CHEMISTRY 3202
UNIT 2: KINETICS AND EQUILIBRIUM
OVERVIEW OF UNIT
● Chemical reactions occur at different rates under different conditions. We group all the concepts related to the rates of chemical reactions under the heading kinetics.
● Some chemical reactions are reversible, that is, reactants become products while products become reactants again. In some systems, the opposing reactions occur at the same time. When the opposing rates of change are equal, a chemical system is said to be in equilibrium.
OVERVIEW OF UNIT
We will:● explain and predict how changes in reaction
conditions affect chemical systems● See how conditions such as temperature and
pressure can be manipulated to favor the formation of products
● Identify the major factors that affect reaction rates
OVERVIEW OF UNIT
● Explore examples of systems at chemical equilibrium and make predictions about how equilibria are affected by changes in reaction conditions
● Explore quantitatively chemical equilibrium
Throughout the unit, we will encounter examples of reactions and equilibria in nature, living systems, and commercial/industrial settings.
DID YOU KNOW
An example of this in real life would be:● Chemical engineers design equipment and
chemical processes that maximize product formation in a cost efficient manner
Part 1: Reaction Kinetics (Chp. 12)
● Reaction Kinetics is the study of the rate of a chemical reaction
Qualitative: ● Reactions may be described as being FAST
or SLOW ● Fast – burning, explosions, precipitation ● Slow – rusting, fermentation
Reaction Kinetics
Quantitative:
The rate of a reaction measures how fast products are formed or how fast reactants are consumed
RATE = CHANGE IN QUANTITY
CHANGE IN TIME
POSSIBLE UNITS??
Measuring Reaction Rate
● The method used to determine reaction rate will depend on the reaction being studied. (p. 466)
Methods:
1. Monitor pH if there is an acid or base in the equation
2. Record gas volume or changes in pressure if there is a gas in the reaction
Measuring Reaction Rate
Methods: (cont’d)
3. record changes in mass if solids are present
4. monitor absorption of light if there is a color change
5. changes in electrical conductivity indicate changes in ion concentration
MC: What could we use to measure the rate of this reaction?
Cu(s) + 2 AgNO3(aq) 2 Ag(s) + Cu(NO3)2(aq)
a) pressure c) gas volume
b) pH d) mass
ANSWER: d) because a solid is present
MC: What could we use to measure the rate of this reaction?
SO3(g) + H2O(l) H2SO4(aq)
a) pressure c) gas volume
b) pH d) mass
ANSWERS:
a) & c) because a gas is present
b) because an acid is being produced
ANSWER:
c) because the pH starts off as fairly acidic (low pH) and the ph gradually becomes less acidic/more basic (pH increases). c) is the only equation that starts off with an acid and as the equation progresses, the acid is slowly used up. Thus the pH increases.
What determines RATE??
● All chemical reactions are bond breaking/bond forming events
● The rate of a reaction depends on how quickly bonds are broken and how rapidly new bonds form.
● Kinetic Molecular Theory (KMT) and Collision Theory are used to explain reaction rates.
Kinetic Molecular Theory (KMT)
● Matter is made of particles (atoms, ions, or molecules) in continuous motion
● An increase in temperature: increases the speed of particles reduces the forces of attraction between
particles
Kinetic Molecular Theory (KMT)
● KMT is supported by:
Diffusion – particles of a gas spread to fill their container (‘perfume in a room’) solids dissolve uniformly in liquids over
time.
Pressure – a balloon remains inflated because gas particles are continuously hitting the sides of the balloon
MC: Which observation best supports the Kinetic Molecular Theory?
(A) Acetic acid odour is detected from across the room.
(B) Liquid water freezes at 0°C under standard conditions.
(C) Nitrogen dioxide gas is dark brown in colour.
(D) When burned, butane produces more heat per mole than propane.
ANSWER:
(A) because is the smell is detected from across the room that means that the gas particles of the acetic acid vapor spread to fill out their container (the room). Thus we smell it from across the room.
