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Flashcards in Chemistry 12: Chemical Equilibrium Deck (29):

Physical Equilibria

Equilibrium that does not involve changes to the chemical properties of the substances involved. Must be in a closed system!

Examples: Liquid-Vapour and Solution Equilibrium


Liquid-Vapor Equilibrium (for all volatile/evaporating liquids)

In a closed container, the rate of evaporation equals the rate of condensation

H20(l) H20 (g)

*Rates are same, does not mean that the number of molecules in each state is the same


Solute-solution Equilibrium (for saturated solutions)

In a closed system, the rate of dissolving is equal to the rate of crystallization.

NaCl(s) Na+(aq) + Cl-(aq)

Saturated solution: contains solid crystals and dissolved ions in the solution


What determines the equilibrium point in a physical equilibria?

1. Nature of Reactants
a. Type of liquid (liquid-vapour equilibria) ex: water vs. nail polish remover (faster)
b. Type of solid (solute-solution equilibria)

2. Temperature
a. Liquid-vapour equilibria: increased temperature causes more particles to be in the gas phase
b. Solute-solution equilibria: increased temperature causes more particles to be in the dissolved phase


The Haber Process

N2(g) + 3H2(g) 2NH3(g) + 92 kJ

Purpose: to obtain ammonia, which can easily and quickly be converted to nitrates

To get the highest yield:
• High pressure (favours product side of reaction)
• Low temperature favours exothermic reaction
• An intermediate temperature high enough to allow the reaction to proceed at a reasonable rate, yet not so high as to drive the reaction in the reverse direction, is required.
• Catalyst (speed up reaction)
• Remove NH3 as it forms (lowering product concentration)
- Add reactants (increasing reactant concentration)


Concentration vs Time Graph of Chemical Equilibrium

T = 0
High concentration of reactants, 0 concentration of product. The rate of reaction is at the maximum rate for the forward reaction because of the maximum concentrations. There is no reverse reaction rate at the beginning (product must form first then shift equilibrium position).

As reaction proceeds: reactant concentrations decrease and slope of line becomes less steep; product concentration increases and slope becomes less steep
The forward reaction will start to decrease because the concentration of reactants decrease (being used up) as the products are being formed. As this happens, increase in products result in more collisions, leading the decomposition of products back into its reactants through the reverse reaction.

The rates of the reverse reaction will steadily increase as the concentration of product is increasing.

Eventually, the forward reaction rate becomes equal to the reverse reaction rate. (Dynamic equilibrium)
--> both reactant and concentration lines are horizontal (different concentrations, but constant)


Example of chemical equilibrium (concentration vs time graph)

Nitrogen and hydrogen (in a closed container) react
• Initially, the rate of the forward reaction is fast because of the high concentrations of nitrogen and hydrogen, which produces ammonia at a fast rate.
• As the reaction proceeds and the reactants are used up, the reaction rate decreases
• Eventually the ammonia molecules reach a point where they start to decompose back into its reactants
• After a while, the forward and reverse reactions will reach the same rate = equilibrium
^concentrations are constant


Chemical Equilibrium

Chemical equilibrium: the state of a reaction in which all reactants and products have reached constant concentrations in a close system (horizontal line, no change in concentration)
• Closed system

At a start of a chemical reaction, there is no equilibrium. Concentration of reactants decreases as concentrations of product increases. An increase in the product concentration leads to more likely collisions, causing a reverse reaction to occur. Then, the rates become equal and establish a new equilibrium
• Equilibrium can be reached starting with a forward and reverse reaction


State Le Chatelier's Principle

Whenever a stress is placed on a system at equilibrium, it will shift to off set the stress applied

(When a chemical system at equilibrium is disturbed by a change in a property, the system adjusts in a way that opposes the change.)

FORWARD: decrease in reactants, increase in products
REVERSE: increase in reactants, decrease in products


Stresses that can be placed on a system at equilibrium

1. CHANGE in concentration→gases, aq, or two liquids
2. CHANGE in temperature/energy
3. CHANGE in Volume or Pressure (gases)
o A change in pressure changes all individual partial pressures
4. Addition of Catalyst (no change)
5. Addition of an Inert Gas (no change)


What happens when stress is placed on a system at equilibrium?

