gaseous molecules present in the system
When the volume of an equilibrium system is increased, the system responds by increasing its own volume. This is achieved by shiftingthe equilibrium position towards the side containing the greater
number of molecules thereby increasing the total number of gaseous molecules present in the system
N2(g) + 3H2(g) 2NH3(g)
If the volume decreases, the system responds by reducing its own volume by shifting the equilibrium position towards the side containing the fewer
number of molecules (the right)
Changing TemperatureWhen we change the temperature of a system at equilibrium, the
equilibrium constant changes
If the temperature of an equilibrium system is increased, the equilibrium position shifts to remove the heat
If the temperature of an equilibrium system is decreased, the equilibrium position shifts to replace the heat
Endothermic Reactions (∆H > 0)Heat is treated as a reactant:
aA(g) + bB(g) + heat cC(g) + dD(g)
Increase Temperature: Equilibrium shifts towards products
K increases
Decrease Temperature: Equilibrium shifts towards reactants
K decreases
Exothermic Reactions (∆H < 0)Heat is treated as a product:
aA(g) + bB(g) cC(g) + dD(g) + heat
Increase Temperature: Equilibrium shifts towards reactants
K decreases
Decrease Temperature: Equilibrium shifts towards products
K increases
CatalystsSpeed up the rate of both forward and reverse reactions making the
time required to reach equilibrium shorter
However, have no effect on the equilibrium position