Collision Theory
● states that reacting particles must collide with one another in order for a reaction to occur
However:● Particles must collide with proper orientation● The collisions must have enough intensity to
break old bonds and allow new bonds to form
Collision Theory
● to increase reaction rate you must increase the number of successful collisions between reactant molecules
● http://phet.colorado.edu/en/simulation/reactions-and-rates
Reaction Rates Simulation
Correct Orientation
Why is the correct orientation correct?● Essentially atoms must come in such a way
such that positive charge comes together with negative charge (opposites attract) and in such a way where there is enough room for it to happen.
Correct Orientation
Sufficient Collision Intensity
● Basically, unless particles collide with each other with enough intensity, then bonds will not break and thus reform. Therefore no reaction will occur, even if the orientation of the collision was correct.
Sufficient Collision Intensity
Paper, plastic and many other materials burn in air (oxygen gas). Oxygen molecules in the air are in a state of constant motion (at 20°C they move at about 1650 km/h) and that they are likely colliding with your notebook right now.
● So why isn't your notebook on fire (i.e. reacting with oxygen in the air)?
Sufficient Collision Intensity
● Basically, the collisions between the carbon rich cellulose fibers in your notebook and oxygen gas in the air are not intense enough to rearrange the chemical bonds in these substances.
MC: Which must occur for a chemical reaction to take place?
(A) addition of a catalyst
(B) addition of energy
(C) collisions between reacting particles
(D) formation of a reaction intermediate
ANSWER:
C) because this is what the Collision Theory states and if there are no collisions, then there are no bond breaking and bond reforming moments. So no reaction would happen
Factors Affecting Reaction Rate
1. Concentration (only for (aq) or (g) reactants)
– an increase in the concentration of a reactant usually increases the rate of a chemical reaction
- the rate increases because there are:
- more particles resulting in
- more collisions between particles &
- more successful collisions.
Factors affecting reaction rates
2. Temperature
- an increase in the temperature increases the rate of a chemical reaction
- the higher temperature results in:
- faster moving particles
- more collisions between particles
- more intense collisions
NOTE: A temperature increase of 10 ºC usually causes reaction rate to double.
More successfulcollisions
FasterRate
Factors Affecting Reaction Rate
3. Nature of Reactants – compounds with fewer bonds to break will react more rapidly than compounds with many bonds
eg. propane (C3H8) burns faster than candlewax (C25H52) because it has fewer bonds
Factors affecting reaction ratesPropane Candlewax
Factors Affecting Reaction Rate
3. Nature of Reactants
- compounds with weak bonds react more rapidly than compounds with strong bonds
– ions will react more rapidly than atoms and molecules
Factors Affecting Reaction Rate
4. Surface Area
- crushing a solid to produce a powder, or changing a substance to the gas phase, exposes more particles for collision
-if more particles are available for collision there will be:
- more collisions
- more successful collisions
FASTER RATE
Factors affecting reaction rates
For example:
If we did an experiment that involved dropping an alka seltzer tablet, pieces of an alka seltzer tablet and a crushed alka seltzer tablet into their own glass of water at room temperature, we should find that the crushed tablet would react noticeably faster than the other two trials. The reason is because there was more surface area to react with the water when the tablet was crushed.
Which reaction occurs faster at room temperature?
(A) 2H2(g) + O
2(g) → 2H
2O(liq)
(B) 2Ag+(aq) + CrO4
2-(aq) → Ag2CrO
4(aq)
(c) Pb(s) + 2HCl(aq) → PbCl2(s) + H
2(g)
(D) CH4(g) + 2O
2(g) → CO
2(g) + 2H
2O(g)
ANSWER:
(B) because ions are generally the quickest types of species to react since they don't have bonds to break. They just come together to form compounds.
MC: Which factor explains why coal dust is
explosive?