What happens when a chemical reaction undergoes stress:
• Reaction is at a state of dynamic equilibrium
• Stress is placed; no longer at equilibrium
• To re-establish equilibrium, there will be a shift in the equilibrium position to restore equilibrium
• Once dynamic equilibrium is reached again, there is a new equilibrium position and no further change occurs


Stress: Concentration

*only affects substances that are (g) or (aq) or two liquids are present*

Increase concentration of reactants: shift equilibrium FORWARD (produce more product due to increase in number of successful collisions for forward reaction; use up what was added)

Decrease concentration of reactants: shift equilibrium REVERSE (counteract the lower reactant concentration)

Increase concentration of products: shift equilibrium REVERSE (increase in number of successful collisions for the reverse reaction)

Decrease concentration of products: shift equilibrium FORWARD (produce more product to offset stress by replacing the product removed)

*if the amount of a solid is changed, then there is no effect on the equilibrium position


Stress: Temperature

At equilibrium, temperature is constant.
If the system is disturbed by changes in temperature, both the forward and reverse reaction rates will either increase or decrease, however one will be favoured more

If reaction vessel is:
COOLED (decrease in temperature): favour exothermic reaction (shift to produce heat)

HEATED (increase in temperature): favour endothermic reaction (shift to use up the heat and cool down)

Endothermic reactions: heat on reactant side
Increase in temperature: FORWARD (to use heat and cool)
Decrease in temperature: REVERSE (gives off heat and warms)

Exothermic reactions: heat on product side
Increase in temperature: REVERSE (to use heat and cool)
Decrease in temperature: FORWARD (gives off heat and warms)

Endo forward = exo reverse


Stress: Pressure/Volume

*Only gases are affected by pressure/volume changes (count number of GAS MOLECULES)

Pressure and volume are inversely related

Decrease volume --> increases pressure
Increase volume --> decreases pressure

Pressure change affects all the individual partial pressures/concentrations. This will change BOTH the forward and reverse reaction rats, but the reaction will try to offset the change.

Increase in pressure/decrease in volume: want to decrease pressure, therefore favour the reaction that produces FEWER gas particles

Decrease in pressure/increase in volume: want to increase pressure, therefore favour the reaction that produces MORE gas particles

*if have the same number of gas molecules on both sides of the reaction, then changing the pressure will nave NO EFFECT on the equilibrium position. -->amount of all substances remain constant (no way to offset the stress)


Different states in chemical reaction

Concentration & pressure changes: even though changes in certain states (i.e. solids and liquids) do not contribute to shifts, other shifts still have an impact on their amounts. Ex: forward reaction will produce more solid product


Stress: Catalyst

Increases rate of the forward and reverse reactions equally (does not change the overall energy i.e. enthalpy heat is the same for original and catalyzed reaction)
-therefore, the concentrations of all substances remain constant

DO NOT AFFECT the position of equilibrium

o Reaction reaches equilibrium much faster


Stress: Inert Gas/Noble Gas

Inert gases are unreactive

When added to a system, the total pressure changes, but not he partial pressures/concentrations of each gas.

• Although there are more collisions due to more reactant entities, the collisions involving inert gases will not cause a chemical reaction

*Pressure changes stress the equilibrium only when there are changes in EFFECTIVE CONCENTRATIONS AND EFFECTIVE COLLISIONS (since inert gases do not affect number of collisions, no shift)


Tips when working with equilibrium

-highlight all the solids and liquids in the chemical reactions
-place the heat change in the chemical reaction (endo or exo)


Equilibrium Constant (Kc) Expression (mass action)

Equilibrium Constant (Kc) Expression (mass action): description of a chemical system at equilibrium.

Ratio of the concentrations of products to concentrations of reactants AT EQUILIBRIUM

For all chemical reactions at equilibrium:

Kc = [Products]/[Reactants]
Kc = [C]^c[D]^d/[A]^a[B]^b

Where Kc is the equilibrium constant (no units)
A, B, C, D are chemical entities in gaseous or aqueous states
a, b, c, d are coefficients in a balanced chemical reaction

The equilibrium constant REMAINS THE SAME at equilibrium and a specific temperature, regardless of the initial concentrations

• Kc can be used to determine the equilibrium position of a chemical reaction system (toward products or toward reactants)
• Kc = ratio of the rate constants of the forward and reverse reaction


Characteristics of Kc (equilibrium constant) expressions

The coefficient from he balanced equation become a power in the equilibrium constant expression

Solids and liquids are not included (only gases and aq)
-->S and L cannot be expressed as concentrations →their concentrations are constant (kept at 1)

*if two or more liquids are present anywhere in the equation, one can dilute the other, thus changing its concentration by dilution

Ex: (s) + (g) --> (s)
Kc = 1/[g]

(l) + (g) --> (g)
Kc = [g]/[g]

(l) + (g) --> (l) + (g)
Kc = [l][g]/[l][g]


How to calculate Kc at equilibrium from equilibrium values?

1) write out Kc expression
2) substitute in the values/concentration (don't forget exponents)

-Kc at equilibrium given concentration
-->sometimes have to convert to concentration (c = n/v)
*must be in concentration

Solve for concentration of an entity
-->rearrange the Kc expression (don't forget exponents!)