(A) concentration
(B) pressure
(C) surface area
(D) temperature
ANSWER:
(c) because coal dust has far more surface area susceptible to a reaction that would lead to spontaneous combustion than chunks of coal
http://en.wikipedia.org/wiki/Coal_dust
PRACTICE
p. 484; #’s 1 & 2
p. 486; #’s 1,2, 4, 6, & 7
KINETICS & EQUILIBRIUM SHEET #1
Factors affecting reaction rates
5. Catalysts
- a catalyst increases the reaction rate by providing a different reaction pathway or mechanism with a lower activation energy
- a catalyst IS NOT consumed by a chemical reaction.
Potential Energy Diagrams (p. 473)
● PE diagrams show changes in potential energy (stored chemical energy) during chemical reactions
● Exothermic reactions release more energy than they absorb (eg. burning)
● Endothermic reactions absorb more energy than they release (eg. photosynthesis)
Potential Energy Diagrams
● ΔH represents the heat of reaction or enthalpy of reaction
● ΔH is the difference between the PE of the reactants and the PE of the products
● The minimum energy needed for a chemical reaction to occur is the activation energy
Potential Energy Diagrams
● the activated complex for a reaction is a temporary, unstable, intermediate species that quickly decomposes to products
eg. H2 + I2 → H2I2 → 2 HI
ACTIVATEDCOMPLEX
Endothermic PE Diagram
Exothermic PE Diagram
ΔH & equations ● The energy term may be included in a chemical
equation● eg. CO(g) + 2 H2(g) → CH3OH(g) + 65 kJ
OR written as ΔH to the right of the equation.
eg. CO(g) + 2 H2(g) → CH3OH(g) ΔH = - 65 kJ
Energy is Produced/Released
EXOTHERMIC
ΔH & equations
Another eg.
eg. N2(g) + O2(g) + 90 kJ → 2 NO(g)
OR N2(g) + O2(g) → 2 NO(g) ΔH = + 90 kJ
Energy is REQUIRED ENDOTHERMIC
Formula: (OPTIONAL)
Eaforward - Eareverse = ΔH
This formula is NOT necessary if you prefer using the PE diagram.
Ea (forward Ea (reverse)
ΔH Endo or Exo
25 -30
50 20
150 250
65 28
Sketch a PE diagram for each reaction
06:52 PM
Reaction Progress
PE
∆HEafwd
CO2 + H2O
C6H12O6 + O2
Photosynthesis
Earev
06:52 PM
Reaction Progress
PE
∆H
Eafwd
CO2 + H2O
C6H12O6 + O2
Respiration
Earev
06:52 PM
p. 474
∆H
06:52 PM
∆H
06:52 PM
∆H
p. 475
06:52 PM
Reaction Progress
PE
no catalyst
catalyzed
Effect of a catalyst Endothermic
06:52 PM
Reaction Progress
PE
no catalystEXOTHERMIC
06:52 PM
● Sample problem: p. 475● Questions:
p. 476; #’s 2, 4
p. 484; #’s 1 – 4
06:52 PM
Reaction Mechanisms (pp. 477 – 485)
reaction mechanism – the steps that occur in a chemical reactionelementary reaction - each step in a reaction mechanismreaction intermediate – a molecule, atom or ion formed in one step and consumed in a later step NOTE: reaction intermediates are NOT included in the overall equation
06:52 PM
Reaction Mechanisms
eg. #1
Step #1 NO(g) + O2(g) NO3(g)
Step #2 NO3(g) + NO(g) 2 NO2(g)
Overall
Equation:2 NO(g) + O2(g) 2 NO2(g)
06:52 PM
HBr + O2 → HOOBr fast
HOOBr + HBr → 2 HOBr slow
2 HOBr + 2 HBr → 2 H2O + 2 Br2 fast
p. 478 #’s 5 – 8
06:52 PM
Reaction Mechanisms
rate-determining step (RDS)
- the slowest step in a reaction mechanism
- to increase the rate of a reaction you must speed up the RDS
- increasing the concentration of a reactant will increase the rate ONLY IF the reactant is in the RDS
06:52 PM
Reaction Mechanisms
PE diagrams
- every step in a reaction mechanism has an activation energy which can be drawn on a PE diagram
06:52 PM
Reaction Progress
PE
Reaction Mechanisms
3-step mechanism
#1
#2
#3RDS ??