Magnitudes of Kc

Kc>>1 (large value): equilibrium position is towards the right (AT EQUILIBRIUM, the concentration of products is greater than the concentration of reactants) →Products favoured (complete reaction)

Kc = 1: concentration of products = concentration of reactants

Kc reactants favoured (incomplete reaction)


Calculate Kc for systems not at equilibrium (given experimental data)

Must determine the equilibrium concentrations of reactants and products from experimental data -->used to calculate equilibrium constant

1) write out the chemical reaction (if not already given)
*pay attention to the states and energy

2) Balance the reaction

3) set up the I.C.E Table
I = initial concentration
C = change (- for reactants, + for products)
E = equilibrium concentration

4) Fill known information
- initial reaction concentrations (if not stated, assume products have concentration zero initially unless otherwise stated)
-equilibrium concentration --->used to find change
*must be concentration (convert mols to concentration)

5) Use mole ratio (BEZ: balance equation zone) to calculate change (all stoichiometrically related)

6) Determine the final equilibrium concentrations of all species
Reactants = initial - change
Products = initial + change

7) set up the equilibrium constant expression using equilibrium concentration

8) calculate Kc

*Kc is only determined using equilibrium concentrations ONLY


How to differentiate between systems at equilibrium and systems not at equilibrium?

• Look at words in the question
o Injected, initially, mixed and reacted (NO EQUILIBRIUM) →I.C.E

o In/at equilibrium, given Kc →directly at equilibrium


Steps when given a problem relating to equilibrium and the equilibrium constant

*Make note of units and states
1) identify whether the reaction is at or not at equilibrium (look at how the question is worded)

a) not at equilibrium -->set up ICE table (CONCENTRATION)-->signs of change -->input known values (assume product concentration 0 if not stated)
b) at equilibrium --> form Kc expression (CONCENTRATION, do not include solids or liquids)


Types of questions

1. Given initial concentrations of reactants and one equilibrium concentration, find Kc
-ICE table
-Kc expression

2. Given Kc, concentrations of entities at equilibrium, find known concentration at equilibrium
-rearrange Kc expression (don't forget exponents)

3. Given initial moles in a flask (not 1 L) and one equilibrium concentration (or mole), find Kc
-convert moles to Mol/L (C = n/v)
-ICE table-->find equilibrium concentrations
-Kc expression

4. Given Kc, equilibrium concentrations and flask (not 1 L), find number of moles
-rearrange Kc expression to solve for unknown concentration
-convert concentration to moles (n = cv)

5. Given flask volume (not 1 L) and moles of entities AT EQUILIBRIUM, find Kc
-convert moles to Mol/L (C = n/v)
-Kc expression to find Kc

6. Same conditions, different moles of entities at equilibrium, find moles of known
-convert moles to Mol/L (C = n/v)
-rearrange the Kc expression to solve for unknown concentration
-convert concentration into moles

7. questions about if the equilibrium favours the formation of reactants or products
-find Kc
Kc >> 1 = products


Reaction Quotient

Reaction Quotient (Qc): the product of the concentrations of the product, divided by the product of the concentrations of reactants, for a chemical reaction that is not necessarily at equilibrium.
• When the system is NOT at equilibrium and want to determine which way the reaction needs to shift to reach equilibrium
• Quotient: result of a mathematical division of the products over reactants

Reaction quotient = Qc = [C]^c[D]^d/[A]^a[B]^b
*Looks exactly like Kc, BUT concentrations are not necessarily at equilibrium
• Compare Qc with Kc

If Qc is less than Kc (QcKc): not at equilibrium

• Ratio of the concentrations of products to the concentration of reactants is larger than AT EQUILIBRIUM
• Concentrations of products must decrease (numerator decrease) while the concentrations of reactants must increase (smaller value) so that Qc decreases to Kc →shift reverse until Qc=Kc at equilibrium

Qc = Kc --> at equilibrium (no shifting)

Question example: Given equilibrium constant and following concentrations asked to predict the direction the system will shift to reach equilibrium
• Determine the Qc using concentrations (Products/reactants ← like solving for Kc)
• Compare Qc value to Kc value
• Determine shift:
o Qc>Kc →reverse to use up product
o Qc=Kc→no shift
o Qc forward to produce product


How to interpret Concentration vs. Time Graphs

1) determine the reactants and products
R: initially high concentrations
P: initially low concentrations

2) determine when the reaction reaches equilibrium (concentrations of entities are constant/horizontal line)

3) identify the stresses placed on the system at equilibrium
-spike -->concentration change
-gradual change -->Pressure or temperature change
*calculate the Kc of the equilibrium reached after the gradual change and compare it to the previous Kc. If the new Kc is different from the previous Kc, then temperature change
**Kc only changes with temperature
***FOLLOWS SHIFTING (determine F or R reaction)
ex: concentration of reactant A increases -->spike up of concentration A -->products to increase (Forward), reactants to decrease


Summary of the effect of changes in conditions on the position of equilibrium

Increase concentration -->eqm shifts to opposite side
Decrease concentration -->eqm shifts to same side

Increase pressure -->eqm shifts to side with least moles of gas
Decrease pressure -->eqm shifts to side with most moles of gas

Increase temperature -->eqm shifts in endothermic direction
Decrease temperature -->eqm shifts in exothermic direction
*temperature changes Kc

Add a catalyst: no change

Add an inert/noble gas: no change