69
06:52 PM
Reaction Mechanisms
eg:
Step #1 H2CO2 + H+ H2CO2H+ fast
Step#2 H2CO2H+ HCO+ + H2O slow
Step #3 HCO+ CO + H+ fast
06:52 PM
Reaction Mechanisms
eg:
Overall H2CO2 H2O + CO
Omit H+ - catalyst
Omit H2CO2H+ & HCO+ - reaction intermediates
06:52 PM
Reaction Mechanisms
Reaction Progress
PE
H2CO2 + H+CO + H+
p. 829 # 128
Step 1 H2(g) + NO(g) → H2O(g) + N(g)
Step 2
Step 3 H2(g) + O(g) → H2O(g)
2H2(g) + 2NO(g) → N2(g) + 2H2O(g)
Part 2: Chemical EquilibriumEquilibriumA balancing Act!
Text Ch 13: p 488 - 541
Part 2: Chemical Equilibrium
● All reactions we have done have shown reactants being converted 100% to products
● Many reactions are reversible with some products being converted back to reactants
Part 2: Chemical Equilibrium
● Dynamic equilibrium occurs when 2 opposing processes occur at the same rate
● A chemical equilibrium occurs when two opposing chemical reactions occur at equal rates.
Types of Equilibria
1. Phase EquilibriaAn equilibrium may be established between different phases of a compound in a sealed container
eg. H2O(l) in a sealed container
Types of Equilibria
Initially: H2O(l) changes to H2O(g)
H2O(l) → H2O(g)
Gradually: H2O(g) changes to H2O(l)
H2O(l) ← H2O(g)
Types of Equilibria
Using equilibrium notation:
H2O(l) ⇌H2O(g)
Types of Equilibria
2. Solubility Equilibria occur in saturated solutions when NaCl(s) is placed in water, the initial
rate of dissolving is fast
NaCl(s) NaCl(aq)
Types of Equilibria
as more solid dissolves, the rate of dissolving slows and recrystallization begins.
eg. NaCl(s) NaCl(aq)
Types of Equilibria
when the solution is saturated there are NO VISIBLE CHANGES
At equilibrium, the RATE of dissolving and the RATE of recrystallization are EQUAL.
eg. NaCl(s) NaCl(aq)
equilibrium
Types of Equilibria
3. Chemical Equilibrium Chemical reactions that are reversible usually
result in chemical equilibrium
eg. NO2 gas changing to N2O4
Types of Equilibria
Initially the forward rate is high
eg. 2 NO2(g) N2O4(g)
as more product forms, the reverse reaction begins and increases in rate.
eg. 2 NO2(g) N2O4(g)
Types of Equilibria
eventually the forward rate slows and the reverse rate increases such that the FORWARD AND REVERSE RATES ARE EQUAL
eg. 2 NO2(g) N2O4(g)
http://www.chm.davidson.edu/ronutt/che115/EquKin/EquKin.htm
Conditions for Equilibrium (p. 492)
1. Macroscopic properties are constant
ie. NO OBSERVABLE CHANGE
2. Forward and reverse rates must be equal
3. A CLOSED SYSTEM is required for equilibrium
Conditions for Equilibrium
4. Equilibrium may occur from either directioneg. 2 NO2(g) N2O4(g)
OR N2O4(g) 2 NO2(g)
Kinetics & Equilibrium #4
Shifts in Equilibrium
Equilibrium occurs when the forward rate equals the reverse rate.
Changes in concentration, temperature and pressure/volume can cause the forward or reverse reaction rate to change.
Shifts in Equilibrium
● Eventually, a new equilibrium will be established with different reactant and product concentrations
● Le Châtelier’s Principle (LCP) is used to predict changes in concentrations when a stress (change in C, T, P/V) is applied to a system at equilibrium.
Le Châtelier’s Principle (p.520)
When a stress is applied to a system at equilibrium, the system will adjust or shift to relieve the stress.
– A change(stress) is applied to a system at equilibrium
– Forward or reverse rate will change– New equilibrium established
Le Châtelier’s Principle
1. Changes in concentration
An increase in concentration on one side of an equation favors or drives the reaction to the opposite side.
eg. What will happen if CO is added to this system at equilibrium?
CO(g) + 2 H2(g) --> CH3OH(g)
LCP- the system will ‘shift’ to the right to use the
CO and produce more CH3OH
- some H2 will be used
[CH3OH] will increase
[H2] will decrease
[CO] ‘spikes’ and then drops to a value higher than it was before the change
?? possible graph ??
forward rate increases;reverse rate catches up
mol/L
time
CO
CH3OH
H2
Le Châtelier’s Principle
An equilibrium shifts away from a substance that increases in
concentration
OR
toward a substance that decreases in concentration.
Le Châtelier’s Principle
decrease [CO] → equilibrium shifts to the left
increase [H2] → equilibrium shifts to the right
decrease [CH3OH] → equilibrium shifts to the right
CO(g) + 2 H2(g) → CH3OH(g)
What happens if we:
Le Châtelier’s Principle
IMPORTANT NOTE:
- adding a solid does NOT change molar concentration
- changing the amount of a solid in an equilibrium will NOT cause a shift
Le Châtelier’s Principle
eg.
CaCO3(s) → CaO(s) + CO2(g)
add CaCO3(s) → NO SHIFT
add CO2(g) → Shifts to the left
What happens if we:
Le Châtelier’s Principle 2. Temperature
- Raising the temperature of an exothermic equilibrium favors the formation of reactants.
eg. CO(g) + 2 H2(g) →CH3OH(g) + 65 kJ
What happens if we increase the temperature in this equilibrium?
Le Châtelier’s Principle- Raising the temperature of an endothermic equilibrium favors formation of products.
eg. CaCO3(s) + heat → CaO(s) + CO2(g)
What happens if we increase temperature in this equilibrium?
Le Châtelier’s Principleeg. How would an increase in temperature
affect these equilibria?
N2(g) + 3 H2(g) --> 2 NH3(g) + heat
2 SO3(g) → 2 SO2(g) + O2(g) ΔH = +197 kJ
2 SO3(g) + 197 kJ → 2 SO2(g) + O2(g)
**A change in temperature changes Keq**
Le Châtelier’s PrincipleOR
- Raising the temperature shifts the equilibrium away from the energy term.
- Decreasing the temperature shifts the equilibrium toward the energy term.
CO(g) + 2 H2(g) → CH3OH(g) + 65 kJ
CaCO3(s) + heat → CaO(s) + CO2(g)
Le Châtelier’s Principle
3. Pressure/Volume An increase in pressure of a system at
equilibrium has the same effect as a decrease in the volume of the system.
(inverse relationship)
Le Châtelier’s Principle
Increasing the pressure of a system at equilibrium by reducing volume causes the equilibrium to shift in the direction that reduces pressure
ie. shift to the side with fewer molecules of GAS!!
Note: When the equilibrium is changed in this manner, only GAS atoms/molecules are
considered.
Le Châtelier’s Principle
eg. How would an increase in pressure - caused by a decrease in volume - affect these equilibria?
N2(g) + 3 H2(g) --> 2 NH3(g) (Shift right)
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) (Shift left)
Le Châtelier’s Principle
eg. Why would a change in pressure NOT affect the following equilibria?
H2(g) + I2(g) 2 HI(g)
● (equal # of gas atoms/molecules on both sides)
2 Ag(s) + Zn2+(aq) 2 Ag+
(aq) + Zn(s)
●No gas species present
Le Châtelier’s Principle
4. Catalyst- Does NOT cause a shift in equilibrium- Increases BOTH rates equally so
equilibrium is reached faster
5. Surface Area - same as a catalyst
Le Châtelier’s Principle
p. 533 #’s 1 – 3
Answers on p. 537
WORKSHEET #5
Equilibrium Constant (Keq)
For any system at equilibrium, there is a mathematical relationship between reactant and product concentrations
Equilibrium Constant (K)
See p. 495 eg. N2O4(g) --> 2 NO2(g)
Equilibrium Constant (K)
For the general equilibrium below:
aP + bQ → cR + dS
Keq = [R]c [S]d
[P]a [Q]b
Equilibrium Constant (K)
- a, b, c, & d are coefficients used to balance the equation
- P, Q, R, & S are the reactants and products- Kc is sometimes used instead of Keq when
units are molar concentration
Equilibrium Constant (K)
eg. Write the expression for Keq for:
2 SO2(g) + O2(g) → 2 SO3(g)
2 HCl(g) --> H2(g) + Cl2(g)
][O][SO
][SOK
22
2
23
eq
22
eq [HCl]
]H[][ClK 2
Equilibrium Constant (K)
NOTE!!
Solids or liquids ARE NOT included in the Keq or Kc expression because their concentration is constant
Equilibrium Constant (K)
eg. Write the expression for Kc for:
2 NaCl(s) + H2SO4(aq) → 2 HCl(g) + Na2SO4(aq)
CaCO3(s) → CaO(s) + CO2(g) ]SO[H
]SONa[[HCl]K
42
2
eq42
]CO[Keq 2
Interpreting K
Large vs small K values?
● K much larger than 1 - products favored● [products] > [reactants]
● K much smaller than 1 - reactants favored● [reactants] > [products]
Changing K (p. 497)
● For a given system at equilibrium, the value of the equilibrium constant depends only on temperature
ie. For any equilibrium, the only way to change the actual value of K is to change the temperature
Calculations with K
Types of K calculations:
1. Given equilibrium concentrations, find K
2. Find a missing concentration given K and other concentrations.
3. Given initial concentrations and equilibrium data, find K (ICE tables)
1.Given equilibrium concentrations, find K
eg. Calculate Kc for this equilibrium using the equilibrium concentrations given:
H2(g) + I2(g) → 2 HI(g)
[H2] = 0.22 mol/L[I2] = 0.30 mol/L[HI] = 1.56 mol/L
Ex: Given the equilibrium concentrations;
[CO(g)] = 0.105 mol/L,
[H2(g)] = 0.250 mol/L
[CH3OH(g)] = 0.00261 mol/L,
What is the value of Keq for the equilibrium below?
CO(g) + 2 H2(g) → CH3OH(g)
(A) 0.0994
(B) 0.398
(C) 2.51
(D) 10.0
Be Careful!!The other answers are possible if you mess up
the calculation
Ex: Given the equilibrium concentrations below, what is the value of Keq for
N2(g) + O2(g) → 2 NO(g) ?
[N2(g)] = 0.10 mol/L
[O2(g)] = 0.20 mol/L
[NO(g)] = 0.0030 mol/L
(A) 2.2×10−4
(B) 4.5×10−4
(C) 1.5×10−1
(D) 3.0×10−1
Ex: Find Keq for the equilibrium below:
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
[NH3(g) ] = 0.100 mol/L
[O2(g) ] = 0.200 mol/L
[ NO(g) ] = 0.300 mol/L
[ H2O(g) ] = 0.250 mol/L
2. Find a missing concentration . . .
eg. Find the [HI] in the equilibrium below if Kc = 36.9, [H2] = 0.125 mol/L and [I2] = 2.56 mol/L.
H2(g) + I2(g) 2 HI→ (g)
3. ICE tables
● Keq may be calculated given initial concentrations and at least one equilibrium concentration
● Using the Initial concentration and the Change in concentration we can find the missing Equilibrium concentrations and calculate Keq
ICE tables
eg. 4.30 mol of NH3 was placed in a 1.00 L closed container to establish this equilibrium:
2 NH3(g) →N2(g) + 3 H2(g)
Calculate Kc if the equilibrium concentration of H2(g) = 0.500 mol/L
ICE tables
2 [NH3] → [N2] + 3 [H2]
I
C
E
4.30
0.500
By how much does the [H2] change?
0 0
?? ?
? ?
ICE tables
2 [NH3] --> [N2] 3 [H2]
I
C
E
4.30 mol/L 0 0
-2x +x +3x
4.30 - 2x x 3x
0.500 mol/L
[H2] = 0.500 mol/L
3x = 0.500 mol/L x = 0.167 mol/L
[N2] = x
= 0.167 mol/L
[NH3] = 4.30 - 2x
= 3.97 mol/L
K = (0.500)3 x (0.167)
(3.97)2
= 0.00132
ICE tables
eg. 2.00 mol of N2, 4.00 mol of H2 and 3.00 mol of NH3 were allowed to come to equilibrium in a 1.00 L container
2 NH3(g) → N2(g) + 3 H2(g)
Calculate Kc if the equilibrium concentration of NH3(g) = 3.50 mol/L
ICE tables
2 [NH3] --> [N2] 3 [H2]
I
C
E
3.00 2.00 4.00
+2x -3x
3.00 + 2x 2.00 - x 4.00 - 3x
-x
= 3.50
[H2] = 4.00 mol/L – 3x
= 3.25 mol/L
[N2] = 2.00 mol/L - x
= 1.75 mol/L
[NH3] = 3.50 mol/L
3.00 + 2x = 3.50 mol/L
x = 0.25 mol/L
K = (3.25)3 x (1.75)
(3.50)2
= 4.90
ICE tableseg. The oxidation of ammonia occurs
according to the following expression:
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
0.800 mol of each chemical were placed in a 1.00 L container and there was 0.450 mol of NH3 at equilibrium. Calculate the equilibrium concentrations and Kc
4 NH3(g) 5 O2(g) 4 NO(g) 6 H2O(g)
0.800
-4x
0.800 - 4x
0.450 mol/L
0.800
-5x
0.800 - 5x
0.800
+4x
0.800 + 4x
0.800
+6x
0.800+ 6x
[NH3] = 0.450 mol/L
0.800 – 4x = 0.450 mol/L
x = 0.0875 mol/L
[O2] = 0.3625 mol/L
[NO] = 1.15 mol/L
[H2O] = 1.325 mol/L
K = 37000
ICE tables
eg. 2.30 mol of NH3 was placed in a 2.00 L closed container to establish this equilibrium:
2 NH3(g) → N2(g) + 3 H2(g)
Calculate Kc if 25 % of the NH3(g) reacts.
2 [NH3] --> [N2] 3 [H2]
I
C
E
1.15 0 0
-2x +x +3x
1.15 - 2x x 3x
= (0.75 x 1.15) = 0.8625 mol/L
mol/L1.15L2.00
mol2.30
V
nC
[NH3] = 0.8625 → 75% of 1.15
1.15 - 2x = 0.8625 mol/L
x = 0.1438 mol/L
[N2] = x
= 0.1438 mol/L
[H2] = 3x
= 0.4314 mol/L
K = 0.0155
ICE tables
eg. 0.800 mol of NH3 was placed in a 1.00 L closed container to establish this equilibrium:
2 NH3(g) → N2(g) + 3 H2(g)
Calculate Kc if 70% of the NH3 reacts.
→ Kc = 2.88